Question:
If \( A = \begin{vmatrix} 4 & k & k \\ 0 & k & k \\ 0 & 0 & k \end{vmatrix} \) and \( \det(A) = 256 \), then \( |k| \) equals:
If \( A = \begin{vmatrix} 4 & k & k \\ 0 & k & k \\ 0 & 0 & k \end{vmatrix} \) and \( \det(A) = 256 \), then \( |k| \) equals:
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When you see a matrix with a "triangle" of zeros, don't expand it manually. Just multiply the diagonal entries. It works for both \( \text{det}(A) \) and finding eigenvalues.
Updated On: May 1, 2026
- \( 4 \)
- \( 5 \)
- \( 6 \)
- \( 7 \)
- \( 8 \)
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The Correct Option is
Solution and Explanation
Concept: The determinant of a triangular matrix (either upper or lower) is simply the product of the elements on its main diagonal.
Step 1: Calculating the determinant.
The given matrix is an upper triangular matrix because all elements below the diagonal are zero. \[ \det(A) = 4 \times k \times k = 4k^2 \]
Step 2: Solving for \( |k| \).
We are given \( \det(A) = 256 \): \[ 4k^2 = 256 \] \[ k^2 = \frac{256}{4} = 64 \] Taking the square root: \[ |k| = \sqrt{64} = 8 \]
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