Question
If $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$ show that $A^2-5 A+7 I_2=0$
✓
Answer
$\begin{aligned} & \text { Given } A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right] \\ & A^2=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right] \times\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right] \\ & =\left[\begin{array}{cc}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right] \\ & =\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right] \\ & \text { L.H.S. = A } 2.5 A +7 I _2 \\ & =\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]-5\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]+7\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] \\ & =\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]-\left[\begin{array}{cc}15 & 5 \\ -5 & 10\end{array}\right]+\left[\begin{array}{ll}7 & 0 \\ 0 & 7\end{array}\right] \\ & =\left[\begin{array}{cc}-7 & 0 \\ 0 & -7\end{array}\right]+\left[\begin{array}{cc}7 & 0 \\ 0 & 7\end{array}\right]\end{aligned}$
$
\begin{aligned}
& =\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right] \\
& =0=\text { R.H.S. }
\end{aligned}
$
L.H.S. = R.H.S.
$
\therefore A^2-5 A+7 I_2=0
$
$
\begin{aligned}
& =\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right] \\
& =0=\text { R.H.S. }
\end{aligned}
$
L.H.S. = R.H.S.
$
\therefore A^2-5 A+7 I_2=0
$
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