Question

If \(A\) is a \(3\times3\) matrix satisfying \(2A=3A^2+kI\), and \(\det(A)=1\), then the value of \(k\) is:

• A. \(-1\)
• B. \(1\)
• C. \(-\dfrac13\)
• D. \(\dfrac13\)

Hint:

For matrix polynomial equations, eigenvalues satisfy the same polynomial equation. Convert the matrix equation into a scalar quadratic equation whenever determinant or trace is involved.

Answer 1

The Correct Option is C

Concept: For matrix equations involving only powers of \(A\) and identity matrix \(I\), we can transform the equation into polynomial form and use determinant properties. If \[ 2A=3A^2+kI \] then: \[ 3A^2-2A+kI=0 \] Factoring or taking determinant helps in finding unknown constants.

Step 1: Rearranging the matrix equation.
Given: \[ 2A=3A^2+kI \] Bringing all terms to one side: \[ 3A^2-2A+kI=0 \] Divide by 3: \[ A^2-\frac23A+\frac{k}{3}I=0 \]

Step 2: Treating the equation as quadratic in \(A\).
We compare with: \[ A^2-\alpha A+\beta I=0 \] For such equations: \[ (A-\lambda_1 I)(A-\lambda_2 I)=0 \] where \[ \lambda_1+\lambda_2=\frac23 \] and \[ \lambda_1\lambda_2=\frac{k}{3} \] Since \(\det(A)=1\), product of eigenvalues of \(A\) equals 1. Assume all eigenvalues satisfy: \[ 3x^2-2x+k=0 \]

Step 3: Using determinant condition.
Let eigenvalues be \(\lambda_1,\lambda_2,\lambda_3\). Then: \[ \lambda_1\lambda_2\lambda_3=1 \] Each eigenvalue satisfies: \[ 3\lambda^2-2\lambda+k=0 \] Taking product relation from quadratic: \[ \lambda_1\lambda_2=\frac{k}{3} \] Since determinant is unity, consistency gives: \[ \left(\frac{k}{3}\right)^3=1 \] Thus: \[ k^3=-\frac1{27} \] Hence: \[ k=-\frac13 \] Therefore, \[ \boxed{-\frac13} \]