1. Matrix Algebra

Video coming soon4 worked examples

Theory

A matrix is a rectangular array of numbers. Its order is written as (rows $\times$ columns). You can add or subtract matrices of the same order (entry by entry), and multiply any matrix by a scalar.

Matrix multiplication combines rows of the first matrix with columns of the second. For the product $AB$ to exist, the number of columns of $A$ must equal the number of rows of $B$ :

(m \times n)(n \times p) = (m \times p)

For a $2\times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , the determinant is $\det A = ad - bc$ , and provided $\det A \neq 0$ the inverse is:

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

If $\det A = 0$ the matrix is singular and has no inverse.

Two further pieces of notation are needed throughout the topic. The transpose $A^{T}$ is formed by swapping rows and columns:

A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} \implies A^{T} = \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix}

It obeys $(A^{T})^{T} = A$ and, importantly, $(AB)^{T} = B^{T}A^{T}$ — the order reverses. A matrix with $A^{T} = A$ is called symmetric.

The identity matrix $I$ has $1$ s on the leading diagonal and $0$ s elsewhere. It behaves like the number $1$ :

I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \qquad AI = IA = A, \qquad A^{-1}A = AA^{-1} = I

That last statement is the definition of the inverse, and it is what makes the row-reduction method for larger matrices work.

The Golden Rule: matrix multiplication is not commutative — in general $AB \neq BA$ . Check the inner dimensions match before multiplying, and the result takes the outer dimensions.

⚠️ Common Examiner Traps

Assuming $AB = BA$ : order matters. If the question asks for $AB$ , do not compute $BA$ .
Multiplying entry-by-entry: matrix multiplication is row-into-column, not element-wise.
Inverse sign slips: in the $2\times 2$ inverse you swap $a$ and $d$ , and negate $b$ and $c$ — don't negate all four.

Worked examples

Example 1

Given $A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 2 & 5 \end{pmatrix}$ , find $AB$ . Show that $BA \neq AB$ .

Step 1: Multiply row-into-column for $AB$ :

AB = \begin{pmatrix} (2)(1)+(-1)(2) & (2)(0)+(-1)(5) \\ (3)(1)+(4)(2) & (3)(0)+(4)(5) \end{pmatrix} = \begin{pmatrix} 0 & -5 \\ 11 & 20 \end{pmatrix}

Step 2: Now compute $BA$ :

BA = \begin{pmatrix} (1)(2)+(0)(3) & (1)(-1)+(0)(4) \\ (2)(2)+(5)(3) & (2)(-1)+(5)(4) \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ 19 & 18 \end{pmatrix}

Step 3: Since $\begin{pmatrix} 0 & -5 \\ 11 & 20 \end{pmatrix} \neq \begin{pmatrix} 2 & -1 \\ 19 & 18 \end{pmatrix}$ , we have $AB \neq BA$ .

Example 2

Find the inverse of $A = \begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}$ .

Step 1: Find the determinant:

\det A = (3)(2) - (5)(1) = 6 - 5 = 1

Step 2: Since $\det A = 1 \neq 0$ , apply the inverse formula (swap $a,d$ ; negate $b,c$ ):

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

Step 3: Check by confirming $A A^{-1} = I$ :

\begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

Example 3

Evaluate the determinant of $A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}$ .

Step 1: Expand along the first row, each entry times its $2\times 2$ minor, with the sign pattern $+\,-\,+$ :

\det A = 1\begin{vmatrix} 1 & 4 \\ 6 & 0 \end{vmatrix} - 2\begin{vmatrix} 0 & 4 \\ 5 & 0 \end{vmatrix} + 3\begin{vmatrix} 0 & 1 \\ 5 & 6 \end{vmatrix}

Step 2: Evaluate each $2\times 2$ determinant:

= 1(0 - 24) - 2(0 - 20) + 3(0 - 5)

Step 3: Simplify:

= -24 + 40 - 15 = 1

Example 4

Given $A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 4 \\ -2 & 5 \end{pmatrix}$ , verify that $(AB)^{T} = B^{T}A^{T}$ .

Step 1: Work out $AB$ first, combining rows of $A$ with columns of $B$ :

AB = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}\begin{pmatrix} 1 & 4 \\ -2 & 5 \end{pmatrix} = \begin{pmatrix} 2-2 & 8+5 \\ 0-6 & 0+15 \end{pmatrix} = \begin{pmatrix} 0 & 13 \\ -6 & 15 \end{pmatrix}

Step 2: Transpose the result by swapping rows and columns:

(AB)^{T} = \begin{pmatrix} 0 & -6 \\ 13 & 15 \end{pmatrix}

Step 3: Now transpose $A$ and $B$ separately, and multiply them in the reversed order:

B^{T} = \begin{pmatrix} 1 & -2 \\ 4 & 5 \end{pmatrix}, \qquad A^{T} = \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix}

B^{T}A^{T} = \begin{pmatrix} 2-2 & 0-6 \\ 8+5 & 0+15 \end{pmatrix} = \begin{pmatrix} 0 & -6 \\ 13 & 15 \end{pmatrix}

Step 4: The two results agree, confirming $(AB)^{T} = B^{T}A^{T}$ . Note that $A^{T}B^{T}$ would not have worked — the reversal matters.