Introduction

Mathematics often involves organising information and performing calculations with many values at once. A shop may track sales of several products across different months, a scientist may record measurements from multiple experiments, and a computer may transform thousands of points in a digital image.

One mathematical tool used in these situations is the matrix.

A matrix is a rectangular arrangement of numbers, symbols, or expressions organised into rows and columns. Matrices allow us to represent data efficiently and perform calculations systematically.

For example, consider the sales of three products in two stores:

We can represent this information using a matrix:

If we have another matrix representing sales during a second week, we can add the matrices to find total sales.

We can also multiply matrices by numbers, combine different matrices through multiplication, and use matrices to solve real-world problems involving transformations, production, finance, and scientific measurements.

However, matrix operations follow specific rules. Unlike ordinary numbers, matrices cannot always be added or multiplied, and the order of multiplication can change the result.

Understanding these rules is essential for working confidently with matrices.

What Is a Matrix?

A matrix is a rectangular array of elements arranged in rows and columns.

Matrices are usually represented using capital letters such as A, B, and C.

For example:

This matrix contains:

• 2 rows

• 3 columns

• 6 elements

The order, or dimensions, of the matrix are:

2 × 3

The number of rows is always written first, followed by the number of columns.

Identifying rows and columns

Matrix notation

An element of a matrix can be identified using two subscripts.

For example:

aᵢⱼ

where:

• i identifies the row.

• j identifies the column.

For the matrix A above:

a₁₁ = 2

a₁₂ = 5

a₁₃ = 8

a₂₁ = 1

a₂₂ = 4

a₂₃ = 7

Notice that a₂₃ means the element in row 2, column 3.

It does not mean row 3, column 2.

Types of Matrices

Matrices can be classified according to their dimensions and properties.

Row matrix

A matrix containing one row.

Order: 1 × 3

Column matrix

A matrix containing one column.

Order: 3 × 1

Square matrix

A matrix with the same number of rows and columns.

Order: 2 × 2

Zero matrix

A matrix in which every element is zero.

Identity matrix

A square matrix with 1 along the main diagonal and 0 everywhere else.

For example:

The identity matrix has an important role in matrix multiplication.

For any compatible matrix A:

AI = A and IA = A

The identity matrices used on the two sides may have different orders when A is rectangular.

Adding Matrices

Matrix addition combines corresponding elements from two matrices.

Two matrices can be added only when they have the same dimensions.

For example:

Add corresponding elements:

Therefore:

Visualising matrix addition

General rule

If A and B are both m × n matrices:

(A + B)ᵢⱼ = aᵢⱼ + bᵢⱼ

The result is also an m × n matrix.

Worked Example 1: Adding 2 × 3 Matrices

Given:

Calculate A + B.

Step 1: Check the dimensions.

Both matrices are 2 × 3.

Addition is possible.

Step 2: Add corresponding elements.

Step 3: Simplify.

Properties of Matrix Addition

Matrix addition has several useful properties.

Commutative property

A + B = B + A

The order of addition does not affect the result.

Associative property

(A + B) + C = A + (B + C)

Matrices can be grouped differently when adding three or more matrices.

Additive identity

A + O = A

Adding a zero matrix of the same dimensions does not change a matrix.

Additive inverse

For every matrix A, there is a matrix −A such that:

A + (−A) = O

The additive inverse is formed by changing the sign of every element.

For example:

Then:

Subtracting Matrices

Matrix subtraction follows the same dimensional rule as addition.

Two matrices can be subtracted only if they have the same dimensions.

Subtract corresponding elements.

Worked Example 2: Matrix Subtraction

Given:

Calculate A − B.

Therefore:

Subtraction as addition of the inverse

Matrix subtraction can also be written as:

A − B = A + (−B)

This relationship is useful when simplifying matrix expressions.

Important distinction

Matrix subtraction is generally not commutative.

A − B ≠ B − A

In fact:

B − A = −(A − B)

Scalar Multiplication

A scalar is a single number.

Scalar multiplication means multiplying every element of a matrix by the same number.

For example:

Calculate 3A.

Therefore:

Scalar multiplication does not change the dimensions of the matrix.

Worked Example 3: Multiplying by a Negative Scalar

Given:

Calculate −2B.

Therefore:

Notice that multiplying a negative element by a negative scalar produces a positive result.

Properties of Scalar Multiplication

For scalars k and m and compatible matrices A and B:

Distributive property over matrix addition

k(A + B) = kA + kB

Distributive property over scalar addition

(k + m)A = kA + mA

Associative property

k(mA) = (km)A

Multiplicative identity

1A = A

Multiplication by zero

0A = O

These properties allow matrix expressions to be simplified algebraically.

Worked Example 4: Combining Matrix Operations

Given:

Calculate:

2A − 3B

First:

Next:

Subtract:

Therefore:

Multiplying Matrices

Matrix multiplication is different from matrix addition and scalar multiplication.

We do not simply multiply corresponding elements.

Instead, each element of the resulting matrix is calculated using a row from the first matrix and a column from the second matrix.

The compatibility rule

Suppose:

A has dimensions m × n.

B has dimensions n × p.

Then the product AB is possible, and the result has dimensions:

m × p

The inner dimensions must match.

The outer dimensions determine the size of the result.

How Matrix Multiplication Works

Consider:

To calculate AB, multiply each row of A by each column of B and add the products.

Finding the first element

Use row 1 of A:

[2, 3]

and column 1 of B:

[5, 1]

Calculate:

2(5) + 3(1) = 13

This becomes the element in row 1, column 1 of AB.

Visualising row-by-column multiplication

Calculating the complete product

Top-left:

2(5) + 3(1) = 13

Top-right:

2(2) + 3(6) = 22

Bottom-left:

4(5) + 1(1) = 21

Bottom-right:

4(2) + 1(6) = 14

Therefore:

General formula

For compatible matrices A and B:

cᵢⱼ = Σ aᵢₖbₖⱼ

The sum is taken over all matching positions k in the row of A and column of B.

In words:

Multiply corresponding entries from the row and column, then add the products.

Worked Example 5: Multiplying a 2 × 3 Matrix by a 3 × 2 Matrix

Given:

Calculate AB.

Step 1: Check compatibility.

A is 2 × 3.

B is 3 × 2.

The inner dimensions are both 3.

Multiplication is possible.

The result will be 2 × 2.

Step 2: Calculate each element.

Row 1 × Column 1:

1(7) + 2(9) + 3(11)

= 7 + 18 + 33

= 58

Row 1 × Column 2:

1(8) + 2(10) + 3(12)

= 8 + 20 + 36

= 64

Row 2 × Column 1:

4(7) + 5(9) + 6(11)

= 28 + 45 + 66

= 139

Row 2 × Column 2:

4(8) + 5(10) + 6(12)

= 32 + 50 + 72

= 154

Step 3: Write the result.

Worked Example 6: Multiplying a Matrix by a Column Vector

Given:

and:

Calculate Ax.

Step 1: Check dimensions.

A is 2 × 2.

x is 2 × 1.

Multiplication is possible.

The result will be 2 × 1.

Step 2: Multiply rows by the column.

First element:

2(5) + 3(2) = 16

Second element:

1(5) + 4(2) = 13

Therefore:

This type of multiplication is especially important in linear algebra, coordinate transformations, and systems of equations.

Why Matrix Multiplication Is Not Usually Commutative

Ordinary multiplication is commutative:

3 × 5 = 5 × 3

However, matrix multiplication generally is not.

AB ≠ BA

Consider:

Calculate AB:

Now calculate BA:

Therefore:

AB ≠ BA

The order matters because multiplying rows by columns produces different combinations of elements.

Sometimes AB is defined while BA is not.

For example:

A: 2 × 3

B: 3 × 4

AB is possible and has dimensions 2 × 4.

BA is not possible because its inner dimensions would be 4 and 2.

Properties of Matrix Multiplication

For compatible matrices:

Associative property

(AB)C = A(BC)

The grouping of multiplication can change without changing the result.

Distributive property

A(B + C) = AB + AC

and:

(A + B)C = AC + BC

provided the operations are defined.

Identity property

AI = A

and:

IA = A

using identity matrices of suitable dimensions.

Scalar compatibility

k(AB) = (kA)B = A(kB)

Noncommutative property

In general:

AB ≠ BA

Zero product property

Unlike ordinary real numbers, it is possible for two nonzero matrices to have a zero product.

For example:

Both matrices are nonzero, but:

AB = O

This is an important difference between ordinary arithmetic and matrix algebra.

Determining Whether Operations Are Possible

Before performing any matrix operation, always check the dimensions.

Addition and subtraction

The matrices must have identical dimensions.

For example:

2 × 3 + 2 × 3 is possible.

2 × 3 + 3 × 2 is not possible.

Scalar multiplication

A scalar can multiply a matrix of any dimensions.

For example:

5A is possible for any matrix A.

Matrix multiplication

The number of columns in the first matrix must equal the number of rows in the second matrix.

For example:

(2 × 4)(4 × 3) is possible.

The result is 2 × 3.

However:

(2 × 4)(2 × 3) is not possible.

Quick compatibility guide

Interactive Matrix Operations Explorer

Explore how addition, subtraction, scalar multiplication, and matrix multiplication work.

Matrix Operations in Real-World Problems

Application 1: Combining Sales Data

Two shops sell three products.

Their weekly sales are represented by:

and:

Rows represent shops.

Columns represent products.

The total sales are:

This provides a compact summary of sales across both weeks.

Application 2: Increasing Production

A factory produces:

If production doubles, the new production matrix is:

Scalar multiplication allows every quantity to be adjusted simultaneously.

Application 3: Calculating Revenue

A shop sells three products.

Quantities sold in two stores are:

The prices of the three products are:

The revenue for each store is calculated using:

R = Qp

Therefore:

• Store 1 earns $170.

• Store 2 earns $176.

Matrix multiplication allows the quantities and prices to be combined efficiently.

Matrices and Geometric Transformations

Matrices can describe transformations of points in a coordinate plane.

A point can be represented by a column vector:

Reflection in the x-axis

The transformation matrix is:

For the point (3, 2):

Calculate:

The point (3, 2) is reflected to (3, −2).

Introduction

Matrices allow us to organise numbers, represent transformations, and solve systems of equations. However, matrices also have important properties that cannot be determined simply by looking at their individual elements.

One of these properties is the determinant.

A determinant is a single number calculated from a square matrix. Although it produces only one value, that value reveals important information about the matrix.

For example, the determinant can tell us whether:

• A matrix has an inverse.

• A system of linear equations has a unique solution.

• A geometric transformation enlarges or reduces an area or volume.

• A transformation reverses orientation.

• Points in a coordinate plane lie on the same straight line.

Consider the matrix:

Its determinant is:

det(A) = 3(4) − 2(1) = 10

This value has several interpretations. Geometrically, the transformation represented by A multiplies areas by 10. Algebraically, because the determinant is not zero, the matrix has an inverse.

Determinants therefore connect matrix algebra with geometry, transformations, and systems of equations.

What Is a Determinant?

The determinant is a scalar value associated with a square matrix.

A square matrix has the same number of rows and columns.

For example:

The determinant can be written as:

det(A)

or:

Although vertical bars are also used for absolute value, the determinant is a different operation.

A determinant can be positive, negative, or zero.

Important properties

• Determinants are defined for square matrices.

• The determinant of a 2 × 2 matrix is calculated using four elements.

• The determinant of a 3 × 3 matrix is calculated using nine elements.

• A nonzero determinant indicates that a square matrix is invertible.

• A zero determinant indicates that a square matrix is singular.

• The absolute value of a determinant describes an area or volume scaling factor.

The sign of the determinant provides information about orientation.

Calculating the Determinant of a 2 × 2 Matrix

For a matrix:

the determinant is:

This is one of the most important formulas in introductory matrix algebra.

Visualising the formula

The determinant of a 2 × 2 matrix

The process is:

1. Multiply the top-left element by the bottom-right element.

2. Multiply the top-right element by the bottom-left element.

3. Subtract the second product from the first.

Worked Example 1: Positive Determinant

Calculate:

\[ \begin{vmatrix}4&2\\3&5\end{vmatrix} \]

Using:

det(A) = ad − bc

Substitute:

det(A) = 4(5) − 2(3)

= 20 − 6

det(A) = 14

The determinant is positive and nonzero.

Therefore, the matrix is invertible.

Worked Example 2: Negative Determinant

Calculate:

\[ B=\begin{bmatrix}2&5\\4&3\end{bmatrix} \]

det(B) = 2(3) − 5(4)

= 6 − 20

det(B) = −14

The determinant is negative.

This indicates that the associated linear transformation reverses orientation.

The area scaling factor is |−14| = 14.

Worked Example 3: Zero Determinant

Calculate:

\[ C=\begin{bmatrix}2&4\\3&6\end{bmatrix} \]

det(C) = 2(6) − 4(3)

= 12 − 12

det(C) = 0

The second column is twice the first column.

The columns are linearly dependent.

This matrix is singular and does not have an inverse.

 

Calculating the Determinant of a 3 × 3 Matrix

Consider the general matrix:

\[ A=\begin{bmatrix} a&b&c\\ d&e&f\\ g&h&i \end{bmatrix} \]

The determinant is:

\[ \boxed{\det(A)=a(ei-fh)-b(di-fg)+c(dh-eg)} \]

This formula can be understood through cofactor expansion.

Cofactor expansion along the first row

Each element in the first row is multiplied by the determinant of a smaller 2 × 2 matrix.

The signs follow the pattern:

+ − +

What is a minor?

The minor of an element is the determinant of the smaller matrix formed by removing the row and column containing that element.

For example, for the top-left element a:

\[ \begin{bmatrix} a&b&c\\ d&e&f\\ g&h&i \end{bmatrix} \]

Remove the first row and first column.

The remaining matrix is:

\[ \begin{bmatrix}e&f\\h&i\end{bmatrix} \]

Its determinant is:

ei − fh

This is the minor associated with a.

A cofactor is the minor multiplied by the appropriate positive or negative sign.

 

Worked Example 4: Determinant of a 3 × 3 Matrix

Calculate:

\[ A=\begin{bmatrix} 2&1&3\\ 0&4&5\\ 1&2&6 \end{bmatrix} \]

Expand along the first row.

Step 1: Apply the cofactor expansion formula.

det(A) = 2(4 × 6 − 5 × 2) − 1(0 × 6 − 5 × 1) + 3(0 × 2 − 4 × 1)

Step 2: Calculate the 2 × 2 determinants.

First:

4(6) − 5(2) = 24 − 10 = 14

Second:

0(6) − 5(1) = −5

Third:

0(2) − 4(1) = −4

Step 3: Substitute.

det(A) = 2(14) − 1(−5) + 3(−4)

= 28 + 5 − 12

det(A) = 21

The matrix is invertible because its determinant is nonzero.

 

Worked Example 5: Expanding Along a Row Containing Zeros

Consider:

\[ B=\begin{bmatrix} 3&0&0\\ 2&4&1\\ 5&2&6 \end{bmatrix} \]

Expanding along the first row is particularly efficient because two elements are zero.

det(B) = 3(4 × 6 − 1 × 2)

= 3(24 − 2)

= 3(22)

det(B) = 66

This example demonstrates an important strategy:

When calculating a 3 × 3 determinant, choose a row or column containing the most zeros whenever possible.

 

The Rule of Sarrus

Another method for calculating a 3 × 3 determinant is the Rule of Sarrus.

This method uses diagonal products.

For:

\[ A=\begin{bmatrix} a&b&c\\ d&e&f\\ g&h&i \end{bmatrix} \]

the determinant is:

det(A) = aei + bfg + cdh − ceg − bdi − afh

Visualising Sarrus' Rule

Worked Example 6: Using Sarrus' Rule

Calculate:

\[ C=\begin{bmatrix} 1&2&3\\ 4&5&6\\ 7&8&10 \end{bmatrix} \]

Add the downward diagonal products:

1(5)(10) + 2(6)(7) + 3(4)(8)

= 50 + 84 + 96

= 230

Add the upward diagonal products:

3(5)(7) + 2(4)(10) + 1(6)(8)

= 105 + 80 + 48

= 233

Subtract:

det(C) = 230 − 233

det(C) = −3

 

Properties of Determinants

Determinants follow several important algebraic rules.

Property 1: Identity matrix

The determinant of an identity matrix is 1.

det(I) = 1

For example:

\[ \det\begin{bmatrix}1&0\\0&1\end{bmatrix}=1 \]

Property 2: Interchanging two rows

Swapping two rows changes the sign of the determinant.

If det(A) = 5, then swapping two rows produces a matrix with determinant −5.

Property 3: Identical rows

If two rows are identical, the determinant is zero.

For example:

\[ \begin{vmatrix}2&3\\2&3\end{vmatrix} =2(3)-3(2)=0 \]

Property 4: Proportional rows

If one row is a multiple of another, the determinant is zero.

Property 5: Multiplying one row by a scalar

Multiplying one row by k multiplies the determinant by k.

Property 6: Adding a multiple of one row to another

Adding a multiple of one row to another row does not change the determinant.

This property is especially useful when simplifying larger determinants.

Property 7: Triangular matrices

The determinant of an upper or lower triangular matrix equals the product of its diagonal elements.

For example:

\[ A=\begin{bmatrix} 2&3&4\\ 0&5&6\\ 0&0&7 \end{bmatrix} \]

det(A) = 2 × 5 × 7

det(A) = 70

Property 8: Product of matrices

For square matrices of the same order:

det(AB) = det(A)det(B)

Property 9: Transpose

The determinant does not change when rows and columns are interchanged.

det(Aᵀ) = det(A)

Property 10: Scalar multiplication

For an n × n matrix:

det(kA) = kⁿdet(A)

For example, for a 2 × 2 matrix:

det(3A) = 3²det(A) = 9det(A)

This is different from simply multiplying the determinant by 3.

Try changing the values so that the determinant becomes zero. What relationships do you notice between the rows or columns?

 

Determinants and Invertible Matrices

A square matrix is invertible if another matrix exists that reverses its effect under multiplication.

For an invertible matrix A:

AA⁻¹ = A⁻¹A = I

where:

• A⁻¹ is the inverse matrix.

• I is the identity matrix.

The determinant tells us whether an inverse exists.

Invertibility condition

If det(A) ≠ 0, the matrix is invertible.

If det(A) = 0, the matrix is singular and not invertible.

This is one of the most important applications of determinants.

Formula for the inverse of a 2 × 2 matrix

For:

\[ A=\begin{bmatrix}a&b\\c&d\end{bmatrix} \]

if ad − bc ≠ 0:

\[ \boxed{ A^{-1}=\frac{1}{ad-bc} \begin{bmatrix} d&-b\\ -c&a \end{bmatrix} } \]

Worked Example 7: Finding an Inverse

Given:

\[ A=\begin{bmatrix}4&1\\2&3\end{bmatrix} \]

Step 1: Calculate the determinant.

det(A) = 4(3) − 1(2)

= 12 − 2

= 10

Since det(A) ≠ 0, an inverse exists.

Step 2: Apply the inverse formula.

\[ A^{-1}=\frac{1}{10} \begin{bmatrix} 3&-1\\ -2&4 \end{bmatrix} \]

Therefore:

\[ \boxed{ A^{-1}= \begin{bmatrix} 0.3&-0.1\\ -0.2&0.4 \end{bmatrix} } \]

Worked Example 8: Identifying a Singular Matrix

Given:

\[ B=\begin{bmatrix}6&9\\2&3\end{bmatrix} \]

det(B) = 6(3) − 9(2)

= 18 − 18

= 0

Therefore, B is singular.

B has no inverse.

The first row is three times the second row, so the rows are linearly dependent.

 

Geometric Meaning of Determinants

Determinants have a powerful geometric interpretation.

For a 2 × 2 matrix, the absolute value of the determinant represents the factor by which a linear transformation changes area.

For a 3 × 3 matrix, the absolute value of the determinant represents the factor by which a linear transformation changes volume.

Area scaling in two dimensions

Suppose:

\[ A=\begin{bmatrix}2&0\\0&3\end{bmatrix} \]

This transformation:

• Multiplies x-coordinates by 2.

• Multiplies y-coordinates by 3.

The determinant is:

det(A) = 2(3) − 0(0)

det(A) = 6

Therefore, areas are multiplied by 6.

A square with area 1 becomes a rectangle with area 6.

Positive and negative determinants

A positive determinant means the transformation preserves orientation.

A negative determinant means the transformation reverses orientation.

For example, a reflection in the x-axis has matrix:

\[ R=\begin{bmatrix}1&0\\0&-1\end{bmatrix} \]

Its determinant is:

det(R) = 1(−1) − 0(0)

det(R) = −1

The area is unchanged because |−1| = 1.

However, orientation is reversed.

Zero determinant

If det(A) = 0, a transformation collapses space into a lower-dimensional set.

For example:

\[ A=\begin{bmatrix}1&2\\2&4\end{bmatrix} \]

det(A) = 4 − 4 = 0

The transformation maps the entire plane onto a line.

Areas become zero.

This explains geometrically why the matrix cannot be inverted: different points can be mapped to the same location.

 

Determinants and the Area of a Parallelogram

Suppose two vectors are:

u = (a, c)

v = (b, d)

These vectors form adjacent sides of a parallelogram.

Its area is:

\[ \boxed{\text{Area}=|ad-bc|} \]

This is the absolute value of the determinant:

\[ \begin{vmatrix}a&b\\c&d\end{vmatrix} \]

Worked Example 9: Parallelogram Area

Two vectors are:

u = (3, 1)

v = (2, 4)

Construct the matrix:

\[ A=\begin{bmatrix}3&2\\1&4\end{bmatrix} \]

Calculate:

det(A) = 3(4) − 2(1)

= 12 − 2

= 10

Therefore:

Area = 10 square units

Determinants and the Area of a Triangle

A triangle formed by two vectors has half the area of the parallelogram formed by those vectors.

Therefore:

\[ \boxed{\text{Triangle Area}=\frac12|ad-bc|} \]

Worked Example 10: Triangle Area from Coordinates

Find the area of the triangle with vertices:

A(1, 2)

B(5, 3)

C(3, 7)

Step 1: Form two vectors from A.

AB = (5 − 1, 3 − 2)

AB = (4, 1)

AC = (3 − 1, 7 − 2)

AC = (2, 5)

Step 2: Construct the matrix.

\[ M=\begin{bmatrix}4&2\\1&5\end{bmatrix} \]

Step 3: Calculate the determinant.

det(M) = 4(5) − 2(1)

= 20 − 2

= 18

Step 4: Divide by 2.

Area = 1/2 × |18|

Area = 9 square units

Coordinate formula

For vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃):

Area = 1/2 |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|

This formula can also be written as a 3 × 3 determinant:

\[ \text{Area}=\frac12 \left| \begin{vmatrix} x_1&y_1&1\\ x_2&y_2&1\\ x_3&y_3&1 \end{vmatrix} \right| \]

 

Determinants and Collinear Points

Three points are collinear if they lie on the same straight line.

A determinant can be used to test collinearity.

For points A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃), calculate:

\[ D= \begin{vmatrix} x_1&y_1&1\\ x_2&y_2&1\\ x_3&y_3&1 \end{vmatrix} \]

If D = 0, the three points are collinear.

Worked Example 11: Testing Collinearity

Determine whether the points:

A(1, 2)

B(3, 6)

C(5, 10)

are collinear.

Construct:

\[ D= \begin{vmatrix} 1&2&1\\ 3&6&1\\ 5&10&1 \end{vmatrix} \]

Expand along the first row:

D = 1(6 − 10) − 2(3 − 5) + 1(30 − 30)

= −4 − 2(−2) + 0

= −4 + 4

D = 0

Therefore, the points are collinear.

This agrees with the fact that all three points satisfy:

y = 2x

 

Determinants and Volume in Three Dimensions

A 3 × 3 determinant represents the signed volume scaling factor of a linear transformation in three-dimensional space.

If three vectors form the edges of a parallelepiped, its volume is the absolute value of their determinant.

Worked Example 12: Volume of a Parallelepiped

Consider three vectors:

u = (2, 0, 0)

v = (0, 3, 0)

w = (0, 0, 4)

Construct:

\[ A=\begin{bmatrix} 2&0&0\\ 0&3&0\\ 0&0&4 \end{bmatrix} \]

Since this is a diagonal matrix:

det(A) = 2 × 3 × 4

= 24

Therefore:

Volume = 24 cubic units

For a general parallelepiped, the vectors do not need to be perpendicular. The determinant still gives its signed volume.

 

Determinants and Systems of Linear Equations

Determinants can be used to determine whether a system of linear equations has a unique solution.

Consider:

ax + by = e

cx + dy = f

The coefficient matrix is:

\[ A=\begin{bmatrix}a&b\\c&d\end{bmatrix} \]

If det(A) ≠ 0, the system has exactly one solution.

If det(A) = 0, the system does not have a unique solution. It may have no solutions or infinitely many solutions.

Cramer's Rule

For a system:

ax + by = e

cx + dy = f

Define:

D = ad − bc

Dₓ = ed − bf

Dᵧ = af − ec

If D ≠ 0:

\[ \boxed{x=\frac{D_x}{D},\qquad y=\frac{D_y}{D}} \]

Worked Example 13: Solving a System Using Determinants

Solve:

2x + 3y = 12

4x − y = 5

Step 1: Calculate D.

\[ D=\begin{vmatrix}2&3\\4&-1\end{vmatrix} \]

D = 2(−1) − 3(4)

= −2 − 12

= −14

Step 2: Calculate Dₓ.

Replace the x-coefficient column with the constants.

\[ D_x=\begin{vmatrix}12&3\\5&-1\end{vmatrix} \]

Dₓ = 12(−1) − 3(5)

= −12 − 15

= −27

Step 3: Calculate Dᵧ.

Replace the y-coefficient column with the constants.

\[ D_y=\begin{vmatrix}2&12\\4&5\end{vmatrix} \]

Dᵧ = 2(5) − 12(4)

= 10 − 48

= −38

Step 4: Solve.

x = Dₓ/D = −27/−14

x = 27/14

y = Dᵧ/D = −38/−14

y = 19/7

Step 5: Verify.

First equation:

2(27/14) + 3(19/7)

= 27/7 + 57/7

= 84/7

= 12

Second equation:

4(27/14) − 19/7

= 54/7 − 19/7

= 35/7

= 5

Both equations are satisfied.

 

Solving Determinant Equations

Sometimes the determinant is known, but one matrix element is unknown.

Worked Example 14: Finding an Unknown Element

Suppose:

\[ \begin{vmatrix} x&3\\ 2&5 \end{vmatrix}=14 \]

Find x.

Using the 2 × 2 determinant formula:

5x − 3(2) = 14

5x − 6 = 14

5x = 20

x = 4

Worked Example 15: Finding When a Matrix Is Singular

Determine the values of k for which:

\[ A=\begin{bmatrix}k&2\\8&k\end{bmatrix} \]

is singular.

A matrix is singular when its determinant is zero.

Therefore:

det(A) = 0

k² − 2(8) = 0

k² − 16 = 0

Factor:

(k − 4)(k + 4) = 0

Therefore:

k = 4 or k = −4

For either value, the matrix has no inverse.

 

Real-World Applications of Determinants

3Blue1Brown - Three-dimensional linear transformations

 

 

 

Computer graphics

Determinants help determine how transformations scale areas and volumes, and whether they reverse orientation.

How Industrial Robots Are Revolutionizing Manufacturing

 

 

 

Robotics

Determinants help identify singular configurations in mathematical models of robot movement, where certain motions become difficult or impossible to control independently.

Metalowy most | 004252

 

 

 

Engineering

Determinants help analyse linear systems describing forces, constraints, and structural relationships.

Square feet to acres: how to convert fast | Property Glow

 

 

 

Surveying and geometry

Determinants allow areas of triangles and parallelograms to be calculated directly from coordinates.

William L. Kath - Home

 

 

 

Scientific modelling

Determinants help identify whether linear systems have unique solutions and whether transformations preserve independent directions.

 

Common Mistakes

Mistake 1: Calculating determinants for non-square matrices

Determinants are defined only for square matrices.

Mistake 2: Reversing the subtraction in a 2 × 2 determinant

The correct formula is:

ad − bc

not bc − ad.

Mistake 3: Forgetting the alternating signs in 3 × 3 expansion

The first-row cofactor signs are:

+ − +

Mistake 4: Confusing determinants with absolute values

A determinant may be negative.

Absolute value is used when calculating geometric area or volume.

Mistake 5: Assuming a zero determinant means every matrix element is zero

A matrix can contain many nonzero elements and still have determinant zero.

Mistake 6: Assuming every square matrix has an inverse

Only square matrices with nonzero determinants are invertible.

Mistake 7: Applying Sarrus' Rule to larger matrices

Sarrus' Rule applies specifically to 3 × 3 determinants.

Mistake 8: Multiplying every element by a scalar without accounting for matrix order

For an n × n matrix:

det(kA) = kⁿdet(A)

Mistake 9: Ignoring negative signs in cofactor expansion

Subtracting a negative minor produces a positive contribution.

Mistake 10: Confusing determinant and area

The determinant gives signed area scaling. Geometric area is always nonnegative.

Mistake 11: Using Cramer's Rule when the determinant is zero

Cramer's Rule requires a nonzero coefficient determinant.

Mistake 12: Assuming det(A) = 0 always means no solution

A linear system with determinant zero may have no solutions or infinitely many solutions.

 

Key Terms

• Determinant: Scalar value calculated from a square matrix.

• Square Matrix: Matrix with equal numbers of rows and columns.

• Scalar: Single numerical value.

• Minor: Determinant obtained by deleting a specified row and column.

• Cofactor: Signed minor associated with a matrix element.

• Cofactor Expansion: Method for calculating determinants using minors and cofactors.

• Rule of Sarrus: Diagonal method for calculating 3 × 3 determinants.

• Invertible Matrix: Square matrix with a multiplicative inverse.

• Inverse Matrix: Matrix that produces the identity matrix when multiplied by the original matrix.

• Singular Matrix: Square matrix with determinant zero.

• Nonsingular Matrix: Square matrix with a nonzero determinant.

• Identity Matrix: Square matrix with 1 on the main diagonal and 0 elsewhere.

• Main Diagonal: Diagonal extending from the top-left to the bottom-right of a matrix.

• Triangular Matrix: Square matrix with zeros above or below the main diagonal.

• Linear Transformation: Mapping represented by multiplication by a matrix.

• Orientation: Directional arrangement preserved or reversed by a transformation.

• Area Scaling Factor: Absolute value of a 2 × 2 transformation determinant.

• Volume Scaling Factor: Absolute value of a 3 × 3 transformation determinant.

• Parallelogram: Quadrilateral with opposite sides parallel.

• Parallelepiped: Three-dimensional solid with parallelogram faces.

• Collinear Points: Points lying on the same straight line.

• Linear Dependence: Relationship in which one vector can be expressed using others.

• Cramer's Rule: Method for solving certain linear systems using determinants.

• Coefficient Matrix: Matrix containing the coefficients of a linear system.

• Unique Solution: Exactly one solution satisfying a system of equations.

• Row Operation: Operation performed on rows of a matrix.

• Transpose: Matrix obtained by interchanging rows and columns.

 

Key Takeaways

• Determinants are scalar values associated with square matrices.

• The determinant of a 2 × 2 matrix is ad − bc.

• A 3 × 3 determinant can be calculated using cofactor expansion or Sarrus' Rule.

• Cofactor expansion uses alternating positive and negative signs.

• A determinant can be positive, negative, or zero.

• A square matrix is invertible exactly when its determinant is nonzero.

• A zero determinant indicates a singular matrix.

• The determinant of a triangular matrix is the product of its diagonal elements.

• Swapping two rows reverses the determinant's sign.

• Proportional rows or columns produce a zero determinant.

• For square matrices, det(AB) = det(A)det(B).

• The absolute value of a 2 × 2 determinant gives an area scaling factor.

• The absolute value of a 3 × 3 determinant gives a volume scaling factor.

• Determinants can calculate areas of triangles and parallelograms.

• Determinants can test whether three points are collinear.

• Cramer's Rule uses determinants to solve systems with unique solutions.

• Determinants connect algebra, geometry, transformations, and scientific applications.

 

Check Your Understanding

Questions

Question 1: Basic Determinant

Calculate:

\[ \begin{vmatrix}3&2\\4&5\end{vmatrix} \]

Question 2: Negative Determinant

Calculate:

\[ \begin{vmatrix}1&6\\3&2\end{vmatrix} \]

Question 3: Zero Determinant

Determine whether:

\[ A=\begin{bmatrix}2&4\\5&10\end{bmatrix} \]

is invertible.

Question 4: 3 × 3 Determinant

Calculate:

\[ \begin{vmatrix} 1&2&3\\ 0&4&5\\ 1&0&6 \end{vmatrix} \]

Question 5: Triangular Matrix

Find the determinant of:

\[ \begin{bmatrix} 3&2&1\\ 0&4&5\\ 0&0&6 \end{bmatrix} \]

Question 6: Determinant Properties

If det(A) = 7, what happens to the determinant when two rows are swapped?

Question 7: Scalar Multiplication

If A is a 2 × 2 matrix and det(A) = 5, calculate det(3A).

Question 8: Invertibility

Determine whether:

\[ B=\begin{bmatrix}4&3\\2&5\end{bmatrix} \]

has an inverse.

Question 9: Finding an Unknown

Solve:

\[ \begin{vmatrix}x&2\\3&4\end{vmatrix}=10 \]

Question 10: Singular Matrix

Find k if:

\[ \begin{bmatrix}k&3\\4&6\end{bmatrix} \]

is singular.

Question 11: Parallelogram Area

Find the area of the parallelogram formed by:

u = (4, 1)

v = (2, 5)

Question 12: Triangle Area

Find the area of the triangle with vertices:

A(0, 0), B(6, 0), C(2, 4).

Question 13: Volume

Calculate the volume of the parallelepiped formed by the column vectors of:

\[ A=\begin{bmatrix} 2&0&0\\ 0&5&0\\ 0&0&3 \end{bmatrix} \]

Question 14: Cramer's Rule

Solve using determinants:

3x + 2y = 12

x − y = 1

Question 15: Integrated Determinant Challenge

Given:

\[ A=\begin{bmatrix} 2&1&0\\ 3&4&2\\ 1&0&5 \end{bmatrix} \]

a) Calculate det(A).

b) Determine whether A is invertible.

c) Explain the geometric meaning of the determinant.

d) Calculate det(2A).

 

Fully Worked Answers

Answer 1: Basic Determinant

det(A) = 3(5) − 2(4)

= 15 − 8

Answer: 7

Answer 2: Negative Determinant

det(A) = 1(2) − 6(3)

= 2 − 18

Answer: −16

Answer 3: Zero Determinant

det(A) = 2(10) − 4(5)

= 20 − 20

= 0

The determinant is zero.

Answer: The matrix is singular and not invertible.

Answer 4: 3 × 3 Determinant

\[ A=\begin{bmatrix} 1&2&3\\ 0&4&5\\ 1&0&6 \end{bmatrix} \]

Expand along the first row:

det(A) = 1(4 × 6 − 5 × 0) − 2(0 × 6 − 5 × 1) + 3(0 × 0 − 4 × 1)

= 1(24) − 2(−5) + 3(−4)

= 24 + 10 − 12

Answer: 22

Answer 5: Triangular Matrix

The matrix is upper triangular.

Multiply the diagonal elements:

det(A) = 3 × 4 × 6

Answer: 72

Answer 6: Determinant Properties

Swapping two rows changes the sign of the determinant.

Original determinant:

7

New determinant:

Answer: −7

Answer 7: Scalar Multiplication

For a 2 × 2 matrix:

det(kA) = k²det(A)

Therefore:

det(3A) = 3²(5)

= 9(5)

Answer: 45

Answer 8: Invertibility

\[ B=\begin{bmatrix}4&3\\2&5\end{bmatrix} \]

det(B) = 4(5) − 3(2)

= 20 − 6

= 14

Since det(B) ≠ 0:

Answer: B is invertible.

Answer 9: Finding an Unknown

\[ \begin{vmatrix}x&2\\3&4\end{vmatrix}=10 \]

4x − 6 = 10

4x = 16

Answer: x = 4

Answer 10: Singular Matrix

A singular matrix has determinant zero.

\[ \begin{vmatrix}k&3\\4&6\end{vmatrix}=0 \]

6k − 12 = 0

6k = 12

Answer: k = 2

Answer 11: Parallelogram Area

Vectors:

u = (4, 1)

v = (2, 5)

Area = |4(5) − 2(1)|

= |20 − 2|

Answer: 18 square units

Answer 12: Triangle Area

Vertices:

A(0, 0)

B(6, 0)

C(2, 4)

Vectors:

AB = (6, 0)

AC = (2, 4)

Area = 1/2 |6(4) − 2(0)|

= 1/2 |24|

Answer: 12 square units

Answer 13: Volume

The matrix is diagonal.

det(A) = 2 × 5 × 3

= 30

Volume = |det(A)|

Answer: 30 cubic units

Answer 14: Cramer's Rule

Equations:

3x + 2y = 12

x − y = 1

Step 1: Calculate D.

\[ D=\begin{vmatrix}3&2\\1&-1\end{vmatrix} \]

D = 3(−1) − 2(1)

= −3 − 2

= −5

Step 2: Calculate Dₓ.

\[ D_x=\begin{vmatrix}12&2\\1&-1\end{vmatrix} \]

Dₓ = 12(−1) − 2(1)

= −14

Step 3: Calculate Dᵧ.

\[ D_y=\begin{vmatrix}3&12\\1&1\end{vmatrix} \]

Dᵧ = 3(1) − 12(1)

= −9

Step 4: Solve.

x = Dₓ/D

= −14/−5

x = 14/5

y = Dᵧ/D

= −9/−5

y = 9/5

Verification:

3(14/5) + 2(9/5)

= 42/5 + 18/5

= 60/5 = 12

14/5 − 9/5 = 1

Both equations are satisfied.

Answer 15: Integrated Determinant Challenge

Given:

\[ A=\begin{bmatrix} 2&1&0\\ 3&4&2\\ 1&0&5 \end{bmatrix} \]

Part a: Calculate det(A).

Expand along the first row:

det(A) = 2(4 × 5 − 2 × 0) − 1(3 × 5 − 2 × 1) + 0

= 2(20) − (15 − 2)

= 40 − 13

det(A) = 27

Part b: Determine invertibility.

Since det(A) = 27 ≠ 0:

A is invertible.

Part c: Geometric interpretation.

The matrix represents a linear transformation in three-dimensional space.

Its determinant is 27.

Therefore, volumes are multiplied by:

|27| = 27

Because the determinant is positive, orientation is preserved.

Part d: Calculate det(2A).

For a 3 × 3 matrix:

det(kA) = k³det(A)

Therefore:

det(2A) = 2³(27)

= 8(27)

det(2A) = 216

 

Final Reflection

Determinants are much more than a calculation involving matrix elements.

They reveal fundamental properties of matrices and the transformations they represent.

A nonzero determinant tells us that a matrix is invertible and that the associated linear transformation does not collapse space into a lower dimension. A zero determinant indicates that some information is lost and the transformation cannot be reversed.

Geometrically, determinants connect matrix algebra to area, volume, orientation, and coordinate geometry. Algebraically, they provide tools for solving systems of equations and determining whether unique solutions exist.

By mastering 2 × 2 and 3 × 3 determinants, understanding their properties, and applying them to geometric and algebraic problems, we develop an important foundation for further study of linear algebra.

The central idea is that a determinant is a single number that reveals whether a matrix preserves independent dimensions, how it changes area or volume, and whether its transformation can be reversed.

 

 

 

 

 

Introduction

In ordinary arithmetic, many operations can be reversed.

For example, if a number is multiplied by 5, we can reverse the operation by dividing by 5.

5 × 8 = 40

40 ÷ 5 = 8

Matrices have a similar idea. Under certain conditions, a matrix multiplication can be reversed using an inverse matrix.

Suppose a matrix A transforms a vector x into another vector b:

Ax = b

If A has an inverse, we can use that inverse to recover the original vector:

x = A⁻¹b

Inverse matrices are important because they allow us to solve systems of linear equations, reverse geometric transformations, recover original data, and analyse mathematical models.

However, not every matrix has an inverse. Some transformations lose information, making it impossible to recover the original values uniquely.

In this lesson, we will investigate how inverse matrices are calculated, verified, and applied to mathematical and real-world problems.

What Is an Inverse Matrix?

The inverse matrix of a square matrix A is another matrix, written as A⁻¹, that reverses the effect of multiplication by A.

The defining property is:

AA⁻¹ = A⁻¹A = I

where I is the identity matrix.

For a 2 × 2 matrix, the identity matrix is:

The identity matrix plays a role similar to the number 1 in ordinary multiplication.

For example:

5 × 1 = 5

Similarly:

AI = IA = A

Understanding the relationship

If:

Ax = b

then multiplying both sides on the left by A⁻¹ gives:

A⁻¹Ax = A⁻¹b

Since:

A⁻¹A = I

we obtain:

Ix = A⁻¹b

Therefore:

This relationship is fundamental to solving linear systems using matrices.

Visualising inverse transformations

The inverse matrix undoes the original transformation.

Conditions for an Inverse Matrix to Exist

A matrix must satisfy certain conditions before it can have an inverse.

Condition 1: The matrix must be square

A square matrix has the same number of rows and columns.

Examples include:

• 2 × 2 matrices

• 3 × 3 matrices

• 4 × 4 matrices

A rectangular matrix does not have an ordinary two-sided inverse, although some rectangular matrices have specialised one-sided or generalised inverses.

Condition 2: The determinant must not be zero

A square matrix is invertible exactly when:

det(A) ≠ 0

If:

det(A) = 0

the matrix is called singular and does not have an inverse.

Worked Example 1: Checking Invertibility

Consider:

Calculate the determinant:

det(A) = 3(4) − 2(1)

= 12 − 2

= 10

Since det(A) ≠ 0:

A is invertible.

Now consider:

det(B) = 2(6) − 4(3)

= 12 − 12

= 0

Therefore:

B is singular and has no inverse.

Calculating the Inverse of a 2 × 2 Matrix

For a matrix:

the inverse is:

provided:

ad − bc ≠ 0

The four-step method

1. Calculate the determinant: ad − bc.

2. Check that the determinant is not zero.

3. Exchange the positions of a and d, and change the signs of b and c.

4. Multiply every element by 1/det(A).

Visualising the inverse formula

Finding Inverses Using Row Operations

The 2 × 2 inverse formula is convenient, but larger matrices require more general methods.

One important method is Gauss–Jordan elimination.

This method uses elementary row operations to transform a matrix into the identity matrix.

The augmented matrix method

To find A⁻¹, construct:

[A | I]

Perform row operations until the left side becomes I.

If successful:

[A | I] → [I | A⁻¹]

The right side is then the inverse matrix.

Allowed row operations

• Swap two rows.

• Multiply a row by a nonzero scalar.

• Add a multiple of one row to another row.

Worked Example 5: Inverse Using Gauss–Jordan Elimination

Find the inverse of:

\[ A=\begin{bmatrix}2&1\\1&1\end{bmatrix} \]

Step 1: Construct the augmented matrix.

\[ \left[\begin{array}{cc|cc} 2&1&1&0\\ 1&1&0&1 \end{array}\right] \]

Step 2: Swap the two rows.

\[ \left[\begin{array}{cc|cc} 1&1&0&1\\ 2&1&1&0 \end{array}\right] \]

Step 3: Eliminate the 2 below the first pivot.

R₂ → R₂ − 2R₁

\[ \left[\begin{array}{cc|cc} 1&1&0&1\\ 0&-1&1&-2 \end{array}\right] \]

Step 4: Make the second pivot equal to 1.

R₂ → −R₂

\[ \left[\begin{array}{cc|cc} 1&1&0&1\\ 0&1&-1&2 \end{array}\right] \]

Step 5: Eliminate the 1 above the second pivot.

R₁ → R₁ − R₂

\[ \left[\begin{array}{cc|cc} 1&0&1&-1\\ 0&1&-1&2 \end{array}\right] \]

The left side is now the identity matrix.

Therefore:

\[ \boxed{ A^{-1}=\begin{bmatrix}1&-1\\-1&2\end{bmatrix} } \]

This method works for invertible square matrices of any order.

 

Finding the Inverse of a 3 × 3 Matrix

A 3 × 3 inverse can be calculated using Gauss–Jordan elimination or the adjugate method.

The general formula is:

\[ \boxed{A^{-1}=\frac{1}{\det(A)}\operatorname{adj}(A)} \]

where adj(A) is the adjugate matrix, obtained by transposing the matrix of cofactors.

For larger matrices, Gauss–Jordan elimination is often easier to organise.

Worked Example 6: A 3 × 3 Inverse

Find the inverse of:

\[ A=\begin{bmatrix} 1&1&0\\ 0&1&1\\ 0&0&1 \end{bmatrix} \]

Start with:

\[ \left[\begin{array}{ccc|ccc} 1&1&0&1&0&0\\ 0&1&1&0&1&0\\ 0&0&1&0&0&1 \end{array}\right] \]

Step 1: Eliminate the 1 in row 2, column 3.

R₂ → R₂ − R₃

\[ \left[\begin{array}{ccc|ccc} 1&1&0&1&0&0\\ 0&1&0&0&1&-1\\ 0&0&1&0&0&1 \end{array}\right] \]

Step 2: Eliminate the 1 in row 1, column 2.

R₁ → R₁ − R₂

\[ \left[\begin{array}{ccc|ccc} 1&0&0&1&-1&1\\ 0&1&0&0&1&-1\\ 0&0&1&0&0&1 \end{array}\right] \]

Therefore:

\[ \boxed{ A^{-1}= \begin{bmatrix} 1&-1&1\\ 0&1&-1\\ 0&0&1 \end{bmatrix} } \]

Verification

Multiplying A by A⁻¹ gives:

\[ AA^{-1}= \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} \]

The result is the 3 × 3 identity matrix.

Therefore, the inverse is correct.

 

Solving Matrix Equations Using Inverses

Inverse matrices are especially useful for solving equations of the form:

AX = B

If A is invertible, multiply both sides on the left by A⁻¹:

A⁻¹AX = A⁻¹B

IX = A⁻¹B

Therefore:

\[ \boxed{X=A^{-1}B} \]

Why multiplication order matters

Matrix multiplication is generally not commutative.

Therefore:

A⁻¹B ≠ BA⁻¹

We must multiply on the correct side.

If the original equation is:

XA = B

then:

X = BA⁻¹

provided A is invertible.

 

Worked Example 7: Solving a Matrix Equation

Solve:

\[ AX=B \]

where:

\[ A=\begin{bmatrix}2&1\\1&1\end{bmatrix} \]

and:

\[ B=\begin{bmatrix}7\\4\end{bmatrix} \]

From our earlier example:

\[ A^{-1}=\begin{bmatrix}1&-1\\-1&2\end{bmatrix} \]

Use:

X = A⁻¹B

\[ X= \begin{bmatrix}1&-1\\-1&2\end{bmatrix} \begin{bmatrix}7\\4\end{bmatrix} \]

Calculate the first element:

1(7) + (−1)(4) = 3

Calculate the second element:

(−1)(7) + 2(4) = 1

Therefore:

\[ \boxed{X=\begin{bmatrix}3\\1\end{bmatrix}} \]

Verification

\[ AX= \begin{bmatrix}2&1\\1&1\end{bmatrix} \begin{bmatrix}3\\1\end{bmatrix} \]

\[ AX=\begin{bmatrix}7\\4\end{bmatrix}=B \]

The solution is correct.

 

Solving Systems of Linear Equations

A system of linear equations can be written as a matrix equation.

For example:

2x + 3y = 13

x + 2y = 8

can be written as:

\[ \begin{bmatrix}2&3\\1&2\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}13\\8\end{bmatrix} \]

This has the form:

AX = B

Worked Example 8: Solving a System Using the Inverse

Step 1: Identify the coefficient matrix.

\[ A=\begin{bmatrix}2&3\\1&2\end{bmatrix} \]

Step 2: Calculate the determinant.

det(A) = 2(2) − 3(1)

= 4 − 3

= 1

The inverse exists.

Step 3: Find the inverse.

\[ A^{-1}= \begin{bmatrix} 2&-3\\ -1&2 \end{bmatrix} \]

Step 4: Multiply by the constants.

\[ X= \begin{bmatrix}2&-3\\-1&2\end{bmatrix} \begin{bmatrix}13\\8\end{bmatrix} \]

\[ X= \begin{bmatrix} 26-24\\ -13+16 \end{bmatrix} \]

\[ X=\begin{bmatrix}2\\3\end{bmatrix} \]

Therefore:

x = 2

y = 3

Verification

2(2) + 3(3) = 13

2 + 2(3) = 8

Both equations are satisfied.

 

Solving a System of Three Equations

Inverse matrices can also solve systems with three unknowns.

Consider:

x + y = 5

y + z = 7

z = 4

Write:

\[ A=\begin{bmatrix} 1&1&0\\ 0&1&1\\ 0&0&1 \end{bmatrix} \]

\[ X=\begin{bmatrix}x\\y\\z\end{bmatrix} \]

\[ B=\begin{bmatrix}5\\7\\4\end{bmatrix} \]

From Worked Example 6:

\[ A^{-1}= \begin{bmatrix} 1&-1&1\\ 0&1&-1\\ 0&0&1 \end{bmatrix} \]

Calculate:

\[ X=A^{-1}B \]

\[ X= \begin{bmatrix} 1&-1&1\\ 0&1&-1\\ 0&0&1 \end{bmatrix} \begin{bmatrix}5\\7\\4\end{bmatrix} \]

First element:

5 − 7 + 4 = 2

Second element:

7 − 4 = 3

Third element:

4

Therefore:

\[ \boxed{X=\begin{bmatrix}2\\3\\4\end{bmatrix}} \]

So:

x = 2, y = 3, z = 4

 

Singular Matrices

A singular matrix is a square matrix that has no inverse.

A square matrix is singular exactly when its determinant is zero.

Why do singular matrices occur?

Singular matrices often arise when rows or columns are linearly dependent.

For example:

\[ A=\begin{bmatrix}2&4\\3&6\end{bmatrix} \]

The second column is twice the first column.

The determinant is:

det(A) = 2(6) − 4(3)

= 12 − 12

= 0

Therefore, the matrix is singular.

Geometric interpretation

A singular matrix represents a transformation that collapses space into a lower dimension.

For example, a two-dimensional region may be compressed onto a line.

When information is lost in this way, no inverse transformation can uniquely recover every original point.

 

Singular Matrices and Linear Systems

Consider the system:

2x + 4y = 10

x + 2y = 5

The coefficient matrix is:

\[ A=\begin{bmatrix}2&4\\1&2\end{bmatrix} \]

Its determinant is:

det(A) = 2(2) − 4(1)

= 0

The matrix is singular.

The first equation is twice the second equation.

Therefore, both equations describe the same line.

There are infinitely many solutions.

Now consider:

2x + 4y = 10

x + 2y = 7

The coefficient matrix is still singular.

However, the equations are inconsistent.

The first equation implies:

x + 2y = 5

while the second requires:

x + 2y = 7

These cannot both be true.

Therefore, the system has no solution.

Important: A zero determinant means that a square linear system does not have a unique solution. It does not tell us whether there are no solutions or infinitely many solutions without examining the equations.

 

Properties of Inverse Matrices

Inverse matrices follow several useful algebraic rules.

Property 1: Inverse of the inverse

(A⁻¹)⁻¹ = A

Reversing the inverse returns the original matrix.

Property 2: Inverse of a product

(AB)⁻¹ = B⁻¹A⁻¹

The order is reversed.

This is important because matrix multiplication is not generally commutative.

Property 3: Inverse of a transpose

(Aᵀ)⁻¹ = (A⁻¹)ᵀ

Property 4: Inverse of a scalar multiple

For k ≠ 0:

(kA)⁻¹ = (1/k)A⁻¹

provided A is invertible.

Property 5: Identity matrix

I⁻¹ = I

Property 6: Determinant of an inverse

det(A⁻¹) = 1/det(A)

Property 7: Inverse of a power

For a positive integer n:

(Aⁿ)⁻¹ = (A⁻¹)ⁿ

These properties are useful when simplifying complex matrix expressions.

 

Inverse Matrices and Geometric Transformations

Matrices can represent transformations of points in the coordinate plane.

Inverse matrices reverse these transformations.

Worked Example 9: Reversing a Stretch

Consider:

\[ A=\begin{bmatrix}2&0\\0&3\end{bmatrix} \]

This transformation stretches x-coordinates by a factor of 2 and y-coordinates by a factor of 3.

For the point:

P(2, 4)

\[ P=\begin{bmatrix}2\\4\end{bmatrix} \]

The transformed point is:

\[ AP= \begin{bmatrix}2&0\\0&3\end{bmatrix} \begin{bmatrix}2\\4\end{bmatrix} \]

\[ AP=\begin{bmatrix}4\\12\end{bmatrix} \]

The new point is:

P′(4, 12)

To reverse the transformation, calculate the inverse.

\[ A^{-1}=\begin{bmatrix}1/2&0\\0&1/3\end{bmatrix} \]

Apply the inverse to P′:

\[ A^{-1}P'= \begin{bmatrix}1/2&0\\0&1/3\end{bmatrix} \begin{bmatrix}4\\12\end{bmatrix} \]

\[ A^{-1}P'=\begin{bmatrix}2\\4\end{bmatrix} \]

The original point has been recovered.

Worked Example 10: Reversing a Rotation

A counterclockwise rotation through 90° is represented by:

\[ R=\begin{bmatrix}0&-1\\1&0\end{bmatrix} \]

The inverse is:

\[ R^{-1}=\begin{bmatrix}0&1\\-1&0\end{bmatrix} \]

The inverse represents a clockwise rotation through 90°.

If the original point is:

P(3, 2)

the counterclockwise rotation produces:

P′(−2, 3)

Applying R⁻¹ returns:

P(3, 2)

This demonstrates how inverse matrices can undo geometric transformations.

 

Practical Applications of Inverse Matrices

Application 1: Business and Production Planning

A factory produces two products, A and B.

Each unit of Product A requires:

• 2 hours of machine time.

• 1 hour of labour.

Each unit of Product B requires:

• 1 hour of machine time.

• 3 hours of labour.

The factory has:

• 100 machine hours.

• 120 labour hours.

Let:

x = units of Product A

y = units of Product B

The equations are:

2x + y = 100

x + 3y = 120

Write:

\[ A=\begin{bmatrix}2&1\\1&3\end{bmatrix} \]

\[ X=\begin{bmatrix}x\\y\end{bmatrix} \]

\[ B=\begin{bmatrix}100\\120\end{bmatrix} \]

The determinant is:

det(A) = 2(3) − 1(1)

= 5

The inverse is:

\[ A^{-1}=\frac15\begin{bmatrix}3&-1\\-1&2\end{bmatrix} \]

Calculate:

\[ X=A^{-1}B \]

\[ X=\frac15 \begin{bmatrix}3&-1\\-1&2\end{bmatrix} \begin{bmatrix}100\\120\end{bmatrix} \]

\[ X=\frac15\begin{bmatrix}180\\140\end{bmatrix} \]

\[ X=\begin{bmatrix}36\\28\end{bmatrix} \]

Therefore:

The factory should produce 36 units of Product A and 28 units of Product B to use exactly the available machine and labour hours.

 

Application 2: Recovering Original Data

Suppose two original measurements, x and y, are combined into new measurements:

u = 2x + y

v = x + y

This can be represented as:

\[ \begin{bmatrix}u\\v\end{bmatrix} = \begin{bmatrix}2&1\\1&1\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} \]

If:

u = 11

v = 7

we can recover the original measurements using the inverse.

\[ A^{-1}=\begin{bmatrix}1&-1\\-1&2\end{bmatrix} \]

Therefore:

\[ \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}1&-1\\-1&2\end{bmatrix} \begin{bmatrix}11\\7\end{bmatrix} \]

\[ \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}4\\3\end{bmatrix} \]

The original measurements were:

x = 4, y = 3

This type of calculation appears in signal processing and scientific measurement systems.

 

Application 3: Computer Graphics

三维坐标正交投影到二维平面 | Lixingcong

 

 

 

The Evolution of Robot Manipulation

 

 

 

Signal Processing Toolbox - MATLAB

 

8

 

Inverse matrices are used in computer graphics to reverse transformations.

For example, a point may be transformed from an object's local coordinate system into a world coordinate system.

An inverse transformation can recover its original local coordinates.

Applications include:

• Video game graphics

• Computer animation

• 3D modelling

• Camera positioning

• Robotics

• Engineering simulations

 

Choosing the Best Method for Finding an Inverse

Matrix type	Recommended method
2 × 2 matrix	Direct inverse formula
Simple 3 × 3 matrix	Gauss–Jordan elimination
Larger square matrix	Row reduction or computational methods
Diagonal matrix	Take reciprocals of nonzero diagonal elements
Identity matrix	The inverse is itself
Singular matrix	No inverse exists

For a diagonal matrix:

\[ A=\begin{bmatrix}2&0\\0&5\end{bmatrix} \]

the inverse is:

\[ A^{-1}=\begin{bmatrix}1/2&0\\0&1/5\end{bmatrix} \]

This works because the diagonal entries are nonzero.

 

Common Mistakes

Mistake 1: Assuming every matrix has an inverse

Only square matrices with nonzero determinants have ordinary two-sided inverses.

Mistake 2: Dividing every element by the determinant without rearranging the matrix

For a 2 × 2 inverse, the diagonal entries must be exchanged and the off-diagonal signs changed before multiplying by 1/det(A).

Mistake 3: Confusing A⁻¹ with 1/A

Matrix inversion is not element-by-element division.

Mistake 4: Forgetting to check the determinant

If det(A) = 0, the inverse does not exist.

Mistake 5: Making sign errors

Negative signs in the inverse formula must be handled carefully.

Mistake 6: Multiplying matrices in the wrong order

For AX = B:

X = A⁻¹B

not BA⁻¹.

Mistake 7: Assuming AB = BA

Matrix multiplication is generally not commutative.

Mistake 8: Forgetting to verify an inverse

Multiplying A by A⁻¹ should produce the identity matrix.

Mistake 9: Performing row operations on only one side of an augmented matrix

In Gauss–Jordan elimination, every row operation must be applied to the entire augmented matrix.

Mistake 10: Assuming a singular system always has no solution

A singular coefficient matrix may correspond to no solutions or infinitely many solutions.

Mistake 11: Rounding too early

Premature rounding can introduce errors, especially when the determinant is small.

Mistake 12: Forgetting the context of a solution

A mathematical result should be interpreted in the practical situation being modelled.

 

Key Terms

• Inverse Matrix: Matrix that reverses the effect of multiplication by an invertible matrix.

• Invertible Matrix: Square matrix possessing a multiplicative inverse.

• Identity Matrix: Square matrix with 1 on the main diagonal and 0 elsewhere.

• Determinant: Scalar value used to determine whether a square matrix is invertible.

• Singular Matrix: Square matrix with determinant zero.

• Nonsingular Matrix: Square matrix with nonzero determinant.

• Matrix Multiplication: Operation combining rows and columns through products and sums.

• Coefficient Matrix: Matrix containing coefficients of a linear system.

• Matrix Equation: Equation involving matrices.

• Augmented Matrix: Matrix formed by placing matrices side by side for row reduction.

• Elementary Row Operation: Permitted operation used to transform matrix rows.

• Gauss–Jordan Elimination: Method of row reduction used to solve systems and calculate inverses.

• Pivot: Leading nonzero element used during row reduction.

• Row Reduction: Process of simplifying a matrix using elementary row operations.

• Adjugate Matrix: Transpose of the cofactor matrix.

• Cofactor: Signed minor associated with a matrix element.

• Linear Transformation: Mapping represented by matrix multiplication.

• Inverse Transformation: Transformation that reverses an original transformation.

• Linear Dependence: Relationship in which one row or column can be expressed using others.

• Unique Solution: Exactly one solution to a system of equations.

• Inconsistent System: System with no solutions.

• Dependent System: System with infinitely many solutions when consistent.

• Scalar: Single number.

• Transpose: Matrix formed by interchanging rows and columns.

• Diagonal Matrix: Square matrix whose off-diagonal entries are zero.

 

Key Takeaways

• An inverse matrix reverses the effect of an invertible matrix.

• The defining property is AA⁻¹ = A⁻¹A = I.

• Only square matrices with nonzero determinants have ordinary inverses.

• The inverse of a 2 × 2 matrix can be calculated using a standard formula.

• The determinant must be checked before calculating an inverse.

• Inverse matrices can be verified through matrix multiplication.

• Gauss–Jordan elimination can calculate inverses of larger matrices.

• Matrix equations of the form AX = B can be solved using X = A⁻¹B.

• Matrix multiplication order matters.

• Singular matrices have determinant zero and no inverse.

• A singular coefficient matrix cannot produce a unique solution to a square linear system.

• Inverse matrices can reverse geometric transformations.

• Inverse matrices are useful in business, science, engineering, and computer graphics.

• In practical problems, solutions must be interpreted in context.

 

Check Your Understanding

Questions

Question 1: Invertibility

Determine whether the matrix is invertible:

\[ A=\begin{bmatrix}3&1\\2&4\end{bmatrix} \]

Question 2: Finding an Inverse

Find:

\[ A^{-1} \]

for:

\[ A=\begin{bmatrix}2&1\\1&1\end{bmatrix} \]

Question 3: Negative Determinant

Find the inverse of:

\[ B=\begin{bmatrix}1&3\\2&4\end{bmatrix} \]

Question 4: Singular Matrix

Explain why:

\[ C=\begin{bmatrix}2&6\\1&3\end{bmatrix} \]

has no inverse.

Question 5: Verification

Verify whether:

\[ B=\begin{bmatrix}3&-1\\-2&1\end{bmatrix} \]

is the inverse of:

\[ A=\begin{bmatrix}1&1\\2&3\end{bmatrix} \]

Question 6: Matrix Equation

Solve:

AX = B

where:

\[ A=\begin{bmatrix}2&1\\1&1\end{bmatrix} \]

\[ B=\begin{bmatrix}9\\5\end{bmatrix} \]

Question 7: Linear System

Solve using matrix inverses:

3x + y = 11

x + y = 5

Question 8: Diagonal Matrix

Find the inverse of:

\[ A=\begin{bmatrix}4&0\\0&2\end{bmatrix} \]

Question 9: Inverse Properties

If det(A) = 5, calculate det(A⁻¹).

Question 10: Product of Inverses

Simplify:

(AB)⁻¹

Question 11: Unknown Matrix Element

Find k so that:

\[ A=\begin{bmatrix}k&2\\3&6\end{bmatrix} \]

is singular.

Question 12: Geometric Transformation

A transformation doubles x-coordinates and triples y-coordinates.

Write the transformation matrix and its inverse.

Question 13: Three-Variable System

Use the inverse:

\[ A^{-1}= \begin{bmatrix} 1&-1&1\\ 0&1&-1\\ 0&0&1 \end{bmatrix} \]

to solve:

\[ AX=\begin{bmatrix}6\\8\\5\end{bmatrix} \]

Question 14: Business Application

A factory makes two products.

Product A requires 2 hours of machine time and 1 hour of labour.

Product B requires 1 hour of machine time and 2 hours of labour.

The factory has 80 machine hours and 70 labour hours.

Use inverse matrices to determine the number of units of each product.

Question 15: Integrated Challenge

Given:

\[ A=\begin{bmatrix}5&2\\3&1\end{bmatrix} \]

a) Calculate det(A).

b) Determine whether A is invertible.

c) Find A⁻¹.

d) Verify the inverse.

e) Solve:

\[ AX=\begin{bmatrix}19\\11\end{bmatrix} \]

 

Fully Worked Answers

Answer 1: Invertibility

\[ A=\begin{bmatrix}3&1\\2&4\end{bmatrix} \]

det(A) = 3(4) − 1(2)

= 12 − 2

= 10

Since det(A) ≠ 0:

Answer: A is invertible.

Answer 2: Finding an Inverse

\[ A=\begin{bmatrix}2&1\\1&1\end{bmatrix} \]

det(A) = 2(1) − 1(1)

= 1

Apply the inverse formula:

\[ A^{-1}= \begin{bmatrix} 1&-1\\ -1&2 \end{bmatrix} \]

Answer 3: Negative Determinant

\[ B=\begin{bmatrix}1&3\\2&4\end{bmatrix} \]

det(B) = 1(4) − 3(2)

= 4 − 6

= −2

Therefore:

\[ B^{-1}=\frac{1}{-2} \begin{bmatrix} 4&-3\\ -2&1 \end{bmatrix} \]

\[ \boxed{ B^{-1}= \begin{bmatrix} -2&3/2\\ 1&-1/2 \end{bmatrix} } \]

Answer 4: Singular Matrix

\[ C=\begin{bmatrix}2&6\\1&3\end{bmatrix} \]

det(C) = 2(3) − 6(1)

= 6 − 6

= 0

Therefore, C is singular.

Answer: The matrix has no inverse because its determinant is zero.

Answer 5: Verification

Given:

\[ A=\begin{bmatrix}1&1\\2&3\end{bmatrix} \]

\[ B=\begin{bmatrix}3&-1\\-2&1\end{bmatrix} \]

Calculate AB:

\[ AB= \begin{bmatrix} 1(3)+1(-2)&1(-1)+1(1)\\ 2(3)+3(-2)&2(-1)+3(1) \end{bmatrix} \]

\[ AB=\begin{bmatrix}1&0\\0&1\end{bmatrix} \]

Since AB = I and both matrices are square of the same order:

Answer: B is the inverse of A.

Answer 6: Matrix Equation

Given:

\[ A=\begin{bmatrix}2&1\\1&1\end{bmatrix} \]

\[ B=\begin{bmatrix}9\\5\end{bmatrix} \]

The inverse is:

\[ A^{-1}=\begin{bmatrix}1&-1\\-1&2\end{bmatrix} \]

Calculate:

\[ X=A^{-1}B \]

\[ X= \begin{bmatrix}1&-1\\-1&2\end{bmatrix} \begin{bmatrix}9\\5\end{bmatrix} \]

\[ X=\begin{bmatrix}4\\1\end{bmatrix} \]

Answer: x = 4, y = 1

Answer 7: Linear System

Equations:

3x + y = 11

x + y = 5

Coefficient matrix:

\[ A=\begin{bmatrix}3&1\\1&1\end{bmatrix} \]

det(A) = 3(1) − 1(1)

= 2

Inverse:

\[ A^{-1}=\frac12 \begin{bmatrix} 1&-1\\ -1&3 \end{bmatrix} \]

Calculate:

\[ X=\frac12 \begin{bmatrix} 1&-1\\ -1&3 \end{bmatrix} \begin{bmatrix}11\\5\end{bmatrix} \]

\[ X=\frac12\begin{bmatrix}6\\4\end{bmatrix} \]

\[ X=\begin{bmatrix}3\\2\end{bmatrix} \]

Answer: x = 3, y = 2

Answer 8: Diagonal Matrix

\[ A=\begin{bmatrix}4&0\\0&2\end{bmatrix} \]

Take the reciprocal of each diagonal element:

\[ \boxed{ A^{-1}= \begin{bmatrix} 1/4&0\\ 0&1/2 \end{bmatrix} } \]

Answer 9: Inverse Properties

Given:

det(A) = 5

Use:

det(A⁻¹) = 1/det(A)

Therefore:

Answer: det(A⁻¹) = 1/5

Answer 10: Product of Inverses

The inverse of a product reverses the multiplication order.

\[ \boxed{(AB)^{-1}=B^{-1}A^{-1}} \]

This assumes both A and B are invertible square matrices of the same order.

Answer 11: Unknown Matrix Element

\[ A=\begin{bmatrix}k&2\\3&6\end{bmatrix} \]

For A to be singular:

det(A) = 0

6k − 2(3) = 0

6k − 6 = 0

6k = 6

Answer: k = 1

Answer 12: Geometric Transformation

Doubling x-coordinates and tripling y-coordinates gives:

\[ A=\begin{bmatrix}2&0\\0&3\end{bmatrix} \]

The inverse reverses these stretches:

\[ \boxed{ A^{-1}= \begin{bmatrix} 1/2&0\\ 0&1/3 \end{bmatrix} } \]

Answer 13: Three-Variable System

Given:

\[ A^{-1}= \begin{bmatrix} 1&-1&1\\ 0&1&-1\\ 0&0&1 \end{bmatrix} \]

and:

\[ B=\begin{bmatrix}6\\8\\5\end{bmatrix} \]

Calculate:

\[ X=A^{-1}B \]

First element:

6 − 8 + 5 = 3

Second element:

8 − 5 = 3

Third element:

5

Therefore:

\[ \boxed{ X=\begin{bmatrix}3\\3\\5\end{bmatrix} } \]

Answer 14: Business Application

Let:

x = units of Product A

y = units of Product B

Equations:

2x + y = 80

x + 2y = 70

Write:

\[ A=\begin{bmatrix}2&1\\1&2\end{bmatrix} \]

\[ B=\begin{bmatrix}80\\70\end{bmatrix} \]

Calculate the determinant:

det(A) = 2(2) − 1(1)

= 3

The inverse is:

\[ A^{-1}=\frac13 \begin{bmatrix} 2&-1\\ -1&2 \end{bmatrix} \]

Calculate:

\[ X=A^{-1}B \]

\[ X=\frac13 \begin{bmatrix} 2&-1\\ -1&2 \end{bmatrix} \begin{bmatrix}80\\70\end{bmatrix} \]

\[ X=\frac13 \begin{bmatrix} 160-70\\ -80+140 \end{bmatrix} \]

\[ X=\frac13 \begin{bmatrix}90\\60\end{bmatrix} \]

\[ X=\begin{bmatrix}30\\20\end{bmatrix} \]

Answer: 30 units of Product A and 20 units of Product B.

Answer 15: Integrated Challenge

Given:

\[ A=\begin{bmatrix}5&2\\3&1\end{bmatrix} \]

Part a: Determinant

det(A) = 5(1) − 2(3)

= 5 − 6

det(A) = −1

Part b: Invertibility

Since det(A) ≠ 0:

A is invertible.

Part c: Find the inverse

\[ A^{-1}=\frac{1}{-1} \begin{bmatrix} 1&-2\\ -3&5 \end{bmatrix} \]

Therefore:

\[ \boxed{ A^{-1}= \begin{bmatrix} -1&2\\ 3&-5 \end{bmatrix} } \]

Part d: Verify

\[ AA^{-1}= \begin{bmatrix}5&2\\3&1\end{bmatrix} \begin{bmatrix}-1&2\\3&-5\end{bmatrix} \]

Top-left:

5(−1) + 2(3) = 1

Top-right:

5(2) + 2(−5) = 0

Bottom-left:

3(−1) + 1(3) = 0

Bottom-right:

3(2) + 1(−5) = 1

Therefore:

\[ AA^{-1}=\begin{bmatrix}1&0\\0&1\end{bmatrix} \]

The inverse is verified.

Part e: Solve AX = B

\[ B=\begin{bmatrix}19\\11\end{bmatrix} \]

\[ X=A^{-1}B \]

\[ X= \begin{bmatrix} -1&2\\ 3&-5 \end{bmatrix} \begin{bmatrix}19\\11\end{bmatrix} \]

First element:

−19 + 22 = 3

Second element:

57 − 55 = 2

Therefore:

\[ \boxed{ X=\begin{bmatrix}3\\2\end{bmatrix} } \]

Answer: x = 3, y = 2

Verification:

5(3) + 2(2) = 19

3(3) + 1(2) = 11

Both equations are satisfied.

 

Final Reflection

Inverse matrices provide a powerful way to reverse matrix transformations and solve systems of linear equations.

The key requirement is invertibility. A square matrix with a nonzero determinant has an inverse, while a singular matrix cannot be inverted because its transformation loses information.

The inverse of a 2 × 2 matrix can be calculated directly using a formula. Larger matrices can be inverted using row reduction, and all inverses can be verified by multiplying them with the original matrix to obtain the identity matrix.

Beyond algebra, inverse matrices are used to reverse geometric transformations, recover original measurements, solve production problems, and analyse scientific and technological systems.

The central idea is that an inverse matrix allows us to reverse a matrix operation, provided the original transformation preserves enough information to be uniquely undone.

 

Introduction

Many real-world problems involve several unknown quantities connected by multiple equations.

For example:

• A business may need to determine how many products to manufacture using limited resources.

• A scientist may need to calculate unknown concentrations in a chemical mixture.

• An engineer may need to determine forces acting on a structure.

• A transport company may need to calculate the costs of different delivery methods.

• A mathematician may need to determine where two or more lines intersect.

These situations can often be represented using systems of linear equations.

A system of linear equations is a collection of two or more linear equations involving the same variables. The goal is to find values that satisfy every equation simultaneously.

Consider:

2x + y = 7

x − y = 2

Solving the system gives:

x = 3, y = 1

We can verify this by substituting into both equations:

2(3) + 1 = 7

3 − 1 = 2

Both equations are satisfied.

Although systems can be solved using substitution, elimination, or graphing, matrices provide a systematic and powerful alternative.

Matrix methods become particularly valuable when solving larger systems containing three or more unknowns.

In this lesson, we will investigate how systems are represented using matrices, how matrix methods are used to find solutions, how solutions are interpreted graphically, and how these techniques apply to authentic mathematical problems.

Understanding Systems of Linear Equations

A linear equation is an equation in which each variable appears only to the first power and variables are not multiplied together.

Examples include:

2x + 3y = 12

4x − y = 7

x + 2y + 3z = 15

A system of linear equations consists of two or more such equations considered together.

Systems with two unknowns

A general system of two linear equations can be written as:

a₁x + b₁y = c₁

a₂x + b₂y = c₂

where:

• x and y are unknown variables.

• a₁, a₂, b₁, and b₂ are coefficients.

• c₁ and c₂ are constants.

A solution is an ordered pair (x, y) satisfying both equations.

Systems with three unknowns

A system involving three unknowns may have the form:

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z = d₂

a₃x + b₃y + c₃z = d₃

Its solution is an ordered triple:

(x, y, z)

Why use matrices?

Matrices organise the coefficients and constants into a compact structure.

Instead of repeatedly writing entire equations, we can perform operations directly on their numerical coefficients.

This makes it easier to develop systematic procedures and use computational tools.

Representing Systems Using Matrices

Consider the system:

3x + 2y = 16

x − y = 2

We can represent it using three matrices.

Coefficient matrix

The coefficient matrix contains the coefficients of the variables.

Variable matrix

The variable matrix contains the unknown quantities.

Constant matrix

The constant matrix contains the values on the right side of the equations.

The system can now be written as:

or:

Multiplying the matrices on the left gives:

This reproduces the original system.

Visualising the matrix representation

Important rules for constructing matrix equations

• Keep the variables in the same order in every equation.

• Include zero coefficients when a variable is missing.

• Place constants in the correct corresponding rows.

• Ensure the dimensions are compatible for multiplication.

Worked Example 1: Writing a System in Matrix Form

Write the following system in the form AX = B.

2x + 3y − z = 8

4x − y + 2z = 11

x + 5z = 7

Step 1: Identify the variables.

The variables are:

x, y, z

Step 2: Arrange the coefficients.

Notice that the third equation does not contain y.

Therefore, its y-coefficient is zero.

The coefficient matrix is:

Step 3: Construct the variable matrix.

Step 4: Construct the constant matrix.

Therefore:

The dimensions are:

A: 3 × 3

X: 3 × 1

B: 3 × 1

The multiplication is valid because:

(3 × 3)(3 × 1) = 3 × 1

Augmented Matrices

An augmented matrix combines the coefficient matrix and the constant matrix into one arrangement.

For the system:

3x + 2y = 16

x − y = 2

the augmented matrix is:

The vertical line separates the coefficients from the constants.

An augmented matrix is especially useful for solving systems through row operations.

Why augmented matrices are useful

They allow us to:

• Organise all the numerical information in one place.

• Perform elimination systematically.

• Solve systems with several variables.

• Identify systems with no solution.

• Identify systems with infinitely many solutions.

Solving Systems Using Gaussian Elimination

Gaussian elimination is a matrix method that uses elementary row operations to simplify a system.

The goal is to transform the augmented matrix into row echelon form, from which the variables can be found using back-substitution.

Elementary row operations

There are three allowed operations.

1. Interchange two rows.

2. Multiply a row by a nonzero number.

3. Add a multiple of one row to another row.

These operations preserve the solution set of the system.

Worked Example 2: Gaussian Elimination

Solve:

2x + y = 7

x − y = 2

Step 1: Construct the augmented matrix.

Step 2: Swap the rows.

This places a coefficient of 1 in the top-left position.

R₁ ↔ R₂

Step 3: Eliminate the x-coefficient in row 2.

R₂ → R₂ − 2R₁

Step 4: Solve for y.

The second row represents:

3y = 3

Therefore:

y = 1

Step 5: Substitute into the first equation.

x − y = 2

x − 1 = 2

x = 3

Therefore:

Verification

2(3) + 1 = 7

3 − 1 = 2

Both equations are satisfied.

Gauss–Jordan Elimination

Gauss–Jordan elimination extends Gaussian elimination by continuing the row operations until the coefficient portion is in reduced row echelon form.

For a system with a unique solution, the final matrix has the form:

The solution can be read directly from the final column.

Worked Example 3: Gauss–Jordan Elimination

Solve:

2x + y = 9

x + 2y = 6

Step 1: Write the augmented matrix.

Step 2: Swap the rows.

Step 3: Eliminate the 2 below the first pivot.

R₂ → R₂ − 2R₁

Step 4: Divide row 2 by −3.

R₂ → −(1/3)R₂

Step 5: Eliminate the 2 above the second pivot.

R₁ → R₁ − 2R₂

Therefore:

This method is particularly useful because it provides the solution without requiring a separate back-substitution stage.

Solving Systems Using Inverse Matrices

A system can be written as:

AX = B

If A is square and invertible:

This method is useful when the coefficient matrix has a nonzero determinant.

Worked Example 4: Solving with an Inverse Matrix

Solve:

3x + y = 11

2x + y = 8

Step 1: Write the system in matrix form.

Step 2: Calculate the determinant.

det(A) = 3(1) − 1(2)

= 1

The matrix is invertible.

Step 3: Find the inverse.

Step 4: Calculate X.

First element:

11 − 8 = 3

Second element:

−22 + 24 = 2

Therefore:

x = 3, y = 2

Verification

3(3) + 2 = 11

2(3) + 2 = 8

Both equations are satisfied.

Solving Systems Using Cramer's Rule

Cramer's Rule uses determinants to solve square systems with a nonzero coefficient determinant.

For:

ax + by = e

cx + dy = f

define:

If D ≠ 0:

Worked Example 5: Cramer's Rule

Solve:

2x + 3y = 13

x + y = 5

Step 1: Calculate D.

D = 2(1) − 3(1)

= −1

Step 2: Calculate Dₓ.

Dₓ = 13(1) − 3(5)

= −2

Step 3: Calculate Dᵧ.

Dᵧ = 2(5) − 13(1)

= −3

Step 4: Calculate x and y.

x = Dₓ/D = (−2)/(−1) = 2

y = Dᵧ/D = (−3)/(−1) = 3

Therefore:

Solving Systems with Three Unknowns

Matrix methods are particularly valuable for systems containing three or more variables.

Consider:

x + y + z = 6

2x − y + z = 3

x + 2y − z = 2

Worked Example 6: Three-Variable Gaussian Elimination

Step 1: Write the augmented matrix.

Step 2: Eliminate x from rows 2 and 3.

R₂ → R₂ − 2R₁

R₃ → R₃ − R₁

Step 3: Swap rows 2 and 3.

Step 4: Eliminate y from row 3.

R₃ → R₃ + 3R₂

Step 5: Solve for z.

−7z = −21

z = 3

Step 6: Solve for y.

From row 2:

y − 2z = −4

y − 6 = −4

y = 2

Step 7: Solve for x.

From row 1:

x + y + z = 6

x + 2 + 3 = 6

x = 1

Therefore:

Verification

First equation:

1 + 2 + 3 = 6

Second equation:

2(1) − 2 + 3 = 3

Third equation:

1 + 2(2) − 3 = 2

All three equations are satisfied.

Graphical Interpretation of Linear Systems

Every linear equation in two variables represents a straight line.

When two equations are considered together, their solutions correspond to the intersection points of the lines.

There are three main possibilities.

Case 1: One unique solution

Two nonparallel lines intersect at exactly one point.

For example:

y = 2x + 1

y = −x + 7

Set the equations equal:

2x + 1 = −x + 7

3x = 6

x = 2

Substitute:

y = 2(2) + 1 = 5

The intersection is:

(2, 5)

The coefficient determinant is nonzero, so the system has a unique solution.

Case 2: No solution

Consider:

y = 2x + 1

y = 2x − 3

Both lines have the same gradient but different y-intercepts.

They are parallel and never intersect.

Therefore:

The system has no solution.

The coefficient matrix has determinant zero.

Case 3: Infinitely many solutions

Consider:

2x + y = 6

4x + 2y = 12

The second equation is twice the first.

Both equations describe the same line.

Therefore:

The system has infinitely many solutions.

The coefficient matrix again has determinant zero.

Comparing the three cases

Matrix Interpretation of Different Solution Types

When using row reduction, the final augmented matrix reveals the nature of the system.

Unique solution

This represents:

x = 3

y = 2

There is one unique solution.

No solution

The second row represents:

0x + 0y = 3

or:

0 = 3

This is impossible.

Therefore, the system is inconsistent and has no solution.

Infinitely many solutions

The second row represents:

0 = 0

This provides no new restriction.

The first row gives:

x + 2y = 5

Let:

y = t

Then:

x = 5 − 2t

Therefore, the solutions can be written as:

(x, y) = (5 − 2t, t)

where t can be any real number.

This describes infinitely many solutions.

For a general system, the number of free variables determines the number of parameters needed to describe the solution set.

Systems of Three Equations: Graphical Interpretation

In three dimensions, a linear equation generally represents a plane rather than a line.

Three planes can:

• Intersect at one point.

• Share a common line.

• Coincide.

• Have no common intersection.

A unique solution corresponds to one point where all three planes meet.

A system with infinitely many solutions may have a common line or plane of solutions.

A system with no solution has no point satisfying all three equations simultaneously.

For a square 3 × 3 coefficient matrix:

• det(A) ≠ 0 guarantees one unique solution.

• det(A) = 0 means the system has no unique solution.

The second case must be investigated further to determine whether there are no solutions or infinitely many solutions.

Comparing Matrix and Algebraic Methods

Systems can be solved using several approaches.

Example: Comparing methods

Consider:

x + y = 7

x − y = 1

Algebraic elimination

Add the equations:

2x = 8

x = 4

Then:

y = 3

Matrix method

R₂ → R₂ − R₁

Therefore:

y = 3

x = 4

Both methods produce the same solution.

For a small system, algebraic elimination may be faster. For a system with many variables, matrix methods provide a more organised and scalable procedure.

Real-World Application 1: Ticket Sales

A school sells adult and student tickets for a concert.

Adult tickets cost $12.

Student tickets cost $7.

A total of 150 tickets are sold.

The total revenue is $1,350.

How many adult and student tickets were sold?

Step 1: Define variables

Let:

x = number of adult tickets

y = number of student tickets

Step 2: Construct equations

Total tickets:

x + y = 150

Total revenue:

12x + 7y = 1350

Step 3: Write the matrix equation

Step 4: Solve

The determinant is:

D = 1(7) − 1(12)

= −5

The inverse is:

Multiply:

First component:

(1050 − 1350)/(−5) = 60

Second component:

(−1800 + 1350)/(−5) = 90

Therefore:

60 adult tickets and 90 student tickets were sold.

Verification

60 + 90 = 150

12(60) + 7(90) = 720 + 630 = 1350

Both conditions are satisfied.

Real-World Application 2: Mixture Problems

A laboratory needs 20 litres of a 35% solution.

Two solutions are available:

• Solution A: 20% concentration.

• Solution B: 50% concentration.

How many litres of each solution should be mixed?

Step 1: Define variables

x = litres of 20% solution

y = litres of 50% solution

Step 2: Construct equations

Total volume:

x + y = 20

Amount of dissolved substance:

0.20x + 0.50y = 0.35(20)

Therefore:

0.20x + 0.50y = 7

Step 3: Matrix representation

Step 4: Solve by elimination

Multiply the first equation by 0.20:

0.20x + 0.20y = 4

Subtract from the concentration equation:

0.30y = 3

y = 10

Then:

x + 10 = 20

x = 10

Therefore:

10 litres of the 20% solution and 10 litres of the 50% solution are required.

This model assumes volumes are additive and the solutions mix uniformly.

Real-World Application 3: Production Planning

A workshop manufactures tables and chairs.

Each table requires:

• 4 hours of cutting.

• 2 hours of assembly.

Each chair requires:

• 2 hours of cutting.

• 3 hours of assembly.

The workshop has:

• 100 cutting hours.

• 90 assembly hours.

Determine how many tables and chairs can be manufactured if all available hours are used.

Let:

x = number of tables

y = number of chairs

The equations are:

4x + 2y = 100

2x + 3y = 90

The matrix equation is:

The determinant is:

D = 4(3) − 2(2)

= 8

The inverse is:

Calculate:

Therefore:

The workshop can manufacture 15 tables and 20 chairs.

Verification

Cutting:

4(15) + 2(20) = 100

Assembly:

2(15) + 3(20) = 90

The solution satisfies both resource constraints.

Real-World Application 4: Electrical Circuits

In electrical engineering, systems of linear equations are used to calculate unknown currents in circuits.

Kirchhoff's laws produce equations describing relationships between currents and voltages.

For example, suppose two unknown loop currents satisfy:

5I₁ − 2I₂ = 12

−2I₁ + 4I₂ = 6

The matrix equation is:

The determinant is:

D = 5(4) − (−2)(−2)

= 20 − 4

= 16

The inverse is:

Therefore:

Thus:

I₁ = 3.75 A

I₂ = 3.375 A

The currents satisfy the original equations.

This illustrates how matrix methods can solve systems arising in electrical engineering.

Choosing an Efficient Strategy

When solving a system, begin by examining its structure.

If the system has two equations with simple coefficients:

Substitution or elimination may be quickest.

If the system has three or more variables:

Gaussian elimination is often a good choice.

If the coefficient matrix is square and invertible:

Inverse matrices or Cramer's Rule are possible.

If you need to identify whether a system has no solution or infinitely many solutions:

Row reduction is particularly useful.

If a graphical interpretation is important:

Graph the equations to understand the solution geometrically.

If the system is large:

Computational linear algebra methods are usually more efficient than calculating a full inverse explicitly.

Common Mistakes

Mistake 1: Placing coefficients in the wrong columns

Always keep the variables in the same order.

Mistake 2: Forgetting zero coefficients

If an equation does not contain a variable, include a zero in its column.

Mistake 3: Mixing coefficients and constants

The augmented matrix must preserve the correct relationship between equations and constants.

Mistake 4: Performing row operations on only part of the augmented matrix

Every row operation must apply to the entire row, including the constant column.

Mistake 5: Making sign errors during elimination

Subtracting a negative number produces a positive result.

Mistake 6: Using inverse matrices when the determinant is zero

The inverse does not exist for a singular matrix.

Mistake 7: Assuming a zero determinant means no solution

A singular system may have no solutions or infinitely many solutions.

Mistake 8: Confusing Gaussian and Gauss–Jordan elimination

Gaussian elimination typically uses back-substitution; Gauss–Jordan elimination continues to reduced row echelon form.

Mistake 9: Interpreting every system as having one solution

Systems can have one, none, or infinitely many solutions.

Mistake 10: Forgetting to check answers

Substitute solutions into every original equation.

Mistake 11: Using graphing as though it always gives exact values

Graphical solutions may be approximate unless the intersection is known exactly.

Mistake 12: Ignoring real-world constraints

Negative quantities or fractional numbers of indivisible products may not be meaningful.

Key Terms

• Linear Equation: Equation involving variables only to the first power.

• System of Linear Equations: Collection of linear equations involving common variables.

• Simultaneous Solution: Values satisfying all equations in a system.

• Coefficient: Number multiplying a variable.

• Constant: Fixed numerical value.

• Coefficient Matrix: Matrix containing the coefficients of a linear system.

• Variable Matrix: Column matrix containing the unknown variables.

• Constant Matrix: Column matrix containing the right-hand-side values.

• Matrix Equation: Equation expressed using matrices.

• Augmented Matrix: Matrix combining coefficients and constants.

• Gaussian Elimination: Method using row operations to reach row echelon form.

• Gauss–Jordan Elimination: Method using row operations to reach reduced row echelon form.

• Elementary Row Operation: Allowed operation that preserves the solution set.

• Pivot: Leading nonzero entry used during row reduction.

• Row Echelon Form: Matrix form suitable for back-substitution.

• Reduced Row Echelon Form: Matrix form with leading ones and zeros elsewhere in pivot columns.

• Back-Substitution: Solving variables from the last equation upward.

• Inverse Matrix: Matrix reversing the multiplication effect of an invertible matrix.

• Determinant: Scalar value used to test invertibility of a square matrix.

• Cramer's Rule: Method solving square linear systems using determinants.

• Unique Solution: Exactly one solution satisfying the system.

• Inconsistent System: System with no solution.

• Dependent System: System containing redundant equations.

• Independent Equations: Equations contributing independent constraints.

• Singular Matrix: Square matrix with determinant zero.

• Free Variable: Variable that can take arbitrary values in a system with infinitely many solutions.

• Parameter: Symbol used to describe a family of solutions.

• Intersection Point: Point shared by two or more graphs.

• Parallel Lines: Lines in a plane that never intersect.

• Coincident Lines: Lines that occupy exactly the same position.

• Linear Model: Mathematical representation using linear relationships.

Key Takeaways

• A system of linear equations contains two or more equations involving common variables.

• A solution must satisfy every equation simultaneously.

• Systems can be represented compactly as AX = B.

• The coefficient matrix contains the variable coefficients.

• The augmented matrix combines coefficients and constants.

• Gaussian elimination uses row operations and back-substitution.

• Gauss–Jordan elimination reduces the system to a form where solutions can be read directly.

• Inverse matrices solve systems using X = A⁻¹B when A is invertible.

• Cramer's Rule uses determinants to solve square systems with nonzero determinants.

• A system can have one solution, no solution, or infinitely many solutions.

• In two dimensions, solutions correspond to intersections of lines.

• Parallel distinct lines produce no solution.

• Coincident lines produce infinitely many solutions.

• A nonzero determinant guarantees a unique solution for a square system.

• A zero determinant means a square system does not have a unique solution.

• Different solving methods are useful in different situations.

• Matrix methods are especially valuable for larger systems.

• Real-world applications include business, chemistry, engineering, and electrical circuits.

Check Your Understanding

Questions

Question 1: Matrix Representation

Write the system in matrix form:

2x + 3y = 12

4x − y = 5

Question 2: Augmented Matrix

Construct the augmented matrix for:

x + 2y = 7

3x − y = 8

Question 3: Missing Coefficient

Write the coefficient matrix for:

2x + 3z = 8

x + y − z = 4

3y + 2z = 9

Question 4: Gaussian Elimination

Solve:

x + y = 9

2x − y = 3

Question 5: Gauss–Jordan Elimination

Solve:

2x + y = 8

x + 3y = 9

Question 6: Inverse Matrix Method

Solve using an inverse matrix:

3x + y = 10

x + y = 6

Question 7: Cramer's Rule

Solve:

2x + y = 7

x − y = 2

Question 8: Graphical Interpretation

Determine the number of solutions:

y = 3x + 2

y = 3x − 5

Explain your reasoning.

Question 9: Infinitely Many Solutions

Determine the number of solutions:

2x + 4y = 12

x + 2y = 6

Question 10: Singular Matrix

Determine whether the coefficient matrix of:

3x + 6y = 9

x + 2y = 3

is singular.

Question 11: Three-Variable System

Solve:

x + y + z = 9

x − y + z = 3

2x + y − z = 3

Question 12: Ticket Sales

A cinema sells adult tickets for $15 and student tickets for $10.

A total of 120 tickets are sold.

Revenue is $1,500.

How many of each ticket were sold?

Question 13: Mixture Problem

A laboratory needs 30 litres of a 40% solution.

It has 20% and 50% solutions.

How many litres of each are required?

Question 14: Comparing Methods

Explain why Gaussian elimination may be preferable to graphing for a system with four unknown variables.

Question 15: Integrated Challenge

Consider:

2x + y = 11

3x − 2y = 4

a) Write the matrix equation AX = B.

b) Construct the augmented matrix.

c) Calculate det(A).

d) Solve the system using a matrix method.

e) Explain the graphical meaning of the solution.

f) Verify the solution.

Fully Worked Answers

Answer 1: Matrix Representation

System:

2x + 3y = 12

4x − y = 5

The coefficient matrix is:

The variable matrix is:

The constant matrix is:

Therefore:

Answer 2: Augmented Matrix

System:

x + 2y = 7

3x − y = 8

The augmented matrix is:

Answer 3: Missing Coefficient

System:

2x + 3z = 8

x + y − z = 4

3y + 2z = 9

The missing coefficients are zero.

Therefore:

Answer 4: Gaussian Elimination

System:

x + y = 9

2x − y = 3

Augmented matrix:

R₂ → R₂ − 2R₁

Therefore:

−3y = −15

y = 5

Substitute:

x + 5 = 9

x = 4

Answer: x = 4, y = 5

Answer 5: Gauss–Jordan Elimination

System:

2x + y = 8

x + 3y = 9

Augmented matrix:

Swap rows:

R₂ → R₂ − 2R₁

R₂ → −(1/5)R₂

R₁ → R₁ − 3R₂

Answer: x = 3, y = 2

Answer 6: Inverse Matrix Method

System:

3x + y = 10

x + y = 6

Coefficient matrix:

det(A) = 3(1) − 1(1)

= 2

Inverse:

Calculate:

Answer: x = 2, y = 4

Answer 7: Cramer's Rule

System:

2x + y = 7

x − y = 2

Calculate:

D = 2(−1) − 1(1)

= −3

Dₓ = 7(−1) − 1(2)

= −9

Dᵧ = 2(2) − 7(1)

= −3

Therefore:

x = Dₓ/D = (−9)/(−3) = 3

y = Dᵧ/D = (−3)/(−3) = 1

Answer: x = 3, y = 1

Answer 8: Graphical Interpretation

Equations:

y = 3x + 2

y = 3x − 5

Both lines have gradient 3.

However, their y-intercepts are different.

Therefore, they are distinct parallel lines.

Answer: No solution.

Answer 9: Infinitely Many Solutions

Equations:

2x + 4y = 12

x + 2y = 6

Multiply the second equation by 2:

2x + 4y = 12

This is identical to the first equation.

Therefore, both equations represent the same line.

Answer: Infinitely many solutions.

Answer 10: Singular Matrix

Coefficient matrix:

det(A) = 3(2) − 6(1)

= 6 − 6

= 0

Therefore:

Answer: The coefficient matrix is singular.

The equations are also consistent because the first is three times the second, so the system has infinitely many solutions.

Answer 11: Three-Variable System

System:

x + y + z = 9

x − y + z = 3

2x + y − z = 3

Subtract the second equation from the first:

2y = 6

y = 3

Substitute into the first equation:

x + z = 6

Substitute y = 3 into the third equation:

2x + 3 − z = 3

2x − z = 0

Therefore:

z = 2x

Substitute into x + z = 6:

x + 2x = 6

3x = 6

x = 2

Then:

z = 4

Answer: x = 2, y = 3, z = 4

Answer 12: Ticket Sales

Let:

x = adult tickets

y = student tickets

Equations:

x + y = 120

15x + 10y = 1500

Multiply the first equation by 10:

10x + 10y = 1200

Subtract:

5x = 300

x = 60

Then:

y = 120 − 60

y = 60

Answer: 60 adult tickets and 60 student tickets.

Verification:

15(60) + 10(60) = 1500

Answer 13: Mixture Problem

Let:

x = litres of 20% solution

y = litres of 50% solution

Total volume:

x + y = 30

Concentration equation:

0.20x + 0.50y = 0.40(30)

0.20x + 0.50y = 12

Multiply the volume equation by 0.20:

0.20x + 0.20y = 6

Subtract:

0.30y = 6

y = 20

Then:

x = 30 − 20

x = 10

Answer: 10 litres of 20% solution and 20 litres of 50% solution.

Answer 14: Comparing Methods

Gaussian elimination is preferable for a system with four unknowns because:

• It provides a systematic algebraic procedure.

• It can produce exact solutions.

• It avoids the difficulty of graphing equations in four dimensions.

• It can identify inconsistent or dependent systems.

• It extends naturally to larger systems.

Answer: Gaussian elimination is generally more practical and precise for systems involving several variables.

Answer 15: Integrated Challenge

System:

2x + y = 11

3x − 2y = 4

Part a: Matrix equation

Part b: Augmented matrix

Part c: Determinant

det(A) = 2(−2) − 1(3)

= −4 − 3

det(A) = −7

Since the determinant is nonzero, the system has one unique solution.

Part d: Solve using the inverse matrix

Calculate:

Therefore:

Part e: Graphical interpretation

The two lines intersect at exactly one point:

(26/7, 25/7)

This is the unique solution.

Part f: Verification

First equation:

2(26/7) + 25/7

= 52/7 + 25/7

= 77/7

= 11

Second equation:

3(26/7) − 2(25/7)

= 78/7 − 50/7

= 28/7

= 4

Both equations are satisfied.

Final Reflection

Systems of linear equations provide a mathematical framework for understanding relationships involving several unknown quantities.

Matrix notation allows us to represent these relationships efficiently and solve them using systematic procedures.

Gaussian elimination, Gauss–Jordan elimination, inverse matrices, and Cramer's Rule all provide ways to solve systems, but each method has different advantages.

Graphical interpretation helps us understand why systems may have one solution, no solution, or infinitely many solutions.

The greatest value of these methods becomes clear when they are applied to real-world situations. Business planning, scientific mixtures, engineering systems, and electrical circuits can all be represented using systems of linear equations.

The central idea is that matrix methods transform collections of linear equations into organised numerical structures that can be analysed, solved, and interpreted efficiently.