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Question:

If $A=\left[\begin{array}{rrr}1 & 2 & 0 \\ 3 & -4 & 5 \\ 0 & -1 & 3\end{array}\right]$, compute $A^{2}-4 A+3 / 3$.

Solution:

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

Now,

$A^{2}=A A$

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

$\Rightarrow A^{2}=\left[\begin{array}{ccc}1+6+0 & 2-8-0 & 0+10+0 \\ 3-12+0 & 6+16-5 & 0-20+15 \\ 0-3+0 & 0+4-3 & 0-5+9\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{ccc}7 & -6 & 10 \\ -9 & 17 & -5 \\ -3 & 1 & 4\end{array}\right]$

$A^{2}-4 A+3 I_{3}$

$\Rightarrow A^{2}-4 A+3 I_{3}=\left[\begin{array}{ccc}7 & -6 & 10 \\ -9 & 17 & -5 \\ -3 & 1 & 4\end{array}\right]-4\left[\begin{array}{ccc}1 & 2 & 0 \\ 3 & -4 & 5 \\ 0 & -1 & 3\end{array}\right]+3\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$\Rightarrow A^{2}-4 A+3 I_{3}=\left[\begin{array}{ccc}7 & -6 & 10 \\ -9 & 17 & -5 \\ -3 & 1 & 4\end{array}\right]-\left[\begin{array}{ccc}4 & 8 & 0 \\ 12 & -16 & 20 \\ 0 & -4 & 12\end{array}\right]+\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right]$

$\Rightarrow A^{2}-4 A+3 I_{3}=\left[\begin{array}{ccc}7-4+3 & -6-8+0 & 10-0+0 \\ -9-12+0 & 17+16+3 & -5-20+0 \\ -3-0+0 & 1+4+0 & 4-12+3\end{array}\right]$

$\Rightarrow A^{2}-4 A+3 I_{3}=\left[\begin{array}{ccc}6 & -14 & 10 \\ -21 & 36 & -25 \\ -3 & 5 & -5\end{array}\right]$

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