# if

Question:

If $A=\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]$ and $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$, then prove that $A^{2}-A+2 l=0$

Solution:

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

Now,

$A^{2}=A A$

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

$\Rightarrow A^{2}=\left[\begin{array}{cc}9-8 & -6+4 \\ 12-8 & -8+4\end{array}\right]$

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

$A^{2}-A+2 I=\left[\begin{array}{ll}1 & -2 \\ 4 & -4\end{array}\right]-\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]+2\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

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

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

$\Rightarrow A^{2}-A+2 I=\left[\begin{array}{cc}-2+2 & 0+0 \\ 0+0 & -2+2\end{array}\right]$

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

$\Rightarrow A^{2}-A+2 I=0$

Hence proved.