Question:medium
Let \( A = \begin{pmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{pmatrix} \). If \( |A|^2 = 25 \), then \( |\alpha| \) equals to
Let \( A = \begin{pmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{pmatrix} \). If \( |A|^2 = 25 \), then \( |\alpha| \) equals to
Show Hint
For an upper triangular matrix, the determinant is the product of the diagonal elements. Use the given condition on \( |A|^2 \) to solve for unknowns.
Updated On: Nov 28, 2025
- \( 5^2 \)
- 1
- \( \frac{1}{5} \)
- 5
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