Question

If a matrix \(A\) satisfies the equation \[ A^3-6A^2+11A-6I=0, \] then \(A^{-1}\) can be

• A. \(\dfrac{1}{4}I\)
• B. \(4I\)
• C. \(3I\)
• D. \(\dfrac{1}{3}I\)

Hint:

When a matrix satisfies a polynomial equation, factor the polynomial first. The roots of the polynomial give the possible eigenvalues of the matrix, and the reciprocals of those eigenvalues give the eigenvalues of the inverse matrix.

Answer 1

The Correct Option is D

Step 1: Factorize the given matrix equation.
Given, \[ A^3-6A^2+11A-6I=0 \] Observe that \[ x^3-6x^2+11x-6 \] factorizes as \[ (x-1)(x-2)(x-3). \] Hence, \[ (A-I)(A-2I)(A-3I)=0. \]

Step 2: Find the possible eigenvalues of \(A\).
From \[ (A-I)(A-2I)(A-3I)=0, \] every eigenvalue \(\lambda\) of \(A\) must satisfy \[ (\lambda-1)(\lambda-2)(\lambda-3)=0. \] Therefore, \[ \lambda=1,\;2,\;3. \] Since none of these eigenvalues is zero, \(A\) is invertible.

Step 3: Determine the possible eigenvalues of \(A^{-1}\).
If \(\lambda\) is an eigenvalue of \(A\), then \[ \frac{1}{\lambda} \] is an eigenvalue of \(A^{-1}\). Thus the possible eigenvalues of \(A^{-1}\) are \[ 1,\quad \frac{1}{2},\quad \frac{1}{3}. \]

Step 4: Compare with the given options.
The options are \[ \frac{1}{4}I,\quad 4I,\quad 3I,\quad \frac{1}{3}I. \] Only \[ \frac{1}{3}I \] corresponds to a possible value of \(A^{-1}\). Indeed, if \[ A=3I, \] then \[ A^3-6A^2+11A-6I = 27I-54I+33I-6I = 0, \] so \(A=3I\) satisfies the given equation. Hence, \[ A^{-1}=\frac{1}{3}I. \]

Step 5: Final conclusion.
Therefore, \[ \boxed{A^{-1}=\frac{1}{3}I} \] and the correct option is \[ \boxed{(4)}. \]