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: Concept Overview:
Given a polynomial equation satisfied by matrix \(A\), we aim to find \(A^{-1}\). The Cayley-Hamilton theorem states that every square matrix satisfies its characteristic equation. We leverage this to express \(A^{-1}\) through algebraic manipulation of the provided equation.

Step 2: Methodology:
1. Begin with the given polynomial equation in matrix form.
2. Isolate the identity matrix term, \(I\).
3. Multiply the entire equation by \(A^{-1}\) (either pre- or post-multiplication).
4. Simplify the resulting equation to determine \(A^{-1}\).

Step 3: Step-by-Step Solution:
Starting with the equation:\[ A^3 - 3A^2 + 4A - 6I = 0 \]Isolate the term with \(I\):\[ A^3 - 3A^2 + 4A = 6I \]Pre-multiply both sides by \(A^{-1}\) (since \(A^{-1}\) exists):\[ A^{-1}(A^3 - 3A^2 + 4A) = A^{-1}(6I) \]Apply the distributive property:\[ A^{-1}A^3 - A^{-1}(3A^2) + A^{-1}(4A) = 6A^{-1}I \]Using \(A^{-1}A = I\) and \(A^{-1}I = A^{-1}\):\[ (A^{-1}A)A^2 - 3(A^{-1}A)A + 4(A^{-1}A) = 6A^{-1} \]\[ IA^2 - 3IA + 4I = 6A^{-1} \]\[ A^2 - 3A + 4I = 6A^{-1} \]Solve for \(A^{-1}\):\[ A^{-1} = \frac{1}{6}(A^2 - 3A + 4I) \]
Step 4: Conclusion:
The derived expression for \(A^{-1}\) corresponds to option (D).