If A = begin{bmatrix} alpha & 2 2 & alpha en | vedclass

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(C) Given,$A = \begin{bmatrix} \alpha & 2 \ 2 & \alpha \end{bmatrix}$.First,we calculate the determinant of $A$:$|A| = \begin{vmatrix} \alpha & 2 \

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$\pm 3$

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(C) Given,$A = \begin{bmatrix} \alpha & 2 \ 2 & \alpha \end{bmatrix}$.First,we calculate the determinant of $A$:$|A| = \begin{vmatrix} \alpha & 2 \ 2 & \alpha \end{vmatrix} = \alpha^{2} - 4 \quad (i)$.We are given that $|A^{3}| = 125$.Using the property of determinants $|A^{n}| = |A|^{n}$,we have:$|A|^{3} = 125$.Taking the cube root on both sides:$|A| = \sqrt[3]{125} = 5$.Now,substitute $|A| = 5$ into equation $(i)$:$5 = \alpha^{2} - 4$.$\alpha^{2} = 5 + 4 = 9$.Taking the square root on both sides:$\alpha = \pm 3$.

If $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$ and $|A^{3}| = 125$,then $\alpha$ is equal to

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