Inverse Matrices and Solving Simultaneous Equations Using Matrices (VCE SSCE General Mathematics): Revision Notes

Worked Example: Finding an Inverse Using CAS

Let's work through finding the inverse of matrix $A = \begin{bmatrix} 8 & 4 \\ 3 & 2 \end{bmatrix}$.

Step 1: Enter matrix $A$ into your calculator.

Step 2: Type in $A^{-1}$ and evaluate.

Step 3: To verify your answer is correct, form the product $A \times A^{-1}$. This should give you the identity matrix $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

Step 4: Write your final answer.

For this example, the inverse of $A$ is:

$A^{-1} = \begin{bmatrix} 0.5 & -1 \\ -0.75 & 2 \end{bmatrix}$

Worked Example: Finding an Inverse Using ClassPad

The process on a ClassPad calculator is similar. For the same matrix $A = \begin{bmatrix} 8 & 4 \\ 3 & 2 \end{bmatrix}$:

Step 1: Type in $A^{-1}$ or type $A$ followed by the power of $-1$.

Step 2: Press the EXE button to evaluate.

Step 3: Write your answer.

The inverse is displayed as:

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

Note that this is the same answer as before, just expressed as fractions rather than decimals.

Worked Example: Calculating Determinants

Find the determinant of matrices $A = \begin{bmatrix} 5 & 2 \\ 3 & 6 \end{bmatrix}$ and $B = \begin{bmatrix} -3 & 7 \\ 2 & -2 \end{bmatrix}$.

Solution:

For matrix $A$:

• Identify the values: $a = 5$, $b = 2$, $c = 3$, $d = 6$
• Apply the formula: $\text{det}(A) = a \times d - b \times c$
• Calculate: $\text{det}(A) = 5 \times 6 - 2 \times 3 = 30 - 6 = 24$

For matrix $B$:

• Identify the values: $a = -3$, $b = 7$, $c = 2$, $d = -2$
• Apply the formula: $\text{det}(B) = (-3) \times (-2) - 7 \times 2$
• Calculate: $\text{det}(B) = 6 - 14 = -8$

Exam tip: Be especially careful with negative numbers when calculating determinants.

Worked Example: Solving Simultaneous Equations Using Matrices

Let's solve the equations $5x + 2y = 21$ and $7x + 3y = 29$ using matrices.

Step 1: Write the equations in matrix form.

$\begin{bmatrix} 5x + 2y \\ 7x + 3y \end{bmatrix} = \begin{bmatrix} 21 \\ 29 \end{bmatrix}$

Step 2: Express as a matrix product.

$\begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 21 \\ 29 \end{bmatrix}$

Step 3: Name the matrices. Let $A \times X = C$ where:

• $A = \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}$ (coefficients matrix)
• $X = \begin{bmatrix} x \\ y \end{bmatrix}$ (variables matrix - contains our unknowns)
• $C = \begin{bmatrix} 21 \\ 29 \end{bmatrix}$ (constants matrix)

Step 4: Enter matrices $A$ and $C$ into your calculator.

Step 5: Find matrix $X$ using the formula $X = A^{-1} \times C$.

Step 6: Calculate $A^{-1} \times C$ on your calculator.

Step 7: Read off the solution from matrix $X$.

The calculator shows: $X = \begin{bmatrix} 5 \\ -2 \end{bmatrix}$

Step 8: Write the solutions to the original equations.

Since $X = \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ -2 \end{bmatrix}$, we have:

$x = 5 \text{ and } y = -2$

Step 9: Verify by substituting back into the original equations.

Checking in $5x + 2y = 21$:

$5(5) + 2(-2) = 25 - 4 = 21$ ✓

Checking in $7x + 3y = 29$:

$7(5) + 3(-2) = 35 - 6 = 29$ ✓

Both equations are satisfied, confirming our solution is correct.