Question

If A = $\left[\begin{array}{cc}2& -1\\ 3& 2\end{array}\right]$ and B = $\left[\begin{array}{cc}0& 4\\ -1& 7\end{array}\right]$, find 3A² − 2B + I

Answer 1

$$\begin{gathered} \mathrm{Given}:A=\left[\begin{array}{cc}2& -1\\ 3& 2\end{array}\right] \\ \mathrm{Now}, \\ {A}^2=AA \\ \Rightarrow A^2=\left[\begin{array}{cc}2& -1\\ 3& 2\end{array}\right]\left[\begin{array}{cc}2& -1\\ 3& 2\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}4-3& -2-2\\ 6+6& -3+4\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}1& -4\\ 12& 1\end{array}\right] \\ 3A^2-2B+I \\ \Rightarrow 3A^2-2B+I=3\left[\begin{array}{cc}1& -4\\ 12& 1\end{array}\right]-2\left[\begin{array}{cc}0& 4\\ -1& 7\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \Rightarrow 3A^2-2B+I=\left[\begin{array}{cc}3& -12\\ 36& 3\end{array}\right]-\left[\begin{array}{cc}0& 8\\ -2& 14\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \Rightarrow 3A^2-2B+I=\left[\begin{array}{cc}3-0+1& -12-8+0\\ 36+2+0& 3-14+1\end{array}\right] \\ \Rightarrow 3A^2-2B+I=\left[\begin{array}{cc}4& -20\\ 38& -10\end{array}\right] \end{gathered}$$