Mathematics - Matrices Question with Solution | TestHub

Given,

$A=\left(\begin{array}{ccc}\beta & 0& 1\\ 2& 1& -2\\ 3& 1& -2\end{array}\right)$

So, $det\left(A\right)=-1 ...\left(1\right)$

Now for $A^7-\left(\beta -1\right)A^6-\beta A^5$ to be singular

$\left|A^7-\left(\beta -1\right)A^6-\beta A^5\right|=0$

$\Rightarrow \left|A^5\right|\left|A^2-(\beta -1)A-\beta \right|=0 ...\left(2\right)$

$\Rightarrow \left|A^5\right|\left|\left(A+I\right)\left(A-\beta l\right)\right|=0$

$\therefore \left|A^5\right|\left|A+I\right|\left|A-\beta \right|=0$

As $\left|A\right|\ne 0$

$\left|A+1\right|$ or $\left|A-\beta \right|=0$

$\Rightarrow \left(\begin{array}{ccc}\beta +1& 0& 1\\ 2& 2& -2\\ 3& 1& -1\end{array}\right)=0\Rightarrow \left|A+I\right|=-4, \text{here }\left|A\right|\ne 0 \& \left|A+I\right|\ne 0$

Now solving $\left|A-\beta l\right|=\left|\begin{array}{ccc}0& 0& 1\\ 2& 1-\beta & -2\\ 3& 1& -2-\beta \end{array}\right|=0$

$\Rightarrow 2-3\left(1-\beta \right)=0$ $\Rightarrow 2-3+3\beta =0$

$\Rightarrow \beta =\frac{1}{3}$

$\therefore 9\beta =3$