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જો $A=\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]$ હોય, તો સાબિત કરો કે $A^{2}-5 A+7 I=0$.
Solution
It is given that $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$
$\therefore A^{2}=A \cdot A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$
$=\left[\begin{array}{cc}3(3)+1(-1) & 3(1)+1(2) \\ -1(3)+2(-1) & -1(1)+2(2)\end{array}\right]$
$=\left[\begin{array}{cc}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right]=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]$
$L.H.S.$ $=A^{2}-5 A+7 I$
$=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]-5\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]+7\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]-\left[\begin{array}{cc}15 & 5 \\ -5 & 10\end{array}\right]+\left[\begin{array}{ll}7 & 0 \\ 0 & 7\end{array}\right]$
$=\left[\begin{array}{cc}-7 & 0 \\ 0 & -7\end{array}\right]+\left[\begin{array}{ll}7 & 0 \\ 0 & 7\end{array}\right]$
$=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
$=0=\mathrm{R} . \mathrm{HS}$
$\therefore A^{2}-5 A+7 I=0$