If `A=[{:(1,2,2),(2,1,2),(2,2,1):}]`, then show that `A^(2)-4A-5I_(3)=0`. Hemce find `A^(-1)`.

Answer

To solve the problem, we need to show that \( A^2 - 4A - 5I_3 = 0 \) 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 \( A \) 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, we 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 \( 5I_3 \)
The identity matrix \( I_3 \) is:
\[ I_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
Thus,
\[ 5I_3 = 5 \cdot I_3 = \begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix} \]

### Step 5: Calculate \( A^2 - 4A - 5I_3 \)
Now we can substitute our results into the equation:
\[ A^2 - 4A - 5I_3 = \begin{pmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{pmatrix} - \begin{pmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{pmatrix} - \begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix} \]

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

2. Now subtract \( 5I_3 \):
\[ \begin{pmatrix} 5 - 5 & 0 - 0 & 0 - 0 \\ 0 - 0 & 5 - 5 & 0 - 0 \\ 0 - 0 & 0 - 0 & 5 - 5 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \]

Thus, we have shown that:
\[ A^2 - 4A - 5I_3 = 0 \]

### Step 6: Finding \( A^{-1} \)
From the equation \( A^2 - 4A - 5I_3 = 0 \), we can rearrange it to find \( A^{-1} \):
\[ A^2 - 4A = 5I_3 \]
Multiplying both sides by \( A^{-1} \):
\[ A - 4I_3 = 5A^{-1} \]
Thus,
\[ A^{-1} = \frac{1}{5}(A - 4I_3) \]

Substituting \( A \) and \( I_3 \):
\[ A - 4I_3 = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} - \begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix} = \begin{pmatrix} 1 - 4 & 2 & 2 \\ 2 & 1 - 4 & 2 \\ 2 & 2 & 1 - 4 \end{pmatrix} = \begin{pmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{pmatrix} \]

Finally, we have:
\[ A^{-1} = \frac{1}{5} \begin{pmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{pmatrix} \]

### Final Answer
Thus, the inverse of \( A \) is:
\[ A^{-1} = \begin{pmatrix} -\frac{3}{5} & \frac{2}{5} & \frac{2}{5} \\ \frac{2}{5} & -\frac{3}{5} & \frac{2}{5} \\ \frac{2}{5} & \frac{2}{5} & -\frac{3}{5} \end{pmatrix} \]