Addition and Subtraction

Matrices can only be added or subtracted if they have the same order. You add or subtract corresponding elements.

Example 1

Given $A = \begin{pmatrix} 3 & 5 \\ 1 & 7 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 1 \\ 4 & 6 \end{pmatrix}$

Find A + B

Solution

$A + B = \begin{pmatrix} 3+2 & 5+1 \\ 1+4 & 7+6 \end{pmatrix} = \begin{pmatrix} 5 & 6 \\ 5 & 13 \end{pmatrix}$

Scalar Multiplication

Multiply every element by the scalar (a single number).

Example 2

If $k = 3$ and $A = \begin{pmatrix} 4 & 2 \\ 1 & 5 \end{pmatrix}$, find 3A

Solution

$3A = \begin{pmatrix} 3\times4 & 3\times2 \\ 3\times1 & 3\times5 \end{pmatrix} = \begin{pmatrix} 12 & 6 \\ 3 & 15 \end{pmatrix}$

Matrix Multiplication

To multiply a 2×2 matrix by another 2×2 matrix, use the row-by-column rule:

$\begin{pmatrix} a & c \\ b & d \end{pmatrix} \times \begin{pmatrix} e & g \\ f & h \end{pmatrix} = \begin{pmatrix} ae+cf & ag+ch \\ be+df & bg+dh \end{pmatrix}$

Example 3

Find $A \times B$ where $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix}$

Solution

$A \times B = \begin{pmatrix} 1\times5 + 2\times7 & 1\times6 + 2\times8 \\ 3\times5 + 4\times7 & 3\times6 + 4\times8 \end{pmatrix} = \begin{pmatrix} 5+14 & 6+16 \\ 15+28 & 18+32 \end{pmatrix} = \begin{pmatrix} 19 & 22 \\ 43 & 50 \end{pmatrix}$

The determinant is a special number calculated from a 2×2 matrix. It tells us important information about the matrix.

Formula

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

$\text{det}(A) = |A| = ad - bc$

This is the product of the leading diagonal minus the product of the main diagonal.

Example 4

Find the determinant of $A = \begin{pmatrix} 3 & 7 \\ 2 & 5 \end{pmatrix}$

Solution

$|A| = 3\times5 - 2\times7 = 15 - 14 = 1$

Example 5

Find the value of x if the determinant of $\begin{pmatrix} 2x & 4 \\ 3 & 6 \end{pmatrix}$ equals 6.

Solution

$2x \times 6 - 3 \times 4 = 6$ $12x - 12 = 6$ $12x = 18$ $x = \frac{18}{12} = \frac{3}{2}$

Singular and Non-singular Matrices

• Non-singular matrix: Determinant is NOT zero (inverse exists)
• Singular matrix: Determinant IS zero (no inverse exists)

Example 6

Is the matrix $B = \begin{pmatrix} 4 & 2 \\ 2 & 1 \end{pmatrix}$ singular or non-singular?

Solution

$|B| = 4\times1 - 2\times2 = 4 - 4 = 0$

Since |B| = 0, B is singular (no inverse exists).

The inverse of a matrix A, written A⁻¹, is the matrix that satisfies:

$AA^{-1} = I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

Only non-singular matrices (determinant ≠ 0) have inverses.

Formula

For $A = \begin{pmatrix} a & c \\ b & d \end{pmatrix}$ with $|A| \neq 0$:

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

Notice the pattern: swap a and d, change signs of b and c.

Example 7

Find the inverse of $A = \begin{pmatrix} 4 & 7 \\ 3 & 5 \end{pmatrix}$

Solution

First find the determinant: $|A| = 4\times5 - 3\times7 = 20 - 21 = -1$

Since |A| ≠ 0, the inverse exists.

$A^{-1} = \frac{1}{-1} \begin{pmatrix} 5 & -7 \\ -3 & 4 \end{pmatrix} = \begin{pmatrix} -5 & 7 \\ 3 & -4 \end{pmatrix}$

Verification: $A \times A^{-1} = \begin{pmatrix} 4 & 7 \\ 3 & 5 \end{pmatrix} \times \begin{pmatrix} -5 & 7 \\ 3 & -4 \end{pmatrix} = \begin{pmatrix} 4(-5)+7(3) & 4(7)+7(-4) \\ 3(-5)+5(3) & 3(7)+5(-4) \end{pmatrix} = \begin{pmatrix} -20+21 & 28-28 \\ -15+15 & 21-20 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$

Example 8

Show why $\begin{pmatrix} 2 & 4 \\ 1 & 2 \end{pmatrix}$ has no inverse.

Solution

$|A| = 2\times2 - 1\times4 = 4 - 4 = 0$

Since the determinant is zero, this is a singular matrix and has no inverse.

A transformation moves a point from one position to another. We can use 2×2 matrices to describe these transformations. If point P has coordinates (x, y), we write it as a column vector $\begin{pmatrix} x \\ y \end{pmatrix}$. The image point P' is found by multiplying by a transformation matrix T:

$\begin{pmatrix} x' \\ y' \end{pmatrix} = T \times \begin{pmatrix} x \\ y \end{pmatrix}$

Common Transformation Matrices

Example 9

Find the image of point A(3, 4) after reflection in the x-axis.

Solution

For reflection in x-axis, the transformation matrix is $T = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$

$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \times \begin{pmatrix} 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 1\times3 + 0\times4 \\ 0\times3 + (-1)\times4 \end{pmatrix} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}$

So the image is A'(3, -4).

Example 10

Find the image of point B(2, 5) after reflection in the line y = x.

Solution

For reflection in y = x, the transformation matrix is $T = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \times \begin{pmatrix} 2 \\ 5 \end{pmatrix} = \begin{pmatrix} 0\times2 + 1\times5 \\ 1\times2 + 0\times5 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \end{pmatrix}$

So the image is B'(5, 2). Notice the coordinates are swapped.

In Tanzania, matrices are used in everyday business and technology. For example, when a supermarket manager in Dar es Salaam tracks inventory across multiple branches, they can use matrices to organize data on quantities of different products. If a shop has 3 types of maize flour in 2 stores, this data forms a 2×3 matrix. Matrix multiplication can then calculate total values in Tanzanian shillings. Transformation matrices are also used in mobile phone screens and computer games to rotate, enlarge, or flip images — the same mathematics students learn in this topic!