Question

If \( A = \begin{bmatrix} 3 & 2 \\ 4 & 7 \end{bmatrix} \) and \( f(x) = x^2 - 10x + 13 \), then show that \( f(A) = O \) and using this result find \( A^{-1} \).

Hint:

When finding the inverse from a polynomial \( aA^2 + bA + cI = O \), the inverse is always given by \( A^{-1} = -\frac{1}{c}(aA + bI) \). This avoids calculating the adjoint matrix.

Answer 1

Step 1: Understanding the Concept:
The Cayley-Hamilton theorem or direct calculation can show \( f(A) = O \). To find \( A^{-1} \), we manipulate the equation \( A^2 - 10A + 13I = O \).
Step 2: Detailed Explanation:
Calculate \( A^2 \):
\[ A^2 = \begin{bmatrix} 3 & 2 \\ 4 & 7 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 4 & 7 \end{bmatrix} = \begin{bmatrix} 9+8 & 6+14 \\ 12+28 & 8+49 \end{bmatrix} = \begin{bmatrix} 17 & 20 \\ 40 & 57 \end{bmatrix} \] Verify \( f(A) = O \):
\[ f(A) = \begin{bmatrix} 17 & 20 \\ 40 & 57 \end{bmatrix} - 10 \begin{bmatrix} 3 & 2 \\ 4 & 7 \end{bmatrix} + 13 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] \[ f(A) = \begin{bmatrix} 17-30+13 & 20-20+0 \\ 40-40+0 & 57-70+13 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \] To find \( A^{-1} \):
\[ A^2 - 10A + 13I = O \implies 13I = 10A - A^2 \] Pre-multiply by \( A^{-1} \):
\[ 13A^{-1} = 10I - A = 10 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 3 & 2 \\ 4 & 7 \end{bmatrix} = \begin{bmatrix} 7 & -2 \\ -4 & 3 \end{bmatrix} \] \[ A^{-1} = \frac{1}{13} \begin{bmatrix} 7 & -2 \\ -4 & 3 \end{bmatrix} \].
Step 3: Final Answer:
The inverse is \( A^{-1} = \frac{1}{13} \begin{bmatrix} 7 & -2 \\ -4 & 3 \end{bmatrix} \).