Matrix Inverse, the Determinant, and Matrix Equations Revision Notes for VCE SSCE General Mathematics

Matrix Inverse, the Determinant, and Matrix Equations

Understanding the inverse matrix

You've already learned how to add, subtract, and multiply matrices. But what about dividing them? Matrix division works differently from regular division - it's more complex, similar to how matrix multiplication differs from regular multiplication.

The foundation of matrix division is the inverse matrix. When we have a square matrix $A$ , its inverse is written as $A^{-1}$ .

Key property of inverse matrices

The inverse matrix has a special property that defines it:

$AA^{-1} = A^{-1}A = I$

where $I$ is the identity matrix (a matrix with $1$ s along the main diagonal and $0$ s everywhere else).

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This property is similar to how a number multiplied by its reciprocal equals $1$ . For matrices, when you multiply a matrix by its inverse, you get the identity matrix instead.

Do all matrices have inverses?

Not all matrices have inverses. To understand which matrices have inverses and which don't, we first need to learn about another important concept: the determinant.

Before we get to that, let's see an example of matrices that are inverses of each other.

Demonstrating that two matrices are inverses

To prove that two matrices are inverses, we need to show that when we multiply them together (in both orders), we get the identity matrix.

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Worked Example: Verifying Inverse Matrices

Consider the matrices:

$A = \begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix} \text{ and } B = \begin{bmatrix} 5 & -3 \\ -3 & 2 \end{bmatrix}$

To show these are inverses, we need to verify that $AB = I$ and $BA = I$ .

Step 1: Calculate $AB$

$AB = \begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix} \begin{bmatrix} 5 & -3 \\ -3 & 2 \end{bmatrix}$

Multiply row by column:

$= \begin{bmatrix} 2 \times 5 + 3 \times (-3) & 2 \times (-3) + 3 \times 2 \\ 3 \times 5 + 5 \times (-3) & 3 \times (-3) + 5 \times 2 \end{bmatrix}$

$= \begin{bmatrix} 10 - 9 & -6 + 6 \\ 15 - 15 & -9 + 10 \end{bmatrix}$

$= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$

Step 2: Calculate $BA$

$BA = \begin{bmatrix} 5 & -3 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix}$

$= \begin{bmatrix} 5 \times 2 + (-3) \times 3 & 5 \times 3 + (-3) \times 5 \\ (-3) \times 2 + 2 \times 3 & (-3) \times 3 + 2 \times 5 \end{bmatrix}$

$= \begin{bmatrix} 10 - 9 & 15 - 15 \\ -6 + 6 & -9 + 10 \end{bmatrix}$

$= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$

Conclusion: Since both AB = I and BA = I, we can confirm that $A$ and $B$ are inverses of each other.

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Exam Tip: You can use your calculator to speed up matrix multiplication, but make sure you understand the manual process for simple $2 \times 2$ matrices.

The determinant of a matrix

While the example above shows that some matrices have inverses, many square matrices do not. To understand why, we need to introduce the determinant.

The determinant of a $2 \times 2$ matrix

For a $2 \times 2$ matrix, the determinant is a single number calculated using a simple formula.

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Definition:

If $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , then:

$\det(A) = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$

Memory aid: "Multiply down-right, subtract up-left"

Multiply the elements on the main diagonal: $a \times d$
Subtract the product of the off-diagonal elements: $b \times c$

Calculating determinants - worked examples

Let's calculate the determinants of three different matrices to see what values they can take.

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Worked Example 1:

$A = \begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix}$

$\det(A) = \begin{vmatrix} 2 & 3 \\ 3 & 5 \end{vmatrix} = 2 \times 5 - 3 \times 3 = 10 - 9 = 1$

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Worked Example 2:

$B = \begin{bmatrix} 2 & 3 \\ 2 & 3 \end{bmatrix}$

$\det(B) = \begin{vmatrix} 2 & 3 \\ 2 & 3 \end{vmatrix} = 2 \times 3 - 2 \times 3 = 6 - 6 = 0$

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Worked Example 3:

$C = \begin{bmatrix} 2 & 4 \\ 2 & 3 \end{bmatrix}$

$\det(C) = \begin{vmatrix} 2 & 4 \\ 2 & 3 \end{vmatrix} = 2 \times 3 - 2 \times 4 = 6 - 8 = -2$

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Key observation: The determinant can be positive (like $1$ ), negative (like $-2$ ), or zero. This value is crucial for determining whether a matrix has an inverse.

Why the determinant matters for inverses

For a matrix to have an inverse, its determinant must be non-zero. If $\det(A) = 0$ , the matrix is called singular and does not have an inverse.

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A matrix has an inverse if and only if its determinant is non-zero. Matrices with zero determinant are called singular matrices and cannot be inverted.

Finding the inverse of a $2 \times 2$ matrix

While you'll typically use a calculator to find matrix inverses, understanding the formula helps you see why some matrices don't have inverses.

The formula for a $2 \times 2$ inverse

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Rule:

If $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , then:

$A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = \frac{1}{\det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

provided $ad - bc \neq 0$ , that is, provided $\det(A) \neq 0$

Memory aid for the formula:

Calculate the determinant: $ad - bc$
Swap the positions of $a$ and $d$ (the main diagonal elements)
Change the signs of $b$ and $c$ (the off-diagonal elements)
Divide everything by the determinant

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This formula immediately shows why you can't calculate an inverse when the determinant is zero - you would be dividing by zero!

Using the formula - worked examples

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Worked Example 1: Find the inverse of $A = \begin{bmatrix} 2 & 2 \\ 3 & 4 \end{bmatrix}$

Step 1: Calculate the determinant

$\det(A) = \begin{vmatrix} 2 & 2 \\ 3 & 4 \end{vmatrix} = 2 \times 4 - 3 \times 2 = 8 - 6 = 2$

Step 2: Apply the inverse formula

$A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

$= \frac{1}{2} \begin{bmatrix} 4 & -2 \\ -3 & 2 \end{bmatrix}$

$= \begin{bmatrix} 2 & -1 \\ -1.5 & 1 \end{bmatrix}$

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Worked Example 2: Find the inverse of $B = \begin{bmatrix} 2 & 3 \\ 2 & 3 \end{bmatrix}$

Step 1: Calculate the determinant

$\det(B) = \begin{vmatrix} 2 & 3 \\ 2 & 3 \end{vmatrix} = 2 \times 3 - 2 \times 3 = 0$

Conclusion: Since det(B) = 0, matrix $B$ does not have an inverse. It's a singular matrix.

Using a CAS calculator for larger matrices

For square matrices larger than $2 \times 2$ , the formulas become very complicated. In practice, we use calculators to find determinants and inverses of $n \times n$ matrices.

TI-Nspire CAS method

Here's how to find the determinant and inverse of a matrix using the TI-Nspire CAS calculator.

Example: For matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 1 & 0 \\ 2 & 0 & 2 \end{bmatrix}$ , find $\det(A)$ and $A^{-1}$

Steps:

Press ctrl + N and select Add Calculator
Enter the matrix $A$ into your calculator
To find the determinant:
- Type det(a) and press enter to evaluate
- Alternatively: Press menu > Matrix & Vector > Determinant
- Result: $\det(A) = -20$
To find the inverse:
- Type a^-1 and press enter to evaluate
- For fractional form: Type exact(a^-1) and press enter

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Important notes:

Using exact() gives fractions instead of long decimals
If a matrix has no inverse, you'll get a "Singular matrix" error message

ClassPad method

For the same matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 1 & 0 \\ 2 & 0 & 2 \end{bmatrix}$ :

Steps:

Enter matrix $A$ into your calculator
- Note: Change calculator status to Standard (not Decimal) for fractions to display
To find the determinant:
- Type and highlight $A$ (swipe with stylus)
- Select Interactive from menu bar
- Tap Matrix-Calculation, then tap det

To find the inverse:
- Type A^-1
- Press EXE to evaluate

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If the matrix has no inverse, you'll get an "Undefined" message.

Solving matrix equations

Matrix equations can be solved using techniques similar to those for solving linear equations. However, there are important differences to remember:

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Key Differences from Regular Equations:

Matrix multiplication is not commutative ( $AB \neq BA$ in general)
You must ensure operations are defined (matrices must have compatible dimensions)
You multiply by the inverse rather than dividing

General approach to solving matrix equations

The key principle is to isolate the unknown matrix $X$ by pre-multiplying or post-multiplying by the appropriate inverse matrix.

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The order of multiplication matters! You must multiply in the correct order to maintain the equation's validity.

Worked examples of matrix equations

For these examples, let:

$A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 1 & 3 \\ 1 & 1 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}, \quad C = \begin{bmatrix} 8 & -1 \\ 4 & 6 \end{bmatrix}$

$D = \begin{bmatrix} 2 \\ 3 \end{bmatrix}, \quad E = \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}$

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Worked Example 1: Solve $B + X = C$ for $X$

This is the simplest type - just rearrange:

$B + X = C$

$X = C - B$

$X = \begin{bmatrix} 8 & -1 \\ 4 & 6 \end{bmatrix} - \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}$

$X = \begin{bmatrix} 6 & -2 \\ 1 & 4 \end{bmatrix}$

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Worked Example 2: Solve $BX = C$ for $X$

When $X$ is multiplied on the right by $B$ , we need to pre-multiply both sides by $B^{-1}$ :

$BX = C$

Pre-multiply both sides by $B^{-1}$ :

$B^{-1}BX = B^{-1}C$

Since $B^{-1}B = I$ (the identity matrix):

$X = B^{-1}C$

$X = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 8 & -1 \\ 4 & 6 \end{bmatrix}$

$X = \begin{bmatrix} 12 & -8 \\ -16 & 15 \end{bmatrix}$

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Worked Example 3: Solve $XB = C$ for $X$

When $X$ is multiplied on the left side, we need to post-multiply both sides by $B^{-1}$ :

$XB = C$

Post-multiply both sides by $B^{-1}$ :

$XBB^{-1} = CB^{-1}$

Since $BB^{-1} = I$ :

$X = CB^{-1}$

$X = \begin{bmatrix} 8 & -1 \\ 4 & 6 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}$

$X = \begin{bmatrix} 19 & -10 \\ -10 & 8 \end{bmatrix}$

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Worked Example 4: Solve $BX = D$ for $X$

$BX = D$

$B^{-1}BX = B^{-1}D$

$X = B^{-1}D$

$X = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 2 \\ 3 \end{bmatrix}$

$X = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$

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Worked Example 5: Solve $AX = E$ for $X$

$AX = E$

$A^{-1}AX = A^{-1}E$

$X = A^{-1}E$

(Using calculator to find $A^{-1}$ ):

$X = \begin{bmatrix} -2 & -1 & 5 \\ 1 & 0 & -1 \\ 1 & 1 & -3 \end{bmatrix} \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}$

$X = \begin{bmatrix} -2 \\ 1 \\ 2 \end{bmatrix}$

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Worked Example 6: Solve $BX + \begin{bmatrix} 4 \\ 5 \end{bmatrix} = D$ for $X$

First, rearrange to isolate the $BX$ term:

$BX + \begin{bmatrix} 4 \\ 5 \end{bmatrix} = D$

$BX = D - \begin{bmatrix} 4 \\ 5 \end{bmatrix}$

$X = B^{-1}\left(D - \begin{bmatrix} 4 \\ 5 \end{bmatrix}\right)$

$X = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} -2 \\ -2 \end{bmatrix}$

$X = \begin{bmatrix} -2 \\ 2 \end{bmatrix}$

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Exam Tip: When solving matrix equations:

Always check the dimensions of matrices to ensure multiplication is defined
Remember: pre-multiply when the unknown appears on the right, post-multiply when it appears on the left
Use your calculator for finding inverses of larger matrices, but show your working for the equation solving steps

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Key Points to Remember:

The inverse of a square matrix $A$ , written $A^{-1}$ , has the property that $AA^{-1} = A^{-1}A = I$ , where $I$ is the identity matrix.
The determinant of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is calculated as $\det(A) = ad - bc$ .
A matrix only has an inverse if its determinant is non-zero. Matrices with zero determinant are called singular matrices.
For a $2 \times 2$ matrix, the inverse is: $A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ (swap diagonal elements, negate off-diagonal elements, divide by determinant).
When solving matrix equations, multiply by the inverse matrix in the correct order - pre-multiply if $X$ is on the right, post-multiply if $X$ is on the left.

Matrix Inverse, the Determinant, and Matrix Equations (VCE SSCE General Mathematics): Revision Notes