Question

If $A=\left[\begin{array}{ccc}3& 2& 0\\ 1& 4& 0\\ 0& 0& 5\end{array}\right]$, show that A² − 7A + 10I₃ = O

Answer 1

$$\begin{gathered} \mathrm{Given}:\hspace{0.17em}A=\left[\begin{array}{ccc}3& 2& 0\\ 1& 4& 0\\ 0& 0& 5\end{array}\right] \\ \mathrm{Here}, \\ {A}^2=AA \\ \Rightarrow A^2=\left[\begin{array}{ccc}3& 2& 0\\ 1& 4& 0\\ 0& 0& 5\end{array}\right]\left[\begin{array}{ccc}3& 2& 0\\ 1& 4& 0\\ 0& 0& 5\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{ccc}9+2+0& 6+8+0& 0+0+0\\ 3+4+0& 2+16+0& 0+0+0\\ 0+0+0& 0+0+0& 0+0+25\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{ccc}11& 14& 0\\ 7& 18& 0\\ 0& 0& 25\end{array}\right] \\ \mathrm{Now}, \\ {A}^2-7A+10I_3 \\ \Rightarrow A^2-7A+10I_3=\left[\begin{array}{ccc}11& 14& 0\\ 7& 18& 0\\ 0& 0& 25\end{array}\right]-7\left[\begin{array}{ccc}3& 2& 0\\ 1& 4& 0\\ 0& 0& 5\end{array}\right]+10\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] \\ \Rightarrow A^2-7A+10I_3=\left[\begin{array}{ccc}11& 14& 0\\ 7& 18& 0\\ 0& 0& 25\end{array}\right]-\left[\begin{array}{ccc}21& 14& 0\\ 7& 28& 0\\ 0& 0& 35\end{array}\right]+\left[\begin{array}{ccc}10& 0& 0\\ 0& 10& 0\\ 0& 0& 10\end{array}\right] \\ \Rightarrow A^2-7A+10I_3=\left[\begin{array}{ccc}11-21+10& 14-14+0& 0-0+0\\ 7-7+0& 18-28+10& 0-0+0\\ 0-0+0& 0-0+0& 25-35+10\end{array}\right] \\ \Rightarrow A^2-7A+10I_3=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]=0 \end{gathered}$$