Question:
If \( A = \begin{bmatrix} 5 & 5\alpha & \alpha 0 & \alpha & 5\alpha 0 & 0 & 5 \end{bmatrix} \) and \( \det(A^2) = 25 \), then \( |\alpha| = \)
If \( A = \begin{bmatrix} 5 & 5\alpha & \alpha 0 & \alpha & 5\alpha 0 & 0 & 5 \end{bmatrix} \) and \( \det(A^2) = 25 \), then \( |\alpha| = \)
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Whenever you spot a matrix with a triangle of zeros below or above the main diagonal, do not waste time expanding it row by row! Instantly multiply the diagonal elements together to find its determinant.
Updated On: May 20, 2026
- \( 5 \)
- \( 5^2 \)
- \( 1 \)
- \( 1 / 5 \)
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The Correct Option is D
Solution and Explanation
Concept:
This question relies on two fundamental properties of determinants:
• Determinant of a Power: For any square matrix \(A\) and positive integer \(n\), the determinant of the matrix raised to a power equals the power of its determinant: \[ |A^n| = |A|^n \]
• Determinant of a Triangular Matrix: If a square matrix is upper triangular (all entries below the main diagonal are zero), its determinant is simply the product of its principal diagonal entries: \[ |A| = \prod_{i=1}^{n} a_{ii} \]
Step 1: Evaluating the determinant of matrix \(A\) using its structural property.
Notice that matrix \(A\) is an upper triangular matrix because all elements below the principal diagonal are zero: \[ A = \begin{bmatrix} 5 & 5\alpha & \alpha 0 & \alpha & 5\alpha 0 & 0 & 5 \end{bmatrix} \] Therefore, the determinant \(|A|\) is the product of its diagonal elements: \[ |A| = 5 \cdot \alpha \cdot 5 = 25\alpha \quad \cdots (1) \]
Step 2: Applying the determinant power property to the given equation.
We are given that: \[ \det(A^2) = 25 \] Using the property \(|A^2| = |A|^2\), we can rewrite this as: \[ |A|^2 = 25 \quad \cdots (2) \]
Step 3: Substituting equation (1) into equation (2) and solving for \(|\alpha|\).
Substitute \(|A| = 25\alpha\) into the equation: \[ (25\alpha)^2 = 25 \] \[ 625\alpha^2 = 25 \quad \Rightarrow \quad \alpha^2 = \frac{25}{625} = \frac{1}{25} \] Taking the square root on both sides to find the absolute value \(|\alpha|\): \[ |\alpha| = \sqrt{\frac{1}{25}} = \frac{1}{5} \] This precisely matches option (D).
• Determinant of a Power: For any square matrix \(A\) and positive integer \(n\), the determinant of the matrix raised to a power equals the power of its determinant: \[ |A^n| = |A|^n \]
• Determinant of a Triangular Matrix: If a square matrix is upper triangular (all entries below the main diagonal are zero), its determinant is simply the product of its principal diagonal entries: \[ |A| = \prod_{i=1}^{n} a_{ii} \]
Step 1: Evaluating the determinant of matrix \(A\) using its structural property.
Notice that matrix \(A\) is an upper triangular matrix because all elements below the principal diagonal are zero: \[ A = \begin{bmatrix} 5 & 5\alpha & \alpha 0 & \alpha & 5\alpha 0 & 0 & 5 \end{bmatrix} \] Therefore, the determinant \(|A|\) is the product of its diagonal elements: \[ |A| = 5 \cdot \alpha \cdot 5 = 25\alpha \quad \cdots (1) \]
Step 2: Applying the determinant power property to the given equation.
We are given that: \[ \det(A^2) = 25 \] Using the property \(|A^2| = |A|^2\), we can rewrite this as: \[ |A|^2 = 25 \quad \cdots (2) \]
Step 3: Substituting equation (1) into equation (2) and solving for \(|\alpha|\).
Substitute \(|A| = 25\alpha\) into the equation: \[ (25\alpha)^2 = 25 \] \[ 625\alpha^2 = 25 \quad \Rightarrow \quad \alpha^2 = \frac{25}{625} = \frac{1}{25} \] Taking the square root on both sides to find the absolute value \(|\alpha|\): \[ |\alpha| = \sqrt{\frac{1}{25}} = \frac{1}{5} \] This precisely matches option (D).
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