Let β be a real number. Consider the matrix A $= [[β,0,1],[2,1,-2],[3,1,-2]].$ If $A^{7} - (β-1)A^{6} - βA^{5}$ is a singular matrix, then the value of 9β is ____.

Let β be a real number. Consider the matrix A = β 0 1 2 1 -2 3 1 -2 . If A⁷ - (β - 1)A⁶ - β A⁵ is a singular matrix, then the value of 9β is underlinehspace2cm. A = β 0 1 2 1 -2 3 1 -2

$A^{7} - (\beta-1)A^{6} - \beta A^{5} \text{ is singular}$

$\text{Find } 9\beta$

$1.\ \text{factor out } A^{5}$

$2.\ \text{check whether } A \text{ is invertible}$

$3.\ \det(\cdot) = 0 \Rightarrow \beta$

$A^{7} - (\beta-1)A^{6} - \beta A^{5} = A^{5}\!\left(A^{2}-(\beta-1)A-\beta I\right)$

$\det A = \beta(-2+2) + 1(2-3) = -1 \neq 0 \ \Rightarrow\ A \text{ is invertible}$

$\det\!\left(A^{5}M\right) = \det(A)^{5}\det M = (-1)^{5}\det M$

$\text{singular} \iff \det M = 0$

A² = β²+3 1 β-2 2β-4 -1 4 3β-4 -1 5 M = A²-(β-1)A-β I = 3 1 -1 -2 -2β 2β+2 -1 -β β+3

$\det M = 3(-4\beta) - 1(-4) + (-1)(0) = -12\beta + 4$

$-12\beta+4 = 0 \Rightarrow \beta = \frac{1}{3}$

$9\beta = 9\cdot\frac{1}{3} = 3$

$\boxed{9\beta = 3}$