If `A=[{:(1,2,2),(2,1,2),(2,2,1):}]`, prove that `A^(2)-4A-5I=O` and hence, obtain `A^(-1)`.

Answer

To solve the problem, we need to prove that \( A^2 - 4A - 5I = O \) and then find \( A^{-1} \).

### Step 1: Define the Matrix A
Given the matrix:
\[ A = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} \]

### Step 2: Calculate \( A^2 \)
To find \( A^2 \), we multiply matrix \( A \) by itself:
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} \]

Calculating each element of \( A^2 \):
- First row:
- \( (1 \cdot 1 + 2 \cdot 2 + 2 \cdot 2) = 1 + 4 + 4 = 9 \)
- \( (1 \cdot 2 + 2 \cdot 1 + 2 \cdot 2) = 2 + 2 + 4 = 8 \)
- \( (1 \cdot 2 + 2 \cdot 2 + 2 \cdot 1) = 2 + 4 + 2 = 8 \)

- Second row:
- \( (2 \cdot 1 + 1 \cdot 2 + 2 \cdot 2) = 2 + 2 + 4 = 8 \)
- \( (2 \cdot 2 + 1 \cdot 1 + 2 \cdot 2) = 4 + 1 + 4 = 9 \)
- \( (2 \cdot 2 + 1 \cdot 2 + 2 \cdot 1) = 4 + 2 + 2 = 8 \)

- Third row:
- \( (2 \cdot 1 + 2 \cdot 2 + 1 \cdot 2) = 2 + 4 + 2 = 8 \)
- \( (2 \cdot 2 + 2 \cdot 1 + 1 \cdot 2) = 4 + 2 + 2 = 8 \)
- \( (2 \cdot 2 + 2 \cdot 2 + 1 \cdot 1) = 4 + 4 + 1 = 9 \)

Thus, we have:
\[ A^2 = \begin{pmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{pmatrix} \]

### Step 3: Calculate \( 4A \)
Now, calculate \( 4A \):
\[ 4A = 4 \cdot \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{pmatrix} \]

### Step 4: Calculate \( 5I \)
Where \( I \) is the identity matrix:
\[ 5I = 5 \cdot \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix} \]

### Step 5: Calculate \( A^2 - 4A - 5I \)
Now, we need to compute \( A^2 - 4A - 5I \):
\[ A^2 - 4A - 5I = \begin{pmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{pmatrix} - \begin{pmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{pmatrix} - \begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix} \]

Calculating this step-by-step:
1. \( A^2 - 4A \):
\[ \begin{pmatrix} 9 - 4 & 8 - 8 & 8 - 8 \\ 8 - 8 & 9 - 4 & 8 - 8 \\ 8 - 8 & 8 - 8 & 9 - 4 \end{pmatrix} = \begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix} \]

2. Now subtract \( 5I \):
\[ \begin{pmatrix} 5 - 5 & 0 - 0 & 0 - 0 \\ 0 - 0 & 5 - 5 & 0 - 0 \\ 0 - 0 & 0 - 0 & 5 - 5 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \]

Thus, we have shown that:
\[ A^2 - 4A - 5I = O \]

### Step 6: Obtain \( A^{-1} \)
From the equation \( A^2 - 4A = 5I \), we can rearrange it as:
\[ A^2 - 4A - 5I = 0 \implies A^2 - 4A = 5I \]

Now, multiplying both sides by \( A^{-1} \):
\[ A A^{-1} - 4I = 5A^{-1} \]
\[ I - 4I = 5A^{-1} \]
\[ -3I = 5A^{-1} \]
\[ A^{-1} = -\frac{1}{5} I \]

Now substituting \( A \):
\[ A^{-1} = \frac{1}{5} \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} - 4 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Calculating \( A^{-1} \):
\[ A^{-1} = \frac{1}{5} \begin{pmatrix} 1 - 4 & 2 & 2 \\ 2 & 1 - 4 & 2 \\ 2 & 2 & 1 - 4 \end{pmatrix} = \frac{1}{5} \begin{pmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{pmatrix} \]

Thus, the final result for \( A^{-1} \) is:
\[ A^{-1} = \begin{pmatrix} -\frac{3}{5} & \frac{2}{5} & \frac{2}{5} \\ \frac{2}{5} & -\frac{3}{5} & \frac{2}{5} \\ \frac{2}{5} & \frac{2}{5} & -\frac{3}{5} \end{pmatrix} \]