If A = begin{bmatrix} 5 & 5alpha & alpha 0 & | vedclass

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(D) Given $\operatorname{det}(A^2) = 25$. Using the property $\operatorname{det}(A^n) = (\operatorname{det}(A))^n$, we have $(\operatorname{det}(A))^2

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$\frac{1}{5}$

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(D) Given $\operatorname{det}(A^2) = 25$. Using the property $\operatorname{det}(A^n) = (\operatorname{det}(A))^n$, we have $(\operatorname{det}(A))^2 = 25$. Now, calculate the determinant of matrix $A$: $|A| = \begin{vmatrix} 5 & 5\alpha & \alpha \ 0 & \alpha & 5\alpha \ 0 & 0 & 5 \end{vmatrix}$. Since this is an upper triangular matrix, the determinant is the product of the diagonal elements: $|A| = 5 	imes \alpha 	imes 5 = 25\alpha$. Substituting this into the equation: $(25\alpha)^2 = 25$. $625\alpha^2 = 25$. $\alpha^2 = \frac{25}{625} = \frac{1}{25}$. Taking the square root on both sides: $|\alpha| = \sqrt{\frac{1}{25}} = \frac{1}{5}$.

If $A = \begin{bmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{bmatrix}$ and $\operatorname{det}(A^2) = 25$, then $|\alpha| = $

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