Question

For the following matrices verify the distributivity of matrix multiplication over matrix addition i.e. A (B + C) = AB + AC:
(i) $A=\left[\begin{array}{cc}1& -1\\ 0& 2\end{array}\right],B=\left[\begin{array}{cc}-1& 0\\ 2& 1\end{array}\right]\mathrm{and}C=\left[\begin{array}{cc}0& 1\\ 1& -1\end{array}\right]$
(ii) $A=\left[\begin{array}{cc}2& -1\\ 1& 1\\ -1& 2\end{array}\right],B=\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]\mathrm{and}C=\left[\begin{array}{cc}1& -1\\ 0& 1\end{array}\right].$

Answer 1

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

$$\begin{gathered} \left(\mathrm{ii}\right) \\ A\left(B+C\right)=AB+AC \\ \Rightarrow \left[\begin{array}{cc}2& -1\\ 1& 1\\ -1& 2\end{array}\right]\left(\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]+\left[\begin{array}{cc}1& -1\\ 0& 1\end{array}\right]\right)=\left[\begin{array}{cc}2& -1\\ 1& 1\\ -1& 2\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]+\left[\begin{array}{cc}2& -1\\ 1& 1\\ -1& 2\end{array}\right]\left[\begin{array}{cc}1& -1\\ 0& 1\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}2& -1\\ 1& 1\\ -1& 2\end{array}\right]\left[\begin{array}{cc}0+1& 1-1\\ 1+0& 1+1\end{array}\right]=\left[\begin{array}{cc}0-1& 2-1\\ 0+1& 1+1\\ 0+2& -1+2\end{array}\right]+\left[\begin{array}{cc}2-0& -2-1\\ 1+0& -1+1\\ -1+0& 1+2\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}2& -1\\ 1& 1\\ -1& 2\end{array}\right]\left[\begin{array}{cc}1& 0\\ 1& 2\end{array}\right]=\left[\begin{array}{cc}-1& 1\\ 1& 2\\ 2& 1\end{array}\right]+\left[\begin{array}{cc}2& -3\\ 1& 0\\ -1& 3\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}2-1& 0-2\\ 1+1& 0+2\\ -1+2& 0+4\end{array}\right]=\left[\begin{array}{cc}-1+2& 1-3\\ 1+1& 2+0\\ 2-1& 1+3\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}1& -2\\ 2& 2\\ 1& 4\end{array}\right]=\left[\begin{array}{cc}1& -2\\ 2& 2\\ 1& 4\end{array}\right] \\ \therefore \mathrm{LHS}=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$