Question

If $A=\left(\begin{array}{ccc}2& 3& 4\\ 0& -2& 1\\ 3& -1& 2\end{array}\right),B=\left(\begin{array}{ccc}2& 0& -3\\ 4& 0& -1\\ 3& 4& 5\end{array}\right)$ and $C=\left(\begin{array}{ccc}5& 6& 7\\ -1& 2& 3\\ 4& -5& 4\end{array}\right)$ Prove that $A\left(BC\right)$ and $\left(AB\right)C$.

Answer 1

$A=\left(\stackrel{2}{\underset{3}{0}}\stackrel{3}{\underset{-1}{-2}}\stackrel{4}{\underset{2}{1}}\right);B=\left(\stackrel{2}{\underset{3}{4}}\stackrel{0}{\underset{4}{0}}\stackrel{-3}{\underset{5}{-1}}\right);C=\left(\stackrel{5}{\underset{4}{-1}}\stackrel{6}{\underset{-5}{2}}\stackrel{7}{\underset{4}{3}}\right)$

$BC=\left(\stackrel{-2}{\underset{31}{16}}\stackrel{17}{\underset{1}{29}}\stackrel{2}{\underset{53}{24}}\right)$

$A\left(BC\right)=\left(\stackrel{168}{\underset{40}{-1}}\stackrel{145}{\underset{24}{-57}}\stackrel{288}{\underset{88}{5}}\right)$

Now, $AB=\left(\stackrel{28}{\underset{8}{-5}}\stackrel{16}{\underset{8}{4}}\stackrel{11}{\underset{2}{7}}\right)$

$\left(AB\right)C=\left(\stackrel{168}{\underset{40}{-1}}\stackrel{145}{\underset{24}{-57}}\stackrel{288}{\underset{88}{5}}\right)$

$\therefore$ $A\left(BC\right)=\left(AB\right)C$ Proved.