Question

A is a, \(n \times n\) matrix of real numbers and \(A^3 - 3A^2 + 4A - 6I = 0\), where I is a, \(n \times n\) unit matrix. If \(A^{-1}\) exists, then

• A. \(A^{-1} = A - I\)
• B. \(A^{-1} = A + 6I\)
• C. \(A^{-1} = 3A - 6I\)
• D. \(A^{-1} = \frac{1}{6}(A^2 - 3A + 4I)\)

Hint:

Whenever a matrix \(A\) satisfies a polynomial equation, you can find its inverse (if it exists) by rearranging the equation to the form \(A . P(A) = kI\), where \(P(A)\) is some polynomial in \(A\) and \(k\) is a non-zero scalar. Then, \(A^{-1} = \frac{1}{k}P(A)\). The inverse exists if and only if the constant term in the polynomial is non-zero.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The problem provides a polynomial equation that the matrix \(A\) satisfies. According to the Cayley-Hamilton theorem, every square matrix satisfies its own characteristic equation. We can use the given equation to find an expression for the inverse of \(A\), \(A^{-1}\), by manipulating the equation algebraically.

Step 2: Key Formula or Approach:
1. Start with the given matrix polynomial equation.
2. Isolate the term containing the identity matrix, \(I\).
3. Pre-multiply (or post-multiply) the entire equation by \(A^{-1}\).
4. Simplify the resulting equation to solve for \(A^{-1}\).

Step 3: Detailed Explanation:
We are given the equation: \[ A^3 - 3A^2 + 4A - 6I = 0 \] To find \(A^{-1}\), we can isolate the term with the identity matrix \(I\): \[ A^3 - 3A^2 + 4A = 6I \] Since it is given that \(A^{-1}\) exists, we can pre-multiply both sides of the equation by \(A^{-1}\): \[ A^{-1}(A^3 - 3A^2 + 4A) = A^{-1}(6I) \] Using the distributive property of matrix multiplication: \[ A^{-1}A^3 - A^{-1}(3A^2) + A^{-1}(4A) = 6A^{-1}I \] Using the properties \(A^{-1}A = I\) and \(A^{-1}I = A^{-1}\): \[ (A^{-1}A)A^2 - 3(A^{-1}A)A + 4(A^{-1}A) = 6A^{-1} \] \[ IA^2 - 3IA + 4I = 6A^{-1} \] \[ A^2 - 3A + 4I = 6A^{-1} \] Finally, we can solve for \(A^{-1}\) by dividing by 6: \[ A^{-1} = \frac{1}{6}(A^2 - 3A + 4I) \]
Step 4: Final Answer:
The expression for \(A^{-1}\) matches option (D).