Question

Let $\beta$ be a real number. Consider the matrix $A=\left(\begin{array}{ccc}\beta & 0& 1\\ 2& 1& -2\\ 3& 1& -2\end{array}\right)$ If $A^7-(\beta -1)A^6-\beta A^5$ is a singular matrix, then the value of $9\beta$ is _____

Answer 1

Answer: 3

$A=\left(\begin{array}{ccc}\beta & 0& 1\\ 2& 1& -2\\ 3& 1& -2\end{array}\right)|A|=-1$
$\Rightarrow \left|A^7-(\beta -1)A^6-\beta A^5\right|=0$
$\Rightarrow |A{|}^5\left|A^2-(\beta -1)A-\beta I\right|=0$
$\Rightarrow |A{|}^5\left|\left(A^2-\beta A\right)+A-\beta I\right|=0$
$\Rightarrow |A{|}^5|A(A-\beta I)+I(A-\beta I)|=0$
$|A{|}^5|(A+I)(A-\beta I)|=0$
$A+I=\left(\begin{array}{ccc}\beta +1& 0& 1\\ 2& 2& -2\\ 3& 1& -1\end{array}\right)\Rightarrow |A+I|=-4$,
Here $|A|\ne 0\&|A+I|\ne 0$
$A-\beta I=\left(\begin{array}{ccc}0& 0& 1\\ 2& 1-\beta & -2\\ 3& 1& -2-\beta \end{array}\right)$
$|A-\beta I|=2-3(1-\beta )=3\beta -1=0\Rightarrow \beta =\frac{1}{3}$
$9\beta =3$