Question

If \(A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\), then show that \(A^2 - 5A + 7I = O\). Using this, obtain \(A^{-1}\).

Hint:

Using the Cayley-Hamilton theorem to find the inverse of a matrix is a very efficient method. Once you have the characteristic equation in the form of a matrix polynomial that equals zero, you can easily isolate the \(A^{-1}\) term by multiplying the entire equation by \(A^{-1}\).

Answer 1

Step 1: Understanding the Concept: 
We use the Cayley-Hamilton theorem: a square matrix satisfies its own characteristic equation. First, verify the equation \(A^2 - 5A + 7I = O\). Then, manipulate it to find \(A^{-1}\). 
Step 2: Detailed Explanation: 
Part 1: Verify \(A^2 - 5A + 7I = O\) 
Calculate \(A^2\): \[ A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \] Calculate \(5A\): \[ 5A = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} \] Calculate \(7I\): \[ 7I = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \] Substitute: \[ A^2 - 5A + 7I = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \] Part 2: Find \(A^{-1}\) 
Start from \(A^2 - 5A + 7I = O\). Multiply by \(A^{-1}\): \[ A^{-1}(A^2 - 5A + 7I) = O \implies A - 5I + 7A^{-1} = O \] Solve for \(A^{-1}\): \[ 7A^{-1} = 5I - A \implies A^{-1} = \frac{1}{7}(5I - A) \] Substitute values: \[ A^{-1} = \frac{1}{7} \left( \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \right) = \frac{1}{7} \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix} \] Step 3: Final Answer: 
\[ A^{-1} = \frac{1}{7} \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix} \]