Question

If \[ A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}, \] prove that \[ A^2 - 4A - 5I = 0. \] Hence, find \( A^{-1} \).

Hint:

To find the inverse of a matrix, use properties of matrix operations such as \( A^2 = 4A + 5I \). This can simplify the process.

Answer 1

Step 1: Compute A².

\[ A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \] \[ A^2 = A \cdot A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \] \[ A^2 = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} \] Step 2: Compute 4A and 5I.

\[ 4A = \begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix} \] \[ 5I = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} \] \[ 4A + 5I = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} \] Step 3: Verify the given identity.

\[ A^2 - 4A - 5I = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} - \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} = 0 \] Hence, \[ A^2 - 4A - 5I = 0 \] Step 4: Find A⁻¹.

From the identity, \[ A^2 - 4A - 5I = 0 \] Rearranging, \[ A(A - 4I) = 5I \] Multiplying both sides by \( \frac{1}{5} \), \[ A^{-1} = \frac{1}{5}(A - 4I) \] \[ A - 4I = \begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix} \] Final Answer: \[ \boxed{ A^{-1} = \frac{1}{5} \begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix} } \]