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

Since $A$ is upper triangular, its determinant is the product of diagonal entries. $$|A| = 5 \cdot \alpha \cdot 5 = 25\alpha$$ Using $|A^2| = |A|^2$, $$|A^2| = (25\alpha)^2 = 625\alpha^2$$ Given $|A^2| = 25$, $$625\alpha^2 = 25$$ $$\alpha^2 = \frac{1}{25}$$ $$|\alpha| = \frac{1}{5}$$ This matches option $\text{B}$.