Question

If a matrix A=[2 -1] [-1 2] show that A^2-4A+3I=0​

Answer 1

Given:-$\text { If } \boldsymbol{A}=\left[\begin{array}{cc}2 & -1 \\-1 & 2\end{array}\right] \text {, show that } A^{2}-4 A+3 I=0 \text {. }$

$A^{2}=A \cdot A$

=$\left[\begin{array}{cc}-1 & \overrightarrow{2} & \overrigh \\-1 & 2\end{array}\right]$$\left[\begin{array}{cc}3 & -1 \\-1 & 2\end{array}\right]$

$=\left[\begin{array}{cc}4+1 & -2+(-2) \\-2+(-2) & 1+4\end{array}\right]$

=$\left[\begin{array}{rr}5 & -4 \\-4 & 5\end{array}\right]$

$4 \mathrm{~A}=4\left[\begin{array}{cc}2 & -1 \\-1 & 2\end{array}\right]$

$=\left[\begin{array}{cc}8 & -4 \\-4 & 8\end{array}\right]$

$\text { LHS }=A^{2}-4 A+3 I$

= $\left[\begin{array}{cc}5 & -4 \\-4 & 5\end{array}\right]-\left[\begin{array}{cc}8 & -4 \\-4 & 8\end{array}\right]+\left[\begin{array}{ll}3 & 0 \\0 & 3\end{array}\right]$

=$=\left[\begin{array}{cc}-3 & 0 \\0 & -3\end{array}\right]+\left[\begin{array}{ll}3 & 0 \\0 & 3\end{array}\right]$

$=\left[\begin{array}{ll}0 & 0 \\0 & 0\end{array}\right]=0=\mathbb{R H S}$