If `A=[(1,2,2),(2,1,2),(2,2,1)]` then `A^(2)-4A` is equal to

Options

A. `2l_(3)`
B. `3l_(3)`
C. `4l_(3)`
D. `5l_(3)`

Explanation

To solve the problem, we need to compute \( A^2 - 4A \) for the matrix \( A = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} \).

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

To find \( A^2 \), we multiply matrix \( A \) by itself:

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

Calculating each element of the resulting matrix:

- First row, first column:
\[ 1 \cdot 1 + 2 \cdot 2 + 2 \cdot 2 = 1 + 4 + 4 = 9 \]

- First row, second column:
\[ 1 \cdot 2 + 2 \cdot 1 + 2 \cdot 2 = 2 + 2 + 4 = 8 \]

- First row, third column:
\[ 1 \cdot 2 + 2 \cdot 2 + 2 \cdot 1 = 2 + 4 + 2 = 8 \]

- Second row, first column:
\[ 2 \cdot 1 + 1 \cdot 2 + 2 \cdot 2 = 2 + 2 + 4 = 8 \]

- Second row, second column:
\[ 2 \cdot 2 + 1 \cdot 1 + 2 \cdot 2 = 4 + 1 + 4 = 9 \]

- Second row, third column:
\[ 2 \cdot 2 + 1 \cdot 2 + 2 \cdot 1 = 4 + 2 + 2 = 8 \]

- Third row, first column:
\[ 2 \cdot 1 + 2 \cdot 2 + 1 \cdot 2 = 2 + 4 + 2 = 8 \]

- Third row, second column:
\[ 2 \cdot 2 + 2 \cdot 1 + 1 \cdot 2 = 4 + 2 + 2 = 8 \]

- Third row, third column:
\[ 2 \cdot 2 + 2 \cdot 2 + 1 \cdot 1 = 4 + 4 + 1 = 9 \]

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

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

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

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

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

Calculating each element:
- First row, first column:
\[ 9 - 4 = 5 \]

- First row, second column:
\[ 8 - 8 = 0 \]

- First row, third column:
\[ 8 - 8 = 0 \]

- Second row, first column:
\[ 8 - 8 = 0 \]

- Second row, second column:
\[ 9 - 4 = 5 \]

- Second row, third column:
\[ 8 - 8 = 0 \]

- Third row, first column:
\[ 8 - 8 = 0 \]

- Third row, second column:
\[ 8 - 8 = 0 \]

- Third row, third column:
\[ 9 - 4 = 5 \]

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

### Final Result

The result can be expressed as:
\[ A^2 - 4A = 5I_3 \]
where \( I_3 \) is the \( 3 \times 3 \) identity matrix.