Question

If $A=\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]$, find k such that A² − 8A + kI = 0.

Answer 1

$$\begin{gathered} \mathrm{Given}:\hspace{0.17em}A=\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right] \\ Now, \\ {A}^2=AA \\ \Rightarrow A^2=\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}1-0& 0+0\\ -1-7& 0+49\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}1& 0\\ -8& 49\end{array}\right] \\ {A}^2-8A+kI=0 \\ \Rightarrow \left[\begin{array}{cc}1& 0\\ -8& 49\end{array}\right]-8\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]+k\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=0 \\ \Rightarrow \left[\begin{array}{cc}1& 0\\ -8& 49\end{array}\right]-\left[\begin{array}{cc}8& 0\\ -8& 56\end{array}\right]+\left[\begin{array}{cc}k& 0\\ 0& k\end{array}\right]=0 \\ \Rightarrow \left[\begin{array}{cc}1-8+k& 0-0+0\\ -8+8+0& 49-56+k\end{array}\right]=0 \\ \Rightarrow \left[\begin{array}{cc}-7+k& 0\\ 0& -7+k\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ \mathrm{The}\mathrm{corresponding}\mathrm{elements}\mathrm{of}\mathrm{two}\mathrm{equal}\mathrm{matrices}\mathrm{are}\mathrm{equal}. \\ \therefore -7+k=0 \\ \Rightarrow k=7 \end{gathered}$$