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 .

• A. 3.00
• B. 3
• C. 3.0

Answer 1

Answer: (A), (B), (C)

$A=\left[\begin{array}{ccc}\beta & 0& 1\\ 2& 1& -2\\ 3& 1& -2\end{array}\right]$
$det\left(A\right)=-1\cdots \cdots \left(1\right)$
For $A^7-(\beta -1)A^6-\beta A^5$ to be singular
$|A^5||(A^2-(\beta -1)A-\beta |=0$

$⟹|A^5||(A+I)(A-\beta I)|=0\cdots \cdots \left(2\right)$

$\therefore |A^5||A+I||A-\beta I|=0$
As $|A|\ne 0$
$|A+I|or|A-\beta I|=0$

$⟹\left[\begin{array}{ccc}\beta +1& 0& 1\\ 2& 2& -2\\ 3& 1& -1\end{array}\right]=0\{|A+I|\ne 0\}$
Given, $-1=0\left(Rejected\right)$
$\therefore |A-\beta I|=\left|\begin{array}{ccc}0& 0& 1\\ 2& 1-\beta & -2\\ 3& 1& -2-\beta \end{array}\right|=0$
$⟹2-3(1-\beta )=0$
$⟹2-3+3\beta =0$
$⟹\beta =\frac{1}{3}$
$\therefore 9\beta =3$.