Question

If $A=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$, then show that $A^3=\left[\begin{array}{cc}7& -12\\ 3& -5\end{array}\right]$.

Answer 1

As we know, product of two matrices of order of $mxp$ is possible only if the matrices has order $mxn$ and $nxp$.

So, $A^3=A^2.A$

(Which all have same order $2\times 2$)

$A^3=(A.A).A.$

$=\left(\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]\right)\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$

$=\left[\begin{array}{cc}9-4& -12+4\\ 3-1& -4+1\end{array}\right]\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$

$A^3=\left[\begin{array}{cc}5& -8\\ 2& -3\end{array}\right]\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$

$=\left[\begin{array}{cc}15-8& -20+8\\ 6-3& -8+3\end{array}\right]$

$A^3=\left[\begin{array}{cc}7& -12\\ 3& -5\end{array}\right]$