Determinants, Inverses & Matrix Transformations

Matrices give a compact way to perform several calculations at once. In the optional CSEC section on Vectors and Matrices, you may be asked to use matrices to undo operations, solve simultaneous equations, or transform points on a plane.

The key idea is that matrix work is highly procedural, but it still needs interpretation. A determinant tells you whether an inverse exists; an inverse matrix reverses the effect of another matrix; a transformation matrix moves points in a predictable way. State what the matrix is doing before you start multiplying.

What is an Inverse?¶

An inverse matrix reverses the effect of a matrix multiplication, similar to how division reverses multiplication with ordinary numbers.

The inverse of matrix $A$ is matrix $A^{-1}$ such that: $AA^{-1} = A^{-1}A = I$

Think of it like division: $A^{-1}$ "undoes" what $A$ does.

When Does an Inverse Exist?¶

Only when $\det(A) \neq 0$ (called a non-singular matrix)

Formula for 2×2 Inverse¶

The formula swaps the diagonal entries, changes the signs of the other diagonal, and divides by the determinant. The determinant check comes first because division by zero is impossible.

For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$:

$A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$

Steps:

1. Calculate determinant: $ad - bc$
2. Swap $a$ and $d$
3. Negate $b$ and $c$
4. Divide by determinant

2×2 Inverse Example¶

Find the inverse of $A = \begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix}$

Step 1: Find determinant

$\det(A) = 3(4) - 1(2) = 12 - 2 = 10$

Step 2: Apply formula

$A^{-1} = \frac{1}{10} \begin{pmatrix} 4 & -1 \\ -2 & 3 \end{pmatrix} = \begin{pmatrix} 0.4 & -0.1 \\ -0.2 & 0.3 \end{pmatrix}$

Step 3: Verify (multiply $A \cdot A^{-1}$)

$\begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix} \begin{pmatrix} 0.4 & -0.1 \\ -0.2 & 0.3 \end{pmatrix}$

Position (1,1): $3(0.4) + 1(-0.2) = 1.2 - 0.2 = 1$ ✓

Position (1,2): $3(-0.1) + 1(0.3) = -0.3 + 0.3 = 0$ ✓

Position (2,1): $2(0.4) + 4(-0.2) = 0.8 - 0.8 = 0$ ✓

Position (2,2): $2(-0.1) + 4(0.3) = -0.2 + 1.2 = 1$ ✓

Result: $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ ✓

No Inverse When Determinant = 0¶

$A = \begin{pmatrix} 2 & 4 \\ 1 & 2 \end{pmatrix}$

$\det(A) = 2(2) - 4(1) = 4 - 4 = 0$

Since determinant is 0, this matrix has no inverse. (Its columns are proportional—row 1 is exactly 2× row 2.)

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Part 10: Solving Systems Using Matrices¶

Converting a System to Matrix Form¶

Matrix form separates coefficients, unknowns, and constants. This makes a pair of simultaneous equations look like one compact equation: $AX = B$.

System: $\begin{aligned} 2x + 3y &= 8 \\ x - y &= 1 \end{aligned}$

Matrix Form: $AX = B$

$\begin{pmatrix} 2 & 3 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 8 \\ 1 \end{pmatrix}$

Where:

• $A$ = coefficient matrix
• $X$ = variable column vector
• $B$ = constants column vector

Solving Using Matrix Inverse¶

If $AX = B$, multiplying by $A^{-1}$ undoes the coefficient matrix. The result is the unknown vector $X$.

If $AX = B$, then:

$X = A^{-1}B$

(Multiply both sides by $A^{-1}$ on the left)

Complete Example¶

Problem: Solve $\begin{pmatrix} 2 & 3 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 8 \\ 1 \end{pmatrix}$

Step 1: Find $\det(A)$

$\det \begin{pmatrix} 2 & 3 \\ 1 & -1 \end{pmatrix} = 2(-1) - 3(1) = -2 - 3 = -5$

Step 2: Find $A^{-1}$

$A^{-1} = \frac{1}{-5} \begin{pmatrix} -1 & -3 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 0.2 & 0.6 \\ 0.2 & -0.4 \end{pmatrix}$

Step 3: Multiply $A^{-1}B$

$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0.2 & 0.6 \\ 0.2 & -0.4 \end{pmatrix} \begin{pmatrix} 8 \\ 1 \end{pmatrix}$

$x = 0.2(8) + 0.6(1) = 1.6 + 0.6 = 2.2$

$y = 0.2(8) - 0.4(1) = 1.6 - 0.4 = 1.2$

Answer: $x = 2.2$, $y = 1.2$

Check: $2(2.2) + 3(1.2) = 4.4 + 3.6 = 8$ ✓ and $2.2 - 1.2 = 1$ ✓

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Part 11: Transformation Matrices¶

What Transformations Do Matrices Represent?¶

Transformation matrices describe movement on the coordinate plane. Each point is written as a column vector, multiplied by the matrix, and changed into its image.

Matrices can represent:

• Rotation: Turn a shape around origin
• Reflection: Mirror a shape
• Scaling: Enlarge or reduce
• Shear: Skew a shape

How it Works: To transform point $(x, y)$, multiply by transformation matrix

$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$

Rotation Matrix¶

Rotate counterclockwise by angle $\theta$:

$R(\theta) = \begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix}$

Reflection Matrix¶

Reflect across x-axis: $R_x = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$

Reflect across y-axis: $R_y = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$

Reflect across line $y = x$: $R_{y=x} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

Scaling Matrix¶

Scale by factor $k$ (same in both directions): $S(k) = \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$

Scale different in each direction (x by $a$, y by $b$): $S(a,b) = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$

Composite Transformations¶

Composite transformations are performed in sequence. In matrix form, the matrix nearest the point acts first, so the written order must be handled carefully.

Apply multiple transformations by multiplying matrices

To apply transformation $T_2$ after $T_1$: $\text{Result} = T_2 \cdot T_1 \cdot \begin{pmatrix} x \\ y \end{pmatrix}$

(Work right to left: apply $T_1$ first, then $T_2$)

ANIMATION: Composite transformations
- Show original triangle
- Apply first transformation (e.g., rotation)
- Show intermediate result
- Apply second transformation (e.g., scaling)
- Show final result
- Show that multiplying matrices in order gives same result

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Part 12: Non-Commutativity of Matrix Multiplication¶

Matrix Multiplication is Not Commutative¶

Critical Rule: $AB \neq BA$ in general

Unlike multiplication of numbers, matrix multiplication depends on order.

Why It Matters: In transformations, the order you apply them matters completely. Rotating then scaling gives a different result than scaling then rotating.

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Summary: Key Concepts¶