Solve this

Question:

If $A=\left[\begin{array}{lll}3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5\end{array}\right]$, show that $A^{2}-7 A+10 / 3=0$

Solution:

Given : $A=\left[\begin{array}{lll}3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5\end{array}\right]$

Here,

$A^{2}=A A$

$\Rightarrow A^{2}=\left[\begin{array}{lll}3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5\end{array}\right]\left[\begin{array}{lll}3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{ccc}9+2+0 & 6+8+0 & 0+0+0 \\ 3+4+0 & 2+16+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+25\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{ccc}11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25\end{array}\right]$

Now,

$A^{2}-7 A+10 I_{3}$

$\Rightarrow A^{2}-7 A+10 I_{3}=\left[\begin{array}{ccc}11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25\end{array}\right]-7\left[\begin{array}{ccc}3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5\end{array}\right]+10\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$\Rightarrow A^{2}-7 A+10 I_{3}=\left[\begin{array}{ccc}11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25\end{array}\right]-\left[\begin{array}{ccc}21 & 14 & 0 \\ 7 & 28 & 0 \\ 0 & 0 & 35\end{array}\right]+\left[\begin{array}{ccc}10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10\end{array}\right]$

$\Rightarrow A^{2}-7 A+10 I_{3}=\left[\begin{array}{ccc}11-21+10 & 14-14+0 & 0-0+0 \\ 7-7+0 & 18-28+10 & 0-0+0 \\ 0-0+0 & 0-0+0 & 25-35+10\end{array}\right]$

$\Rightarrow A^{2}-7 A+10 I_{3}=\left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$ $=0$

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