Question

Show that $A=\left[\begin{array}{c}53\\ -1-2\end{array}\right]$ satisfies the equation $A^2$ - 3A - 7I = 0 and hence find the value of $A^{-1}$.

Answer 1

We have $A=\left[\begin{array}{c}53\\ -1-2\end{array}\right]$

$\therefore A^2$ = A. A =$\left[\begin{array}{c}53\\ -1-2\end{array}\right]\left[\begin{array}{c}53\\ -1-2\end{array}\right]$

$\left[\begin{array}{c}25-315-6\\ -5+2-3+4\end{array}\right]=\left[\begin{array}{c}229\\ -31\end{array}\right]$

3A = 3 $\left[\begin{array}{c}53\\ -1-2\end{array}\right]=\left[\begin{array}{c}159\\ -3-6\end{array}\right]$

and 7I = 7$\left[\begin{array}{c}10\\ 01\end{array}\right]=\left[\begin{array}{c}70\\ 07\end{array}\right]$

$\therefore$$A^2-3A-7I=\left[\begin{array}{c}229\\ -31\end{array}\right]-\left[\begin{array}{c}159\\ -3-6\end{array}\right]-\left[\begin{array}{c}70\\ 07\end{array}\right]$

=$\left[\begin{array}{c}22-15-79-9-0\\ -3+3-01+6+-7\end{array}\right]$

=$\left[\begin{array}{c}00\\ 00\end{array}\right]$

=0

Since, $A^2$ - 3A - 7I =0

$\Rightarrow A^{-1}\left[\right(A^2)-3A-7I]=A^{-1}0$

$\Rightarrow A^{-1}A.A-3A^{-1}A-7A^{-1}I=0$ $[\because A^{-1}0=0]$

$\Rightarrow IA-3I-7A^{-1}=0$ $[\because A^{-1}A=I]$

$\Rightarrow A-3I-7A^{-1}=0$ $[\because A^{-1}I=A^{-1}]$

$\Rightarrow -7A^{-1}=-A+3I$

$\left[\begin{array}{c}-5-3\\ 12\end{array}\right]+\left[\begin{array}{c}30\\ 03\end{array}\right]=\left[\begin{array}{c}-2-3\\ 15\end{array}\right]$

$A^{-1}=\frac{-1}{7}\left[\begin{array}{c}-2-3\\ 15\end{array}\right]$