Question

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

Answer 1

Given :
$A=\begin{bmatrix} \beta & 0 & 1 \\\ 2 & 1 & -2 \\\ 3 & 1 & -2 \end{bmatrix}$
det(A)=−1 ……(i)
So, For A7 − (β − 1)A6 − βA5 to be singular
|A5| |(A2 − (β − 1)A − β| = 0
⇒ |A5| |(A + I) (A − βI)| = 0 …..(ii)
∴|A5| |A + I| |A − βI| = 0
As we know, |A| ≠ 0
|A+I| or |A−βI| = 0
$⇒\begin{bmatrix} \beta+1 & 0 & 1 \\\ 2 & 2 & -2 \\\ 3 & 1 & -1 \end{bmatrix}=0$ {|A + I| ≠ 0}
It is Given that , −1=0 (Rejected)
$∴ | A − β I | =\begin{vmatrix} 0 & 0 & 1 \\\ 2 & 1-\beta & -2 \\\ 3 & 1 & -2-\beta \end{vmatrix}=0$
⇒ 2 − 3(1 − β) = 0
⇒ $\beta=\frac{1}{3}$
Therefore, 9β = 3.
So, the correct answer is 3.