Question

If $A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]\mathrm{and}I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$, then find λ so that A² = 5A + λI.

Answer 1

$$\begin{gathered} \mathrm{Given}:A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right] \\ \mathrm{Now}, \\ {A}^2=AA \\ \Rightarrow A^2=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}9-1& 3+2\\ -3-2& -1+4\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right] \\ {A}^2=5A+\lambda I \\ \Rightarrow \left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]=5\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]+\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]=\left[\begin{array}{cc}15& 5\\ -5& 10\end{array}\right]+\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]=\left[\begin{array}{cc}15+\lambda & 5+0\\ -5+0& 10+\lambda \end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]=\left[\begin{array}{cc}15+\lambda & 5\\ -5& 10+\lambda \end{array}\right] \\ \mathrm{The}\mathrm{corresponding}\mathrm{elements}\mathrm{of}\mathrm{two}\mathrm{equal}\mathrm{matrices}\mathrm{are}\mathrm{equal}. \\ \therefore 8=15+\lambda \\ \Rightarrow 8-15=\lambda \\ \Rightarrow -7=\lambda \\ \therefore \lambda =-7 \end{gathered}$$