Question

Let \(A\) and \(B\) be two \(3\times3\) matrices such that \[ A^2-4A+3I=O \] and \[ B=A^{-1}+2A. \] Find the determinant of \(B\) if \(\det(A)=3\).

• A. \(9\)
• B. \(27\)
• C. \(81\)
• D. \(3\)

Hint:

Whenever a matrix satisfies a polynomial equation, immediately think of eigenvalues and Cayley--Hamilton type simplifications.

Answer 1

The Correct Option is B

Concept: Higher-order matrix equations are simplified using factorization identities and inverse matrix properties. Determinants of matrix products and scalar multiples are then used carefully.

Step 1: Using the given matrix equation.
Given: \[ A^2-4A+3I=O \] Factorizing: \[ (A-I)(A-3I)=O \] Since \(A\) is invertible (\(\det(A)=3\neq0\)), multiply by \(A^{-1}\): \[ A-4I+3A^{-1}=O \] Thus, \[ 3A^{-1}=4I-A \] \[ A^{-1}=\frac{4I-A}{3} \]

Step 2: Finding matrix \(B\).
\[ B=A^{-1}+2A \] Substituting: \[ B=\frac{4I-A}{3}+2A \] \[ B=\frac{4I-A+6A}{3} \] \[ B=\frac{4I+5A}{3} \]

Step 3: Using eigenvalue relation.
From \[ A^2-4A+3I=0 \] Characteristic roots satisfy: \[ \lambda^2-4\lambda+3=0 \] \[ (\lambda-1)(\lambda-3)=0 \] Hence eigenvalues of \(A\) are \(1\) and \(3\). Since \[ \det(A)=3 \] possible eigenvalues are: \[ 1,1,3 \] Now eigenvalues of \(4I+5A\) are: \[ 4+5(1)=9,\quad 9,\quad 4+5(3)=19 \] Thus, \[ \det(4I+5A)=9\times9\times19 \] Since \[ B=\frac{1}{3}(4I+5A) \] and for a \(3\times3\) matrix, \[ \det\left(\frac{1}{3}M\right)=\frac{1}{27}\det(M) \] Therefore, \[ \det(B)=\frac{1}{27}(9\times9\times19) \] \[ \det(B)=27 \]