Question

Let \[ A = \begin{bmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{bmatrix} \] If \[ |A|^2 = 25 \] then \( |\alpha| \) equals:

• A. \( 5^2 \)
• B. \( 1 \)
• C. \( \frac{1}{5} \)
• D. \( 5 \)

Hint:

For triangular matrices: Determinant = product of diagonal entries. Squared determinant questions often reduce to simple algebra.

Answer 1

The Correct Option is B

Concept: For triangular matrices (upper or lower), determinant equals the product of diagonal elements: \[ |A| = \text{product of diagonal entries} \] Step 1: Identify matrix type. The matrix is upper triangular: \[ A = \begin{bmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{bmatrix} \] So: \[ |A| = 5 \cdot \alpha \cdot 5 = 25\alpha \] Step 2: Use given condition. \[ |A|^2 = 25 \] \[ (25\alpha)^2 = 25 \] \[ 625\alpha^2 = 25 \] Step 3: Solve for \( |\alpha| \). \[ \alpha^2 = \frac{25}{625} = \frac{1}{25} \] \[ |\alpha| = \frac{1}{5} \] Correct choice: \( |\alpha| = \frac{1}{5} \)