Question

For matrix \( A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} \) show that \( A^3 - 6A^2 + 5A + 11I = 0 \) and with the help of this find \( A^{-1} \).

Hint:

Using the Cayley-Hamilton theorem to find the inverse is often computationally faster and less prone to sign errors than the traditional method of finding the determinant and the adjugate matrix, especially for 3x3 matrices where you have already calculated \( A^2 \).

Answer 1

Step 1: Understanding the Concept:
This problem has two parts. First, we need to verify that the matrix \( A \) satisfies the given polynomial equation, which is its characteristic equation (Cayley-Hamilton Theorem). Second, we use this verified equation to find the inverse of matrix \( A \) without using the adjugate method.
Step 2: Verifying the Matrix Equation \( A^3 - 6A^2 + 5A + 11I = 0 \):
First, we calculate \( A^2 \) and \( A^3 \).
\[ A^2 = A \cdot A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix} \] \[ A^3 = A^2 \cdot A = \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} = \begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix} \] Now, substitute \( A^3, A^2, A \) and \( I \) into the equation: \[ A^3 - 6A^2 + 5A + 11I = \begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix} - 6 \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix} + 5 \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} + 11 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \] \[ = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = 0 \] Thus, the equation is verified.
Step 3: Finding \( A^{-1} \) using the equation:
Starting with the verified equation: \[ A^3 - 6A^2 + 5A + 11I = 0 \] Multiply the entire equation by \( A^{-1} \): \[ A^{-1}(A^3 - 6A^2 + 5A + 11I) = 0 \] \[ A^2 - 6A + 5I + 11A^{-1} = 0 \] Isolating \( A^{-1} \): \[ 11A^{-1} = -A^2 + 6A - 5I \] \[ A^{-1} = \frac{1}{11}(-A^2 + 6A - 5I) \] Substitute the matrices: \[ A^{-1} = \frac{1}{11} \left( -\begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix} + 6\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} - 5\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \right) \] \[ A^{-1} = \frac{1}{11} \begin{bmatrix} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{bmatrix} \] Step 4: Final Answer:
The matrix equation is verified, and the inverse is: \[ A^{-1} = \frac{1}{11} \begin{bmatrix} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{bmatrix} \]