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

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(C) Given the matrix $A = \begin{bmatrix} 5 & 5\alpha & \alpha \ 0 & \alpha & 5\alpha \ 0 & 0 & 5 \end{bmatrix}$.Since $A$ is an upper triangular ma

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$\frac{1}{5}$

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(C) Given the matrix $A = \begin{bmatrix} 5 & 5\alpha & \alpha \ 0 & \alpha & 5\alpha \ 0 & 0 & 5 \end{bmatrix}$.Since $A$ is an upper triangular matrix, its determinant $|A|$ is the product of its diagonal elements.$|A| = 5 	imes \alpha 	imes 5 = 25\alpha$.Given that $|A|^2 = 25$, we substitute the value of $|A|$:$(25\alpha)^2 = 25$.$625\alpha^2 = 25$.$\alpha^2 = \frac{25}{625} = \frac{1}{25}$.Taking the square root on both sides, we get $|\alpha| = \sqrt{\frac{1}{25}} = \frac{1}{5}$.

Let $A = \begin{bmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{bmatrix}$. If $|A|^2 = 25$, then $|\alpha|$ is equal to:

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