Question

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

Answer 1

Compute the characteristic polynomial of
`A`.

`p(\lambda) = \det(A - \lambda I) \\\\ ~~~~~~~~ = \begin{vmatrix} \beta - \lambda & 0 & 1 \\ 2 & 1 - \lambda & -2 \\ 3 & 1 & -2-\lambda \end{vmatrix} \\\\ ~~~~~~~~ = (\beta-\lambda)\begin{vmatrix}1-\lambda&-2\\1&-2-\lambda\end{vmatrix} + \begin{vmatrix}2&1-\lambda \\ 3 & 1 \end{vmatrix} \\\\ ~~~~~~~~ = -\lambda^3 + (\beta-1) \lambda^2 + (\beta + 3) \lambda - 1`

By the Cayley-Hamilton theorem,

`p(A) = -A^3 + (\beta - 1) A^2 + (\beta + 3) A - I = 0`

Note that

`A^7 - (\beta - 1) A^6 - \beta A^5 \\\\ ~~~~~~~~ = -(-A^3 + (\beta - 1) A^2 + \beta A) A^4 \\\\ ~~~~~~~~ = -(\underbrace{-A^3 + (\beta - 1) A^2 + (\beta + 3) A - I}_(p(A)=0) - 3A + I) A^4 \\\\ ~~~~~~~~ = (3A - I) A^4`

Recall that
`\det(AB) = \det(A) \det(B)`. Note that
`A` is not singular, and hence
`A^n` is not singular for any natural
`n`.

`\det(A) = \begin{vmatrix} \beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2 \end{vmatrix} \\\\ ~~~~~~~~ = \beta \begin{vmatrix}1&-2\\1&-2\end{vmatrix} + \begin{vmatrix}2&1\\3&1\end{vmatrix} \\\\ ~~~~~~~~ = -1 \\eq 0`

Then
`3A-I` must be singular. So

`\det(3A-I) = \begin{vmatrix} 3\beta - 1 & 0 & 3 \\ 6 & 2 & -6 \\ 9 & 3 & -7 \end{vmatrix} \\\\ ~~~~~~~~ = (3\beta-1) \begin{vmatrix}2&-6\\3&-7\end{vmatrix} + 3 \begin{vmatrix}6&2\\9&3\end{vmatrix} \\\\ ~~~~~~~~ = 4 (3\beta-1)`

`4(3\beta-1) = 0 \implies \beta = \frac13 \implies 9\beta = \boxed{3}`