Question:
Consider the matrix \(A=\begin{bmatrix}2& 3x& y\end{bmatrix}\). If the eigen values of \(A\) are \(4\) and \(8\), then
Consider the matrix \(A=\begin{bmatrix}2& 3x& y\end{bmatrix}\). If the eigen values of \(A\) are \(4\) and \(8\), then
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For a \(2\times2\) matrix, sum of eigenvalues equals trace and product equals determinant.
Updated On: Jul 3, 2026
- \(x=4,\ y=10\)
- \(x=5,\ y=8\)
- \(x=-3,\ y=9\)
- \(x=-4,\ y=10\)
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The Correct Option is D
Solution and Explanation
Concept:
For a square matrix, the sum of eigenvalues is equal to the trace, and the product of eigenvalues is equal to the determinant. Given matrix: \[ A=\begin{bmatrix}2& 3x& y\end{bmatrix} \] Eigenvalues are \[ 4,\quad 8 \]
Step 1: Use trace. Sum of eigenvalues: \[ 4+8=12 \] Trace of \(A\): \[ 2+y \] Therefore, \[ 2+y=12 \] \[ y=10 \]
Step 2: Use determinant. Product of eigenvalues: \[ 4\cdot 8=32 \] Determinant of \(A\): \[ \det A=2y-3x \] Substitute \(y=10\): \[ 2(10)-3x=32 \] \[ 20-3x=32 \] \[ -3x=12 \] \[ x=-4 \]
Step 3: Final answer. \[ \boxed{x=-4,\ y=10} \]
For a square matrix, the sum of eigenvalues is equal to the trace, and the product of eigenvalues is equal to the determinant. Given matrix: \[ A=\begin{bmatrix}2& 3x& y\end{bmatrix} \] Eigenvalues are \[ 4,\quad 8 \]
Step 1: Use trace. Sum of eigenvalues: \[ 4+8=12 \] Trace of \(A\): \[ 2+y \] Therefore, \[ 2+y=12 \] \[ y=10 \]
Step 2: Use determinant. Product of eigenvalues: \[ 4\cdot 8=32 \] Determinant of \(A\): \[ \det A=2y-3x \] Substitute \(y=10\): \[ 2(10)-3x=32 \] \[ 20-3x=32 \] \[ -3x=12 \] \[ x=-4 \]
Step 3: Final answer. \[ \boxed{x=-4,\ y=10} \]
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