Operation of Matrices

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Class 9 Mathematics – Matrix Operations & Properties | New Millennium Academy

Class 9 – Matrix Operations & Properties

Optional Mathematics  |  Unit: Matrices  |  Class Notes

Course Overview

Subject
Optional Mathematics
Class
9
Unit
Matrices

INTRO Learning Objectives

In this session, we will explore the fundamental operations performed on matrices and observe the algebraic properties that govern matrix addition and subtraction using numerical examples.

1 Operations of Matrices

a) Multiplication by Scalar

When a matrix is multiplied by a scalar (a real number), every individual element inside the matrix is multiplied by that number.

Example

Let scalar be $3$ and Matrix $A = \begin{pmatrix} 2 & -1 \\ 4 & 5 \end{pmatrix}$. Then:

$$3 \cdot A = 3 \begin{pmatrix} 2 & -1 \\ 4 & 5 \end{pmatrix} = \begin{pmatrix} 3 \times 2 & 3 \times (-1) \\ 3 \times 4 & 3 \times 5 \end{pmatrix} = \begin{pmatrix} 6 & -3 \\ 12 & 15 \end{pmatrix}$$

b) Addition and Subtraction of Matrices

Two matrices can be added or subtracted if and only if they have the same order. The operation is performed by adding or subtracting the corresponding elements.

Example

If $A = \begin{pmatrix} 5 & 2 \\ 1 & 6 \end{pmatrix}$ and $B = \begin{pmatrix} 3 & 4 \\ 0 & 2 \end{pmatrix}$:

$$A + B = \begin{pmatrix} 5 + 3 & 2 + 4 \\ 1 + 0 & 6 + 2 \end{pmatrix} = \begin{pmatrix} 8 & 6 \\ 1 & 8 \end{pmatrix}$$ $$A - B = \begin{pmatrix} 5 - 3 & 2 - 4 \\ 1 - 0 & 6 - 2 \end{pmatrix} = \begin{pmatrix} 2 & -2 \\ 1 & 4 \end{pmatrix}$$

2 Properties of Matrices (Numerical Proofs)

Let us define three $2 \times 2$ matrices to observe the following properties:

$$A = \begin{pmatrix} 2 & 5 \\ 1 & 4 \end{pmatrix}, \quad B = \begin{pmatrix} 3 & -1 \\ 0 & 2 \end{pmatrix}, \quad C = \begin{pmatrix} -2 & 4 \\ 5 & 1 \end{pmatrix}$$

1. Closure Property (Addition and Subtraction)

Statement: If $A$ and $B$ are matrices of the order $2 \times 2$, their sum $(A+B)$ and difference $(A-B)$ will also be a matrix of the order $2 \times 2$.

Proof for Addition:

$$A + B = \begin{pmatrix} 2 & 5 \\ 1 & 4 \end{pmatrix} + \begin{pmatrix} 3 & -1 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 2 + 3 & 5 + (-1) \\ 1 + 0 & 4 + 2 \end{pmatrix} = \begin{pmatrix} 5 & 4 \\ 1 & 6 \end{pmatrix}$$
Observation: The result is a $2 \times 2$ matrix. Therefore, matrix addition is closed.

Proof for Subtraction:

$$A - B = \begin{pmatrix} 2 & 5 \\ 1 & 4 \end{pmatrix} - \begin{pmatrix} 3 & -1 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 2 - 3 & 5 - (-1) \\ 1 - 0 & 4 - 2 \end{pmatrix} = \begin{pmatrix} -1 & 6 \\ 1 & 2 \end{pmatrix}$$
Observation: The result is also a $2 \times 2$ matrix. Therefore, matrix subtraction is closed.

2. Commutative Property (Addition)

Statement: $A + B = B + A$

Proof:

First, calculate $\text{L.H.S.} (A + B)$:

$$A + B = \begin{pmatrix} 2 & 5 \\ 1 & 4 \end{pmatrix} + \begin{pmatrix} 3 & -1 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 5 & 4 \\ 1 & 6 \end{pmatrix}$$

Next, calculate $\text{R.H.S.} (B + A)$:

$$B + A = \begin{pmatrix} 3 & -1 \\ 0 & 2 \end{pmatrix} + \begin{pmatrix} 2 & 5 \\ 1 & 4 \end{pmatrix} = \begin{pmatrix} 3 + 2 & -1 + 5 \\ 0 + 1 & 2 + 4 \end{pmatrix} = \begin{pmatrix} 5 & 4 \\ 1 & 6 \end{pmatrix}$$
Observation: Since $\text{L.H.S.} = \text{R.H.S.}$, matrix addition is commutative.

3. Associative Property (Addition)

Statement: $(A + B) + C = A + (B + C)$

Proof:

Calculate $\text{L.H.S. } ((A + B) + C)$:

We already know $A + B = \begin{pmatrix} 5 & 4 \\ 1 & 6 \end{pmatrix}$. Now add $C$:

$$(A + B) + C = \begin{pmatrix} 5 & 4 \\ 1 & 6 \end{pmatrix} + \begin{pmatrix} -2 & 4 \\ 5 & 1 \end{pmatrix} = \begin{pmatrix} 5 - 2 & 4 + 4 \\ 1 + 5 & 6 + 1 \end{pmatrix} = \begin{pmatrix} 3 & 8 \\ 6 & 7 \end{pmatrix}$$

Calculate $\text{R.H.S. } (A + (B + C))$:

First find $(B + C)$:

$$B + C = \begin{pmatrix} 3 & -1 \\ 0 & 2 \end{pmatrix} + \begin{pmatrix} -2 & 4 \\ 5 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ 5 & 3 \end{pmatrix}$$

Now add $A$ to $(B + C)$:

$$A + (B + C) = \begin{pmatrix} 2 & 5 \\ 1 & 4 \end{pmatrix} + \begin{pmatrix} 1 & 3 \\ 5 & 3 \end{pmatrix} = \begin{pmatrix} 2 + 1 & 5 + 3 \\ 1 + 5 & 4 + 3 \end{pmatrix} = \begin{pmatrix} 3 & 8 \\ 6 & 7 \end{pmatrix}$$
Observation: Since $\text{L.H.S.} = \text{R.H.S.}$, matrix addition is associative.

4. Additive Inverse

Statement: For every matrix $A$, there exists a matrix $-A$ such that $A + (-A) = \mathbf{O}$ (Zero Matrix).

Proof:

Given $A = \begin{pmatrix} 2 & 5 \\ 1 & 4 \end{pmatrix}$, its negative matrix is $-A = \begin{pmatrix} -2 & -5 \\ -1 & -4 \end{pmatrix}$.

$$A + (-A) = \begin{pmatrix} 2 & 5 \\ 1 & 4 \end{pmatrix} + \begin{pmatrix} -2 & -5 \\ -1 & -4 \end{pmatrix} = \begin{pmatrix} 2 - 2 & 5 - 5 \\ 1 - 1 & 4 - 4 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = \mathbf{O}$$
Observation: Hence, $-A$ is the additive inverse of $A$.

5. Additive Identity

Statement: The zero matrix ($\mathbf{O}$) acts as the additive identity, meaning $A + \mathbf{O} = A$.

Proof:

Let $\mathbf{O} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ and $A = \begin{pmatrix} 2 & 5 \\ 1 & 4 \end{pmatrix}$.

$$A + \mathbf{O} = \begin{pmatrix} 2 & 5 \\ 1 & 4 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 2 + 0 & 5 + 0 \\ 1 + 0 & 4 + 0 \end{pmatrix} = \begin{pmatrix} 2 & 5 \\ 1 & 4 \end{pmatrix} = A$$
Observation: Similarly, adding $A$ to $\mathbf{O}$ will also result in $A$. Hence, the zero matrix is the additive identity.
New Millennium Academy  |  Birauta, Pokhara-17, Kaski, Nepal  |  Class 9 – Optional Mathematics – Matrices
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Course material curated by Mr. Nripendraswar Acharya