Operations on matrices

The order of a matrix or size of a matrix is given by the number of its rows and the number of its columns. So if A has m rows and n columns, then the order of matrix is m x n. It is important to note that the order of any matrix is given by stating the number of its rows first and then the number of its columns.

For example $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ is a $2 \times 2$ matrix or the order of matrix A is $2 \times 2$, and

$B = \begin{pmatrix} a & b & c \\ d & e & f \end{pmatrix}$ is a $2 \times 3$ matrix while $C = \begin{pmatrix} a & b \\ c & d \\ e & f \end{pmatrix}$ is a $3 \times 2$ matrix.

NB: $m \times n \neq n \times m$ since an $m \times n$ is a matrix with $m$ rows and $n$ columns while $n \times m$ is a matrix with $n$ rows and $m$ columns.

The following are the common types of matrices:

1. Zero matrix;

A zero matrix is the matrix whose elements are all zeros.

e.g.

$A = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \quad B = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \quad C = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \quad \text{or} \quad D = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$

2. Square matrix:

Is a matrix whose number of rows is equal to the number of columns.

For example

$A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \quad \text{or} \quad B = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 6 & 8 \\ 5 & 7 & 9 \end{pmatrix}$

3. Identity Matrix:

Is the square matrix whose elements in the leading diagonal are ones and zeros elsewhere.

$\text{e.g.} \quad \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \quad \text{or} \quad \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

4. Column matrix:

Is the matrix with only one column.

$\text{e.g.} \quad A = \begin{pmatrix} 2 \\ 1 \end{pmatrix} \quad \text{or} \quad B = \begin{pmatrix} 1 \\ 6 \\ 9 \end{pmatrix}$

5. Row matrix:

This is a matrix with only one row.

$\text{e.g.} \quad D = \begin{pmatrix} 1 & 10 \end{pmatrix} \quad \text{or} \quad E = \begin{pmatrix} 3 & -2 & 17 \end{pmatrix}$

When adding or subtracting one matrix from another, the corresponding elements (entities) are added or subtracted respectively. This being the case, we can only perform addition and subtraction of matrices with the same orders.

Example 1

Given that

Solution:

$$\mathbf{A} + \mathbf{B} = \begin{pmatrix} a & c \\ b & d \end{pmatrix} + \begin{pmatrix} e & f \\ g & h \end{pmatrix} = \begin{pmatrix} a+e & c+f \\ b+g & d+h \end{pmatrix}$$

Example 2

Given that

$A = \begin{pmatrix} a & c \\ b & d \end{pmatrix}$ and $B = \begin{pmatrix} e & f \\ g & h \end{pmatrix}$, find $A - B$

$A - B = \begin{pmatrix} a-e & c-f \\ b-g & d-h \end{pmatrix}$

Solution: $A - B = \begin{pmatrix} a & c \\ b & d \end{pmatrix} - \begin{pmatrix} e & f \\ g & h \end{pmatrix} = \begin{pmatrix} a-e & c-f \\ b-g & d-h \end{pmatrix}$

Example 3

$$\begin{pmatrix} x & 2 \\ 1 & 0 \end{pmatrix} - \begin{pmatrix} 3 & z \\ 0 & y \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix}$$

Solution:

$$\begin{pmatrix} x & 2 \\ 1 & 0 \end{pmatrix} - \begin{pmatrix} 3 & z \\ 0 & y \end{pmatrix} = \begin{pmatrix} x-3 & 2-z \\ 1-0 & 0-y \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix}$$

Now x - 3 = 2, 2 - z = 0 and 0 - y = 3. So x = 5, y = -3 and z = 2.

Solve for x, y and z in the following matrix equation:

Additive identity matrix.

If M is any square matrix, that is a matrix with order m×m or n×n and Z is another matrix with the same order as M such that M + Z = Z + M = M then Z is the additive identity matrix.

The additive inverse of a matrix.

If A and B are any matrices with the same order such that A + B = Z, then it means that either A is an additive inverse of B or B is an additive inverse of A, that is B = -A or A = -B.

Example 4

Find the additive inverse of A, if $A = \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix}$.

Solution: The additive inverse of $A$ is $-A = -\begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix}$

$\therefore$ The additive inverse of $A$ is $\begin{pmatrix} -2 & -3 \\ -4 & -5 \end{pmatrix}$.

Example 5

Find the additive identity of B if B is a 3×3 matrix.

Solution:

The additive identity of any $n \times n$ matrix is the $n \times n$ zero matrix.

So $Z = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$

A matrix can be multiplied by a constant number (scalar) or by another matrix.

Scalar multiplication of matrices:

Rule: If A is a matrix with elements say a, b, c and d, or

Example 6

Given that

$A = \begin{pmatrix} 7 & 8 \\ 5 & 9 \end{pmatrix}$

Find (a) 2A (b) -5A

Solution:

(a) $A = \begin{pmatrix} 7 & 8 \\ 5 & 9 \end{pmatrix}$, $2A = ?$

$2A = 2 \begin{pmatrix} 7 & 8 \\ 5 & 9 \end{pmatrix}$

$= \begin{pmatrix} 2 \times 7 & 2 \times 8 \\ 2 \times 5 & 2 \times 9 \end{pmatrix}$

$= \begin{pmatrix} 14 & 16 \\ 10 & 18 \end{pmatrix}$

$\therefore$ $2A = \begin{pmatrix} 14 & 16 \\ 10 & 18 \end{pmatrix}$

(b) $-5A = ?$

$-5A = -5 \begin{pmatrix} 7 & 8 \\ 5 & 9 \end{pmatrix}$

$= \begin{pmatrix} -5 \times 7 & -5 \times 8 \\ -5 \times 5 & -5 \times 9 \end{pmatrix}$

$= \begin{pmatrix} -35 & -40 \\ -25 & -45 \end{pmatrix}$

$\therefore -5A = \begin{pmatrix} -35 & -40 \\ -25 & -45 \end{pmatrix}$

Example 7

Given,

$B = \begin{pmatrix} 2 & 4 \\ 5 & 14 \end{pmatrix}, \text{ Find } B + B + B.$

Solution:

$$\begin{aligned} B + B + B &= 3B \\ 3B &= 3 \begin{pmatrix} 2 & 4 \\ 5 & 14 \end{pmatrix} \\ &= \begin{pmatrix} 3 \times 2 & 3 \times 4 \\ 3 \times 5 & 3 \times 14 \end{pmatrix} \\ &= \begin{pmatrix} 6 & 12 \\ 15 & 42 \end{pmatrix} \end{aligned}$$ $$\therefore B + B + B = \begin{pmatrix} 6 & 12 \\ 15 & 42 \end{pmatrix}$$

Multiplication of matrix by another matrix:

AB is the product of matrices A and B while BA is the product of matrix B and A. In AB, matrix A is called a pre-multiplier because it comes first while matrix B is called the post multiplier because it comes after matrix A.

Rules of finding the product of matrices:

1. The pre-multiplier matrix is divided row wise, that is it is divided according to its rows.
2. The post multiplier is divided according to its columns.
3. Multiplication is done by taking an element from the row and multiplied by an element from the column.
4. In rule (iii) above, the left most element of the row is multiplied by the top most element of the column and the right most element from the row is multiplied by the bottom most element of the column and their sums are taken:

$$\text{Now if } A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \text{and } B = \begin{pmatrix} e & f \\ g & h \end{pmatrix}$$ $$\text{Then } A \times B = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} e & f \\ g & h \end{pmatrix}$$ $$A \times B = \begin{pmatrix} ae + bg & af + bh \\ ce + dg & cf + dh \end{pmatrix}$$ $$\text{Similarly } BA = \begin{bmatrix} e & f \\ g & h \end{bmatrix} \begin{bmatrix} e & f \\ g & h \end{bmatrix}$$ $$= \begin{pmatrix} ea + fc & eb + fd \\ ga + hc & gb + hd \end{pmatrix}$$

Therefore it can be concluded that matrix by matrix multiplication is only possible if the number of columns in the pre-multiplier is equal to the number of rows in the post multiplier.

Example 8

Given that;

$$\mathbf{A} = \begin{bmatrix} 9 & 7 \\ 8 & 6 \end{bmatrix} \quad \text{and} \quad \mathbf{B} = \begin{bmatrix} 6 & -1 \\ -2 & 5 \end{bmatrix} \quad \text{find,} \quad \text{(a) } \mathbf{AB} \quad \text{(b) } \mathbf{BA}$$

Solution

(a) $\mathbf{AB} = \begin{bmatrix} 9 & 7 \\ 8 & 6 \end{bmatrix} \times \begin{bmatrix} 6 & -1 \\ -2 & 5 \end{bmatrix}$

$$= \begin{bmatrix} 9 & 7 \\ 8 & 6 \end{bmatrix} \times \begin{bmatrix} 6 & -1 \\ -2 & 5 \end{bmatrix}$$ $$\begin{aligned} &= \begin{pmatrix} 9 \times 6 + 7 \times (-2) & 9 \times (-1) + 7 \times 5 \\ 8 \times 6 + 6 \times (-2) & 8 \times (-1) + 6 \times 5 \end{pmatrix} \\ &= \begin{pmatrix} 54 - 14 & -9 + 35 \\ 48 - 12 & -8 + 30 \end{pmatrix} \\ \therefore AB &= \begin{pmatrix} 40 & 26 \\ 36 & 22 \end{pmatrix} \end{aligned}$$

(b) $BA = \begin{pmatrix} 6 & -1 \\ -2 & 5 \end{pmatrix} \times \begin{pmatrix} 9 & 7 \\ 8 & 6 \end{pmatrix}$

$$\begin{aligned} &= \begin{pmatrix} 6 \times 9 + (-1) \times 8 & 6 \times 7 + (-1) \times 6 \\ (-2) \times 9 + 5 \times 8 & (-2) \times 7 + 5 \times 6 \end{pmatrix} \\ &= \begin{pmatrix} 54 - 8 & 42 - 6 \\ -18 + 40 & -14 + 30 \end{pmatrix} \\ &= \begin{pmatrix} 46 & 36 \\ 22 & 16 \end{pmatrix} \\ \therefore BA &= \begin{pmatrix} 46 & 36 \\ 22 & 16 \end{pmatrix} \end{aligned}$$

From the above example it can be noted that AB ≠ BA, therefore matrix by matrix multiplication does not obey commutative property except when the multiplication involves an identity matrix i.e. AI = IA = A.

Example 9

Let,

$P = \begin{pmatrix} 4 & 2 \\ 3 & 1 \end{pmatrix} \text{ and } Q = \begin{pmatrix} 5 \\ 6 \end{pmatrix} \quad \text{Find } PQ$

Solution:

$PQ = \begin{pmatrix} 4 & 2 \\ 3 & 1 \end{pmatrix} \times \begin{pmatrix} 5 \\ 6 \end{pmatrix} = \begin{pmatrix} 4 \times 5 + 2 \times 6 \\ 3 \times 5 + 1 \times 6 \end{pmatrix}$

$PQ = \begin{pmatrix} 20 + 12 \\ 15 + 6 \end{pmatrix} = \begin{pmatrix} 32 \\ 21 \end{pmatrix}$

$\therefore$ $PQ = \begin{pmatrix} 32 \\ 21 \end{pmatrix}$

Example 10

Find C × D if

$$C = \begin{pmatrix} 2 & 4 & 6 \\ 8 & 1 & 0 \\ 3 & 5 & 7 \end{pmatrix} \quad \text{and} \quad D = \begin{pmatrix} 3 & 4 \\ 10 & 6 \\ 1 & -2 \end{pmatrix}$$

Solution:

$$C \times D = \begin{pmatrix} 2 & 4 & 6 \\ 8 & 1 & 0 \\ 3 & 5 & 7 \end{pmatrix} \times \begin{pmatrix} 3 & 4 \\ 10 & 6 \\ 1 & -2 \end{pmatrix}$$ $$= \begin{pmatrix} 2 \times 3 + 4 \times 10 + 6 \times 1 & 2 \times 4 + 4 \times 6 + 6 \times (-2) \\ 8 \times 3 + 1 \times 10 + 0 \times 1 & 8 \times 4 + 1 \times 6 + 0 \times (-2) \\ 3 \times 3 + 5 \times 10 + 7 \times 1 & 3 \times 4 + 5 \times 6 + 7 \times (-2) \end{pmatrix}$$ $$= \begin{pmatrix} 6 + 40 + 6 & 8 + 24 + (-12) \\ 24 + 10 + 0 & 32 + 6 + 0 \\ 9 + 50 + 7 & 12 + 30 + (-14) \end{pmatrix}$$ $$= \begin{pmatrix} 52 & 20 \\ 34 & 38 \\ 66 & 28 \end{pmatrix}$$ $$\therefore CD = \begin{pmatrix} 52 & 20 \\ 34 & 38 \\ 66 & 28 \end{pmatrix}$$

Product of a matrix and an identity matrix:

If A is any square matrix and I is an identity matrix with the same order as A, then AI = IA = A.

Example 11

Given;

$A = \begin{pmatrix} 1 & 3 \\ 4 & 5 \end{pmatrix} \text{and } I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

Find (a) AI (b) IA

Solution:

$AI = \begin{pmatrix} 1 & 3 \\ 4 & 5 \end{pmatrix} \times \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 \times 1 + 3 \times 0 & 1 \times 0 + 3 \times 1 \\ 4 \times 1 + 5 \times 0 & 4 \times 0 + 5 \times 1 \end{pmatrix}$

$AI = \begin{pmatrix} 1 + 0 & 0 + 3 \\ 4 + 0 & 0 + 5 \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ 4 & 5 \end{pmatrix}$

$\therefore AI = \begin{pmatrix} 1 & 3 \\ 4 & 5 \end{pmatrix} = A$

(b) $IA = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \times \begin{pmatrix} 1 & 3 \\ 4 & 5 \end{pmatrix}$

$IA = \begin{pmatrix} 1 \times 1 + 0 \times 4 & 1 \times 3 + 0 \times 5 \\ 0 \times 1 + 1 \times 4 & 0 \times 3 + 1 \times 5 \end{pmatrix}$

$\therefore IA = \begin{pmatrix} 1 + 0 & 3 + 0 \\ 0 + 4 & 0 + 5 \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ 4 & 5 \end{pmatrix} = A$