Give an example of matrices $A, B$ and $C$, such that $A B=A C$, where $A$ is non-zero matrix but $B \neq C$.
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 $B \neq C$.
If $A=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right], B=\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]$ and $C=\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]$, verify
(i) $(A B) C=A(B C)$.
(ii) $A(B+C)=A B+A C$.
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 $$ A C=\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \end{array}\right]=\left[\begin{array}{cc} 1-2 & 0 \\ -2-1 & 0 \end{array}\right]=\left[\begin{array}{ll} -1 & 0 \\ -3 & 0 \end{array}\right]$$
$$\begin{aligned} &\begin{array}{ll} \therefore & A B+A C=\left[\begin{array}{cc} 8 & -5 \\ -1 & -10 \end{array}\right]+\left[\begin{array}{ll} -1 & 0 \\ -3 & 0 \end{array}\right] \\ \Rightarrow & A B+A C=\left[\begin{array}{cc} 7 & -5 \\ -4 & -10 \end{array}\right]\quad\text{.... (iv)} \end{array}\\ &\text { From Eqs. (iii) and (iv), }\\ &A(B+C)=A B+A C \end{aligned}$$
If $P=\left[\begin{array}{ccc}x & 0 & 0 \\ 0 & \mathrm{y} & 0 \\ 0 & 0 & \mathrm{z}\end{array}\right]$ and $Q=\left[\begin{array}{ccc}\mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c}\end{array}\right]$, then prove that $P Q=\left[\begin{array}{ccc}x \mathrm{a} & 0 & 0 \\ 0 & \mathrm{yb} & 0 \\ 0 & 0 & \mathrm{zc}\end{array}\right]=Q P$
$$ \begin{aligned} & P Q=\left[\begin{array}{lll} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}\right]\left[\begin{array}{lll} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array}\right]=\left[\begin{array}{ccc} x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c \end{array}\right] \quad\text{.... (i)}\\ & \text { and } Q P=\left[\begin{array}{lll} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array}\right]\left[\begin{array}{ccc} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}\right]=\left[\begin{array}{ccc} a x & 0 & 0 \\ 0 & b y & 0 \\ 0 & 0 & z c \end{array}\right]\quad\text{.... (ii)} \end{aligned}$$
Thus, we see that, $P Q=Q P$ [usings Eq. (i) and (ii)] Hence proved.
If $\left[\begin{array}{lll}2 & 1 & 3\end{array}\right]\left[\begin{array}{ccc}-1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{c}1 \\ 0 \\ -1\end{array}\right]=A$, then find the value of $A$.
We have, $\left[\begin{array}{lll}2 & 1 & 3\end{array}\right]\left[\begin{array}{ccc}-1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{c}1 \\ 0 \\ -1\end{array}\right]=A$ $$ \begin{aligned} \therefore \quad\left[\begin{array}{lll} 2 & 3 \end{array}\right]\left[\begin{array}{ccc} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right] & =\left[\begin{array}{lll} -2-1+0 & 0+1+3 & -2+0+3 \end{array}\right] \\ & =\left[\begin{array}{lll} -3 & 4 & 1 \end{array}\right] \end{aligned}$$
Now,
$$\begin{aligned} {\left[\begin{array}{lll} -3 & 4 & 1 \end{array}\right]\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right] } & =A \\ \therefore\quad A & =\left[\begin{array}{lll} -3 & 4 & 1 \end{array}\right]\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right] \\ & =[-3+0-1]=[-4] \end{aligned}$$
If $A=[21], B=\left[\begin{array}{lll}5 & 3 & 4 \\ 8 & 7 & 6\end{array}\right]$ and $C=\left[\begin{array}{ccc}-1 & 2 & 1 \\ 1 & 0 & 2\end{array}\right]$, then verify that $A(B+C)=(A B+A C)$.
We have to verify that, $A(B+C)=A B+A C$
We have, $$A=\left[\begin{array}{ll} 2 & 1 \end{array}\right], B=\left[\begin{array}{lll} 5 & 3 & 4 \\ 8 & 7 & 6 \end{array}\right] \text { and } C=\left[\begin{array}{ccc} -1 & 2 & 1 \\ 1 & 0 & 2 \end{array}\right]$$
$$\begin{aligned} \therefore \quad A(B+C) & =\left[\begin{array}{ll} 2 & 1 \end{array}\right]\left[\begin{array}{lll} 5-1 & 3+2 & 4+1 \\ 8+1 & 7+0 & 6+2 \end{array}\right] \\ & =\left[\begin{array}{ll} 2 & 1 \end{array}\right]\left[\begin{array}{lll} 4 & 5 & 5 \\ 9 & 7 & 8 \end{array}\right] \\ & =\left[\begin{array}{lll} 8+9 & 10+7 & 10+8 \end{array}\right] \\ & =\left[\begin{array}{lll} 17 & 17 & 18 \end{array}\right]\quad\text{.... (i)} \end{aligned}$$
$$\begin{aligned} \text { Also, } \quad A B & =\left[\begin{array}{ll} 2 & 1 \end{array}\right]\left[\begin{array}{lll} 5 & 3 & 4 \\ 8 & 7 & 6 \end{array}\right] \\ & =\left[\begin{array}{lll} 10+8 & 6+7 & 8+6 \end{array}\right]=\left[\begin{array}{lll} 18 & 13 & 14 \end{array}\right] \end{aligned}$$
$$\begin{aligned} \text{and}\quad A C & =\left[\begin{array}{ll} 2 & 1 \end{array}\right]\left[\begin{array}{ccc} -1 & 2 & 1 \\ 1 & 0 & 2 \end{array}\right] \\ & =\left[\begin{array}{lll} -2+1 & 4+0 & 2+2 \end{array}\right]=\left[\begin{array}{lll} -1 & 4 & 4 \end{array}\right] \\ \therefore\quad A B+A C & =\left[\begin{array}{lll} 18 & 13 & 14 \end{array}\right]+\left[\begin{array}{lll} -1 & 4 & 4 \end{array}\right] \\ & =\left[\begin{array}{lll} 17 & 17 & 18 \end{array}\right] \quad\text{.... (ii)}\\ \therefore\quad A(B+C) & =(A B+A C)\quad\text{[using Eqs. (i) and (ii)]} \end{aligned}$$
Hence proved.