Question

If $A = \begin{bmatrix} \alpha^2 & 5 \\ 5 & -\alpha \end{bmatrix}$ and $\det(A^{10}) = 1024$, then $\alpha =$ ?

• A. $-2$
• B. $-1$
• C. $-3$
• D. $0$

Hint:

Use the identity $\det(A^n) = (\det A)^n$ for square matrices.

Answer 1

The Correct Option is C

Let us first find $\det(A)$: \[ \det(A) = (\alpha^2)(-\alpha) - (5)(5) = -\alpha^3 - 25 \] Given $\det(A^{10}) = 1024$, then: \[ \det(A)^{10} = 1024 \Rightarrow \det(A) = \sqrt[10]{1024} = 2 \] So, \[ -\alpha^3 - 25 = 2 \Rightarrow -\alpha^3 = 27 \Rightarrow \alpha = -3 \]