Question

If $A=(\begin{array}{cc}3& 2\\ -1& 4\end{array}),B=(\begin{array}{cc}-2& 5\\ 6& 7\end{array})$ and $C=(\begin{array}{cc}1& 1\\ -5& 3\end{array})$ verify that $A(B+C)=AB+AC$

Answer 1

Now, $B+C=(\begin{array}{cc}-2& 5\\ 6& 7\end{array})+(\begin{array}{cc}1& 1\\ -5& 3\end{array})=(\begin{array}{cc}-1& 6\\ 1& 10\end{array})$
Thus, $A(B+C)=(\begin{array}{cc}3& 2\\ -1& 4\end{array})(\begin{array}{cc}-1& 6\\ 1& 10\end{array})=(\begin{array}{cc}-1& 38\\ 5& 34\end{array})$ ............ (1)
Now, $AB+AC=(\begin{array}{cc}3& 2\\ -1& 4\end{array})(\begin{array}{cc}-2& 5\\ 6& 7\end{array})+(\begin{array}{cc}3& 2\\ -1& 4\end{array})(\begin{array}{cc}1& 1\\ -5& 3\end{array})$
$=(\begin{array}{cc}-6+12& 15+14\\ 2+24& -5+28\end{array})+(\begin{array}{cc}3-10& 3+6\\ -1-20& -1+12\end{array})$
$=(\begin{array}{cc}6& 29\\ 26& 23\end{array})+(\begin{array}{cc}-7& 9\\ -21& 11\end{array})$
$=(\begin{array}{cc}-1& 38\\ 5& 34\end{array})$ ............. (2)
From (1) and (2), we have $A(B+C)=AB+AC$.