We have,

$$ \begin{aligned} & 2 X+3 Y=\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right] \quad\text{.... (i)}\\ \text{and}\quad & 3 X+2 Y=\left[\begin{array}{cr} -2 & 2 \\ 1 & -5 \end{array}\right]\quad\text{.... (ii)} \end{aligned}$$

$$\begin{aligned} &\text { On subtracting Eq. (i) from Eq. (ii), we get }\\ &\begin{aligned} \therefore \quad(3 X+2 Y)-(2 X+3 Y) & =\left[\begin{array}{cc} -2-2 & 2-3 \\ 1-4 & -5-0 \end{array}\right] \\ (X-Y) & =\left[\begin{array}{l} -4-1 \\ -3-5 \end{array}\right]\quad\text{.... (iii)} \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { On adding Eqs. (i) and (ii), we get }\\ &\begin{aligned} (5 X+5 Y) & =\left[\begin{array}{rr} 0 & 5 \\ 5 & -5 \end{array}\right] \\ (X+Y) & =\frac{1}{5}\left[\begin{array}{rr} 0 & 5 \\ 5 & -5 \end{array}\right]=\left[\begin{array}{rr} 0 & 1 \\ 1 & -1 \end{array}\right]\quad\text{.... (iv)} \end{aligned} \end{aligned}$$

On adding Eqs. (iii) and (iv), we get

$$\begin{aligned} (X-Y)+(X+Y) & =\left[\begin{array}{rr} -4 & 0 \\ -2 & -6 \end{array}\right] \\ \Rightarrow\quad 2 X & =2\left[\begin{array}{rr} -2 & 0 \\ -1 & -3 \end{array}\right] \\ \therefore\quad X & =\left[\begin{array}{rr} -2 & 0 \\ -1 & -3 \end{array}\right] \end{aligned}$$

From Eq. (iv),

$$\begin{aligned} & {\left[\begin{array}{rr} -2 & 0 \\ -1 & -3 \end{array}\right]+Y }=\left[\begin{array}{rr} 0 & 1 \\ 1 & -1 \end{array}\right] \\ \therefore\quad & Y=\left[\begin{array}{ll} 2 & 1 \\ 2 & 2 \end{array}\right] \text { and } X=\left[\begin{array}{lr} -2 & 0 \\ -1 & -3 \end{array}\right] \end{aligned}$$

We have, $A=\left[\begin{array}{ll}3 & 5\end{array}\right]_{1 \times 2}$ and $B=\left[\begin{array}{ll}7 & 3\end{array}\right]_{1 \times 2}$

Let $C=\left[\begin{array}{l}x \\ y\end{array}\right]_{2 \times 1}$ is a non-zero matrix of order $2 \times 1$.

$\begin{array}{ll}\therefore & A C=\left[\begin{array}{ll}3 & 5\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=[3 x+5 y] \\ \text { and } & B C=\left[\begin{array}{ll}7 & 3\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=[7 x+3 y]\end{array}$

For $A C=B C$,

$$[3 x+5 y]=[7 x+3 y]$$

On using equality of matrix, we get

$$\begin{array}{lrl} & 3 x+5 y & =7 x+3 y \\ \Rightarrow & 4 x & =2 y \\ \Rightarrow & x & =\frac{1}{2} y \\ \therefore & y & =2 x \\ \Rightarrow & C & =\left[\begin{array}{c} x \\ 2 x \end{array}\right] \end{array}$$

We see that on taking $C$ of order $2 \times 1,2 \times 2,2 \times 3, \ldots$, we get

$$\mathrm{C}=\left[\begin{array}{c} x \\ 2 x \end{array}\right] \cdot\left[\begin{array}{cc} x & x \\ 2 x & 2 x \end{array}\right] \cdot\left[\begin{array}{ccc} x & x & x \\ 2 x & 2 x & 2 x \end{array}\right] \ldots$$

In general,

$$C=\left[\begin{array}{c} k \\ 2 k \end{array}\right] \cdot\left[\begin{array}{cc} k & k \\ 2 k & 2 k \end{array}\right] \text { etc... }$$

where, $k$ is any real number.

Let

$$A=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right], B=\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right] \text { and } C=\left[\begin{array}{ll} 2 & 3 \\ 4 & 4 \end{array}\right]\quad [\because B \neq C]$$

$$\begin{array}{ll} \therefore & A B=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right]=\left[\begin{array}{ll} 2 & 3 \\ 0 & 0 \end{array}\right] \quad\text{.... (i)}\\ \text { and } & A C=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right] \cdot\left[\begin{array}{ll} 2 & 3 \\ 4 & 4 \end{array}\right]=\left[\begin{array}{ll} 2 & 3 \\ 0 & 0 \end{array}\right]\quad\text{.... (ii)} \end{array}$$

Thus, we see that $\quad A B=A C\quad$ [using Eqs. (i) and (ii)]

where, $A$ is non-zero matrix but $B \neq C$.

We have, $A=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right], B=\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]$ and $C=\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]$

(i) $(A B)=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]=\left[\begin{array}{cc}2+6 & 3-8 \\ -4+3 & -6-4\end{array}\right]=\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]$

$$\begin{aligned} \text { and } \quad (A B) C & =\left[\begin{array}{cc} 8 & -5 \\ -1 & -10 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \end{array}\right] \\ & =\left[\begin{array}{cc} 8+5 & 0 \\ -1+10 & 0 \end{array}\right]=\left[\begin{array}{cc} 13 & 0 \\ 9 & 0 \end{array}\right]\quad\text{.... (i)} \end{aligned}$$

$$\begin{aligned} \text { Again, } \quad (B C) & =\left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \end{array}\right] \\ & =\left[\begin{array}{ll} 2-3 & 0 \\ 3+4 & 0 \end{array}\right]=\left[\begin{array}{cc} -1 & 0 \\ 7 & 0 \end{array}\right] \end{aligned}$$

$$\begin{aligned} \text{and}\quad A(B C) & =\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{cc} -1 & 0 \\ 7 & 0 \end{array}\right] \\ & =\left[\begin{array}{cc} -1+14 & 0 \\ +2+7 & 0 \end{array}\right]=\left[\begin{array}{cc} 13 & 0 \\ 9 & 0 \end{array}\right]\quad\text{.... (ii)} \end{aligned}$$

$\therefore \quad(A B) C=A(B C) \quad$ [using Eqs. (i) and (ii)]

$$\begin{aligned} &\text { (ii) }\\ &\begin{aligned} & (B+C)=\left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]+\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \end{array}\right]=\left[\begin{array}{cc} 3 & 3 \\ 2 & -4 \end{array}\right] \\ & \text { and } \\ & A \cdot(B+C)=\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{cc} 3 & 3 \\ 2 & -4 \end{array}\right] \end{aligned} \end{aligned}$$

$=\left[\begin{array}{cc}3+4 & 3-8 \\ -6+2 & -6-4\end{array}\right]=\left[\begin{array}{cc}7 & -5 \\ -4 & -10\end{array}\right]\quad\text{.... (iii)}$

$$\begin{aligned} &\text { Also, }\\ &\begin{aligned} A B & =\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right] \cdot\left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right] \\ & =\left[\begin{array}{cc} 2+6 & 3-8 \\ -4+3 & -6-4 \end{array}\right]=\left[\begin{array}{cc} 8 & -5 \\ -1 & -10 \end{array}\right] \end{aligned} \end{aligned}$$

and $$ A C=\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \end{array}\right]=\left[\begin{array}{cc} 1-2 & 0 \\ -2-1 & 0 \end{array}\right]=\left[\begin{array}{ll} -1 & 0 \\ -3 & 0 \end{array}\right]$$

$$\begin{aligned} &\begin{array}{ll} \therefore & A B+A C=\left[\begin{array}{cc} 8 & -5 \\ -1 & -10 \end{array}\right]+\left[\begin{array}{ll} -1 & 0 \\ -3 & 0 \end{array}\right] \\ \Rightarrow & A B+A C=\left[\begin{array}{cc} 7 & -5 \\ -4 & -10 \end{array}\right]\quad\text{.... (iv)} \end{array}\\ &\text { From Eqs. (iii) and (iv), }\\ &A(B+C)=A B+A C \end{aligned}$$

$$ \begin{aligned} & P Q=\left[\begin{array}{lll} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}\right]\left[\begin{array}{lll} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array}\right]=\left[\begin{array}{ccc} x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c \end{array}\right] \quad\text{.... (i)}\\ & \text { and } Q P=\left[\begin{array}{lll} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array}\right]\left[\begin{array}{ccc} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}\right]=\left[\begin{array}{ccc} a x & 0 & 0 \\ 0 & b y & 0 \\ 0 & 0 & z c \end{array}\right]\quad\text{.... (ii)} \end{aligned}$$

Thus, we see that, $P Q=Q P$ [usings Eq. (i) and (ii)] Hence proved.