If A = begin{bmatrix} alpha^2 & 5 5 & -alpha | vedclass

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(C) Given $A = \begin{bmatrix} \alpha^2 & 5 \ 5 & -\alpha \end{bmatrix}$.The determinant of $A$ is $\det(A) = (\alpha^2)(-\alpha) - (5)(5) = -\alpha^

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$-3$

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(C) Given $A = \begin{bmatrix} \alpha^2 & 5 \ 5 & -\alpha \end{bmatrix}$.The determinant of $A$ is $\det(A) = (\alpha^2)(-\alpha) - (5)(5) = -\alpha^3 - 25$.We are given $\det(A^{10}) = 1024$.Using the property $\det(A^n) = (\det A)^n$,we have $(\det A)^{10} = 1024$.Since $1024 = 2^{10}$,we have $(\det A)^{10} = 2^{10}$,which implies $\det A = 2$ or $\det A = -2$.Case $1$: $-\alpha^3 - 25 = 2 \Rightarrow -\alpha^3 = 27 \Rightarrow \alpha^3 = -27 \Rightarrow \alpha = -3$.Case $2$: $-\alpha^3 - 25 = -2 \Rightarrow -\alpha^3 = 23 \Rightarrow \alpha^3 = -23 \Rightarrow \alpha = -\sqrt[3]{23}$.Comparing with the given options,$\alpha = -3$ is the correct value.

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

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