If $A = \left[ {\begin{array}{cccccccccccccccccccc}3&{ - 2}\\4&{ - 2}\end{array}} \right]$ and $I = \left[ {\begin{array}{cccccccccccccccccccc}1&0\\0&1\end{array}} \right]$, find k so that ${A^2} = kA = 2I.$

.:

We are given that, $A = \left[ {\begin{array}{cccccccccccccccccccc}3&{ - 2}\\4&{ - 2}\end{array}} \right]$ and $I = \left[ {\begin{array}{cccccccccccccccccccc}1&0\\0&1\end{array}} \right]$

Also, A2 = kA $-$ 2I

Substituting the values of A and I from above, we get

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}3&{ - 2}\\4&{ - 2}\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}3&{ - 2}\\4&{ - 2}\end{array}} \right] = k\left[ {\begin{array}{cccccccccccccccccccc}3&{ - 2}\\4&{ - 2}\end{array}} \right] - 2\left[ {\begin{array}{cccccccccccccccccccc}1&0\\0&1\end{array}} \right]$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}{3 \times 3 + ( - 2) \times 4}&{3 \times ( - 2) + ( - 2) \times ( - 2)}\\{4 \times 3 + ( - 2) \times 4}&{4 \times ( - 2) \times ( - 2) \times ( - 2)}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{3k}&{ - 2k}\\{4k}&{ - 2k}\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}2&0\\0&2\end{array}} \right]$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}{9 - 8}&{ - 6 + 4}\\{12 - 8}&{ - 8 + 4}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{3k}&{ - 2k}\\{4k}&{ - 2k}\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}2&0\\0&2\end{array}} \right]$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}1&{ - 2}\\4&{ - 4}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{3k - 2}&{ - 2k}\\{4k}&{ - 2k - 2}\end{array}} \right]$

$\Rightarrow$ $4k = 4 \Rightarrow k = 1$
Hence, $k = 1$