If `A=[[1, 2, 2], [2, 1, 2], [2, 2, 1]] and A^(-1)` exist and not equal to 0, then `(A^(2)-4A)A^(-1)=`

Options

A. `[[-3, 2, 2], [2, -3, 2], [2, 2, -3]]`
B. `[[3, 2, 2], [2, 3, 2], [2, 2, 3]]`
C. `[[5, 2, 0], [2, 5, 0], [0, 2,5]]`
D. `[[5, 2, 5], [2, 5, 5], [5, 5, 2]]`

Answer

To solve the problem, we need to find the expression \((A^2 - 4A)A^{-1}\) given the matrix \(A\) and the conditions specified. Let's go through the steps systematically.

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

First, we need to compute \(A^2\), which is \(A \times A\).

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

Calculating \(A^2\):
\[ A^2 = A \times A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \]

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:
\[ A^2 = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} \]

### Step 2: Calculate \(4A\)

Next, we compute \(4A\):
\[ 4A = 4 \times \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix} \]

### Step 3: Calculate \(A^2 - 4A\)

Now we subtract \(4A\) from \(A^2\):
\[ A^2 - 4A = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} - \begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix} \]

Calculating each element:
- First row:
- \(9 - 4 = 5\)
- \(8 - 8 = 0\)
- \(8 - 8 = 0\)

- Second row:
- \(8 - 8 = 0\)
- \(9 - 4 = 5\)
- \(8 - 8 = 0\)

- Third row:
- \(8 - 8 = 0\)
- \(8 - 8 = 0\)
- \(9 - 4 = 5\)

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

### Step 4: Calculate \((A^2 - 4A)A^{-1}\)

Now we need to multiply \((A^2 - 4A)\) by \(A^{-1}\). Since \(A^{-1}\) exists, we can denote:
\[ A^{-1} = \text{(some matrix)} \]

Using the property that \(A \cdot A^{-1} = I\) (identity matrix), we can express:
\[ (A^2 - 4A)A^{-1} = 5I \]

Where \(I\) is the identity matrix:
\[ I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]

Thus,
\[ (A^2 - 4A)A^{-1} = 5 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} \]

### Final Answer
\[ (A^2 - 4A)A^{-1} = 5I = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} \]