If `A=[(2,-1,1),(-1,2,-1),(1,-1,2)]` show that `A^(2)-5A+4I=0`
Hence find `A^(-1)`

Answer

To solve the problem, we need to show that \( A^2 - 5A + 4I = 0 \) for the given matrix \( A \) and then find \( A^{-1} \).

### Step 1: Calculate \( A^2 \)

Given:
\[ A = \begin{pmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{pmatrix} \]

To find \( A^2 \), we perform the matrix multiplication \( A \times A \).

\[ A^2 = A \times A = \begin{pmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{pmatrix} \times \begin{pmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{pmatrix} \]

Calculating the elements of \( A^2 \):

- First row:
- \( (2 \cdot 2) + (-1 \cdot -1) + (1 \cdot 1) = 4 + 1 + 1 = 6 \)
- \( (2 \cdot -1) + (-1 \cdot 2) + (1 \cdot -1) = -2 - 2 - 1 = -5 \)
- \( (2 \cdot 1) + (-1 \cdot -1) + (1 \cdot 2) = 2 + 1 + 2 = 5 \)

- Second row:
- \( (-1 \cdot 2) + (2 \cdot -1) + (-1 \cdot 1) = -2 - 2 - 1 = -5 \)
- \( (-1 \cdot -1) + (2 \cdot 2) + (-1 \cdot -1) = 1 + 4 + 1 = 6 \)
- \( (-1 \cdot 1) + (2 \cdot -1) + (-1 \cdot 2) = -1 - 2 - 2 = -5 \)

- Third row:
- \( (1 \cdot 2) + (-1 \cdot -1) + (2 \cdot 1) = 2 + 1 + 2 = 5 \)
- \( (1 \cdot -1) + (-1 \cdot 2) + (2 \cdot -1) = -1 - 2 - 2 = -5 \)
- \( (1 \cdot 1) + (-1 \cdot -1) + (2 \cdot 2) = 1 + 1 + 4 = 6 \)

Thus, we have:
\[ A^2 = \begin{pmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{pmatrix} \]

### Step 2: Calculate \( 5A \)

Now we calculate \( 5A \):
\[ 5A = 5 \times \begin{pmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{pmatrix} = \begin{pmatrix} 10 & -5 & 5 \\ -5 & 10 & -5 \\ 5 & -5 & 10 \end{pmatrix} \]

### Step 3: Calculate \( 4I \)

The identity matrix \( I \) for a 3x3 matrix is:
\[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
Thus,
\[ 4I = 4 \times \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix} \]

### Step 4: Calculate \( A^2 - 5A + 4I \)

Now we combine the results:
\[ A^2 - 5A + 4I = \begin{pmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{pmatrix} - \begin{pmatrix} 10 & -5 & 5 \\ -5 & 10 & -5 \\ 5 & -5 & 10 \end{pmatrix} + \begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix} \]

Calculating each element:
- First row:
- \( 6 - 10 + 4 = 0 \)
- \( -5 + 5 + 0 = 0 \)
- \( 5 - 5 + 0 = 0 \)

- Second row:
- \( -5 + 5 + 0 = 0 \)
- \( 6 - 10 + 4 = 0 \)
- \( -5 + 5 + 0 = 0 \)

- Third row:
- \( 5 - 5 + 0 = 0 \)
- \( -5 + 5 + 0 = 0 \)
- \( 6 - 10 + 4 = 0 \)

Thus, we have:
\[ A^2 - 5A + 4I = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \]

This proves that \( A^2 - 5A + 4I = 0 \).

### Step 5: Find \( A^{-1} \)

From the equation \( A^2 - 5A + 4I = 0 \), we can rearrange it to find \( A^{-1} \):
\[ A^2 = 5A - 4I \]
Multiplying both sides by \( A^{-1} \):
\[ A = 5I - 4A^{-1} \]
Rearranging gives:
\[ 4A^{-1} = 5I - A \]
Thus,
\[ A^{-1} = \frac{1}{4}(5I - A) \]

Calculating \( 5I - A \):
\[ 5I - A = \begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix} - \begin{pmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{pmatrix} = \begin{pmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{pmatrix} \]

Thus,
\[ A^{-1} = \frac{1}{4} \begin{pmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{pmatrix} = \begin{pmatrix} \frac{3}{4} & \frac{1}{4} & -\frac{1}{4} \\ \frac{1}{4} & \frac{3}{4} & \frac{1}{4} \\ -\frac{1}{4} & \frac{1}{4} & \frac{3}{4} \end{pmatrix} \]

### Final Answer
\[ A^{-1} = \begin{pmatrix} \frac{3}{4} & \frac{1}{4} & -\frac{1}{4} \\ \frac{1}{4} & \frac{3}{4} & \frac{1}{4} \\ -\frac{1}{4} & \frac{1}{4} & \frac{3}{4} \end{pmatrix} \]