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Question:

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

Solution:

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

Now,

$A^{2}=A A$c

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

$\Rightarrow A^{2}=\left[\begin{array}{cc}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]$

$A^{2}-5 A+7 I_{2}$

$\Rightarrow A^{2}-5 A+7 I_{2}=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]-5\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]+7\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$\Rightarrow A^{2}-5 A+7 I_{2}=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]-\left[\begin{array}{cc}15 & 5 \\ -5 & 10\end{array}\right]+\left[\begin{array}{ll}7 & 0 \\ 0 & 7\end{array}\right]$

$\Rightarrow A^{2}-5 A+7 I_{2}=\left[\begin{array}{cc}8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7\end{array}\right]$

$\Rightarrow A^{2}-5 A+7 I_{2}=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

$\Rightarrow A^{2}-5 A+7 I_{2}=0$

Hence proved.