For the following matrices verify the associativity of matrix multiplication i.e. (AB) C = A (BC):

Solution:

(i)

$(A B) C=A(B C)$

$\Rightarrow\left(\left[\begin{array}{ccc}1 & 2 & 0 \\ -1 & 0 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 2 \\ 0 & 3\end{array}\right]\right)\left[\begin{array}{c}1 \\ -1\end{array}\right]=\left[\begin{array}{ccc}1 & 2 & 0 \\ -1 & 0 & 1\end{array}\right]\left(\left[\begin{array}{cc}1 & 0 \\ -1 & 2 \\ 0 & 3\end{array}\right]\left[\begin{array}{c}1 \\ -1\end{array}\right]\right)$

$\Rightarrow\left(\left[\begin{array}{cc}1-2+0 & 0+4+0 \\ -1 & -0+0 & 0+0+3\end{array}\right]\right)\left[\begin{array}{c}1 \\ -1\end{array}\right]=\left[\begin{array}{ccc}1 & 2 & 0 \\ -1 & 0 & 1\end{array}\right]\left(\left[\begin{array}{c}1-0 \\ -1-2 \\ 0-3\end{array}\right]\right)$

$\Rightarrow\left[\begin{array}{ll}-1 & 4 \\ -1 & 3\end{array}\right]\left[\begin{array}{c}1 \\ -1\end{array}\right]=\left[\begin{array}{ccc}1 & 2 & 0 \\ -1 & 0 & 1\end{array}\right]\left[\left[\begin{array}{c}1 \\ -3 \\ -3\end{array}\right]\right]$

$\Rightarrow\left[\begin{array}{l}-1-4 \\ -1-3\end{array}\right]=\left[\begin{array}{c}1-6-0 \\ -1-0-3\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}-5 \\ -4\end{array}\right]=\left[\begin{array}{l}-5 \\ -4\end{array}\right]$

$\therefore \mathrm{LHS}=\mathrm{RHS}$

Hence proved.

(ii) $(A B) C=A(B C)$

$\Rightarrow\left(\left[\begin{array}{lll}4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}1 & -1 & 1 \\ 0 & 1 & 2 \\ 2 & -1 & 1\end{array}\right]\right)\left[\begin{array}{ccc}1 & 2 & -1 \\ 3 & 0 & 1 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1\end{array}\right]\left(\left[\begin{array}{ccc}1 & -1 & 1 \\ 0 & 1 & 2 \\ 2 & -1 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 2 & -1 \\ 3 & 0 & 1 \\ 0 & 0 & 1\end{array}\right]\right)$

$\Rightarrow\left(\left[\begin{array}{ccc}4+0+6 & -4+2-3 & 4+4+3 \\ 1+0+4 & -1+1-2 & 1+2+2 \\ 3+0+2 & -3+0-1 & 3+0+1\end{array}\right]\right)\left[\begin{array}{ccc}1 & 2 & -1 \\ 3 & 0 & 1 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1\end{array}\right]\left(\left[\begin{array}{cc}1-3+0 & 2-0+0 & -1-1+1 \\ 0+3+0 & 0+0+0 & 0+1+2 \\ 2-3+0 & 4-0+0 & -2-1+1\end{array}\right]\right)$

$\Rightarrow\left[\begin{array}{ccc}10 & -5 & 11 \\ 5 & -2 & 5 \\ 5 & -4 & 4\end{array}\right]\left[\begin{array}{ccc}1 & 2 & -1 \\ 3 & 0 & 1 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}-2 & 2 & -1 \\ 3 & 0 & 3 \\ -1 & 4 & -2\end{array}\right]$

$\Rightarrow\left[\begin{array}{ccc}10-15+0 & 20-0+0 & -10-5+11\end{array}\right]$

$\Rightarrow\left[\begin{array}{ccc}5-6+0 & 10-0+0 & -5-2+5 \\ 5-12+0 & 10-0+0 & -5-4+4\end{array}\right]=\left[\begin{array}{ccc}-8+6-3 & 8+0+12 & -4+6-6 \\ -2+3-2 & 2+0+8 & -1+3-4 \\ -6+0-1 & 6+0+4 & -3+0-2\end{array}\right]$

$\Rightarrow\left[\begin{array}{lll}-5 & 20 & -4 \\ -1 & 10 & -2 \\ -7 & 10 & -5\end{array}\right]=\left[\begin{array}{ccc}-1 & 10 & -2 \\ -7 & 10 & -5\end{array}\right]$

$\therefore$ LHS $=$ RHS

Hence proved.