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 $B = \begin{pmatrix} 3 & 4 \\ 0 & 2 \end{pmatrix}$:

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

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 $2 \times 2$.

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

Observation: The result is a $2 \times 2$ matrix. Therefore, matrix addition is closed.

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

Observation: The result is also a $2 \times 2$ matrix. Therefore, matrix subtraction is closed.

2. Commutative Property (Addition)

Statement: $A + B = B + A$

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

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

Since $\text{L.H.S.} = \text{R.H.S.}$, Matrix addition is commutative.

3. Associative Property (Addition)

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

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 $(B + C)$:

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

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).

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

$$A + (-A) = \begin{pmatrix} 2 & 5 \\ 1 & 4 \end{pmatrix} + \begin{pmatrix} -2 & -5 \\ -1 & -4 \end{pmatrix} = \begin{pmatrix} 2 - 2 & 5 - 5 \\ 1 - 1 & 4 - 4 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = \mathbf{O}$$

Hence, $-A$ is the additive inverse of $A$.

5. Additive Identity

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

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

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

Similarly, adding $A$ to $\mathbf{O}$ will also result in $A$. Hence, the zero matrix is the additive identity.