Question

For the following matrices verify the associativity of matrix multiplication i.e. (AB) C = A (BC):
(i) $A=\left[\begin{array}{ccc}1& 2& 0\\ -1& 0& 1\end{array}\right],B=\left[\begin{array}{cc}1& 0\\ -1& 2\\ 0& 3\end{array}\right]\mathrm{and}C=\left[\begin{array}{c}1\\ -1\end{array}\right]$

(ii) $A=\left[\begin{array}{ccc}4& 2& 3\\ 1& 1& 2\\ 3& 0& 1\end{array}\right],B=\left[\begin{array}{ccc}1& -1& 1\\ 0& 1& 2\\ 2& -1& 1\end{array}\right]\mathrm{and}C=\left[\begin{array}{ccc}1& 2& -1\\ 3& 0& 1\\ 0& 0& 1\end{array}\right]$.

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right) \\ \left(AB\right)C=A\left(BC\right) \\ \Rightarrow \left(\left[\begin{array}{ccc}1& 2& 0\\ -1& 0& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ -1& 2\\ 0& 3\end{array}\right]\right)\left[\begin{array}{c}1\\ -1\end{array}\right]=\left[\begin{array}{ccc}1& 2& 0\\ -1& 0& 1\end{array}\right]\left(\left[\begin{array}{cc}1& 0\\ -1& 2\\ 0& 3\end{array}\right]\left[\begin{array}{c}1\\ -1\end{array}\right]\right) \\ \Rightarrow \left(\left[\begin{array}{cc}1-2+0& 0+4+0\\ -1-0+0& 0+0+3\end{array}\right]\right)\left[\begin{array}{c}1\\ -1\end{array}\right]=\left[\begin{array}{ccc}1& 2& 0\\ -1& 0& 1\end{array}\right]\left(\left[\begin{array}{c}1-0\\ -1-2\\ 0-3\end{array}\right]\right) \\ \Rightarrow \left[\begin{array}{cc}-1& 4\\ -1& 3\end{array}\right]\left[\begin{array}{c}1\\ -1\end{array}\right]=\left[\begin{array}{ccc}1& 2& 0\\ -1& 0& 1\end{array}\right]\left[\left[\begin{array}{c}1\\ -3\\ -3\end{array}\right]\right] \\ \Rightarrow \left[\begin{array}{c}-1-4\\ -1-3\end{array}\right]=\left[\begin{array}{c}1-6-0\\ -1-0-3\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}-5\\ -4\end{array}\right]=\left[\begin{array}{c}-5\\ -4\end{array}\right] \\ \therefore \mathrm{LHS}=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$

$$\begin{gathered} \left(ii\right)\left(AB\right)C=A\left(BC\right) \\ \Rightarrow \left(\left[\begin{array}{ccc}4& 2& 3\\ 1& 1& 2\\ 3& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& -1& 1\\ 0& 1& 2\\ 2& -1& 1\end{array}\right]\right)\left[\begin{array}{ccc}1& 2& -1\\ 3& 0& 1\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}4& 2& 3\\ 1& 1& 2\\ 3& 0& 1\end{array}\right]\left(\left[\begin{array}{ccc}1& -1& 1\\ 0& 1& 2\\ 2& -1& 1\end{array}\right]\left[\begin{array}{ccc}1& 2& -1\\ 3& 0& 1\\ 0& 0& 1\end{array}\right]\right) \\ \Rightarrow \left(\left[\begin{array}{ccc}4+0+6& -4+2-3& 4+4+3\\ 1+0+4& -1+1-2& 1+2+2\\ 3+0+2& -3+0-1& 3+0+1\end{array}\right]\right)\left[\begin{array}{ccc}1& 2& -1\\ 3& 0& 1\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}4& 2& 3\\ 1& 1& 2\\ 3& 0& 1\end{array}\right]\left(\left[\begin{array}{ccc}1-3+0& 2-0+0& -1-1+1\\ 0+3+0& 0+0+0& 0+1+2\\ 2-3+0& 4-0+0& -2-1+1\end{array}\right]\right) \\ \Rightarrow \left[\begin{array}{ccc}10& -5& 11\\ 5& -2& 5\\ 5& -4& 4\end{array}\right]\left[\begin{array}{ccc}1& 2& -1\\ 3& 0& 1\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}4& 2& 3\\ 1& 1& 2\\ 3& 0& 1\end{array}\right]\left[\begin{array}{ccc}-2& 2& -1\\ 3& 0& 3\\ -1& 4& -2\end{array}\right] \\ \Rightarrow \left[\begin{array}{ccc}10-15+0& 20-0+0& -10-5+11\\ 5-6+0& 10-0+0& -5-2+5\\ 5-12+0& 10-0+0& -5-4+4\end{array}\right]=\left[\begin{array}{ccc}-8+6-3& 8+0+12& -4+6-6\\ -2+3-2& 2+0+8& -1+3-4\\ -6+0-1& 6+0+4& -3+0-2\end{array}\right] \\ \Rightarrow \left[\begin{array}{ccc}-5& 20& -4\\ -1& 10& -2\\ -7& 10& -5\end{array}\right]=\left[\begin{array}{ccc}-5& 20& -4\\ -1& 10& -2\\ -7& 10& -5\end{array}\right] \\ \therefore \mathrm{LHS}=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$