Let and a = 4, b = –2.

Solution:

(a) $A+(B+C)=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]+\left[\begin{array}{ll}6 & 0 \\ 2 & 3\end{array}\right]=\left[\begin{array}{ll}7 & 2 \\ 1 & 6\end{array}\right]$

And,

$(A+B)+C=\left[\begin{array}{ll}5 & 2 \\ 0 & 8\end{array}\right]+\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]=\left[\begin{array}{ll}7 & 2 \\ 1 & 6\end{array}\right]=A+(B+C)$

(b) $(B C)=\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]=\left[\begin{array}{cc}8 & 0 \\ 7 & -10\end{array}\right]$

And, $\quad A(B C)=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]\left[\begin{array}{cc}8 & 0 \\ 7 & -10\end{array}\right]=\left[\begin{array}{cc}8+14 & 0-20 \\ -8+21 & 0-30\end{array}\right]=\left[\begin{array}{cc}22 & -20 \\ 13 & -30\end{array}\right]$

Also,

$A B=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right] \cdot\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]=\left[\begin{array}{cc}4+2 & 0+10 \\ -4+3 & 0+15\end{array}\right]=\left[\begin{array}{cc}6 & 10 \\ -1 & 15\end{array}\right]$

Thus, $(A B) C=\left[\begin{array}{cc}6 & 10 \\ -1 & 15\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]=\left[\begin{array}{cc}22 & -20 \\ 13 & -30\end{array}\right]=A(B C)$

Hence proved.

(c) $(a+b) B=(4-2)\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]$ $[\because$ given $a=4, b=-2]$

$=\left[\begin{array}{cc}8 & 0 \\ 2 & 10\end{array}\right]$

Also,

$a B+b B=4 B-2 B=\left[\begin{array}{cc}16 & 0 \\ 4 & 20\end{array}\right]-\left[\begin{array}{cc}8 & 0 \\ 2 & 10\end{array}\right]=\left[\begin{array}{cc}8 & 0 \\ 2 & 10\end{array}\right]=(a+b) B$

Hence proved.

(d) $C-A=\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]-\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]=\left[\begin{array}{cc}1 & -2 \\ 2 & -5\end{array}\right]$

And, $\quad a(C-A)=4(C-A)$

$=\left[\begin{array}{cc}4 & -8 \\ 8 & -20\end{array}\right]$

Also, $a C-a A=4 C-4 A$

$=\left[\begin{array}{cc}8 & 0 \\ 4 & -8\end{array}\right]-\left[\begin{array}{cc}4 & 8 \\ -4 & 12\end{array}\right]=\left[\begin{array}{cc}4 & -8 \\ 8 & -20\end{array}\right]=a(C-A)$

Hence proved.

(e) $A^{T}=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]^{T}=\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]$

Thus,

$\left(A^{T}\right)^{T}=\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]^{T}=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]=A$

Hence proved.

(f) $(b A)^{T}=\left[\begin{array}{cc}-2 & -4 \\ 2 & -6\end{array}\right]^{T}$ $[\because b=-2]$

$=\left[\begin{array}{cc}-2 & 2 \\ -4 & -6\end{array}\right]$

And,

$A^{T}=\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]$

Thus, $\quad b A^{T}=\left[\begin{array}{cc}-2 & 2 \\ -4 & -6\end{array}\right]=(b A)^{T}$

Hence proved.

(g) $A B=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]=\left[\begin{array}{cc}4+2 & 0+10 \\ -4+3 & 0+15\end{array}\right]=\left[\begin{array}{cc}6 & 10 \\ -1 & 15\end{array}\right]$

So,

$(A B)^{T}=\left[\begin{array}{cc}6 & -1 \\ 10 & 15\end{array}\right]$

Now, $B^{T} A^{T}=\left[\begin{array}{cc}4 & 1 \\ 0 & 5\end{array}\right]\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]=\left[\begin{array}{cc}4+2 & -4+3 \\ 0+10 & 0+15\end{array}\right]=\left[\begin{array}{cc}6 & -1 \\ 10 & 15\end{array}\right]=(A B)^{T}$

Hence proved.

(h) $(A-B)=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]-\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]=\left[\begin{array}{cc}1-4 & 2-0 \\ -1-1 & 3-5\end{array}\right]=\left[\begin{array}{cc}-3 & 2 \\ -2 & -2\end{array}\right]$

Thus, $(A-B) C=\left[\begin{array}{cc}-3 & 2 \\ -2 & -2\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]=\left[\begin{array}{cc}-4 & -4 \\ -6 & 4\end{array}\right]$

Now,

$A C=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]=\left[\begin{array}{cc}4 & -4 \\ 1 & -6\end{array}\right]$

And,

$B C=\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]=\left[\begin{array}{cc}8 & 0 \\ 7 & -10\end{array}\right]$

Therefore,

$A C-B C=\left[\begin{array}{cc}4-8 & -4-0 \\ 1-7 & -6+10\end{array}\right]=\left[\begin{array}{cc}-4 & -4 \\ -6 & 4\end{array}\right]=(A-B) C$

Hence proved.

(i) $(A-B)^{T}=\left[\begin{array}{cc}1-4 & 2-0 \\ -1-1 & 3-5\end{array}\right]^{T}=\left[\begin{array}{cc}-3 & 2 \\ -2 & -2\end{array}\right]^{T}=\left[\begin{array}{cc}-3 & -2 \\ 2 & -2\end{array}\right]$

Now, $A^{T}-B^{T}=\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]-\left[\begin{array}{ll}4 & 1 \\ 0 & 5\end{array}\right]=\left[\begin{array}{cc}-3 & -2 \\ 2 & -2\end{array}\right]=(A-B)^{T}$

Hence proved.