If `A=[(1,2,2),(2,1,2),(2,2,1)]` then `A^(2)-4A-5I=`………….

Answer

To solve the problem \( A^2 - 4A - 5I \) where \( A = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} \), we will follow these steps:

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

- 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 calculate \( 4A \):

\[ 4A = 4 \times \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{pmatrix} \]

### Step 3: Calculate \( 5I \)

Now, we calculate \( 5I \) where \( I \) is the identity matrix of the same order as \( A \):

\[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
\[ 5I = 5 \times \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix} \]

### Step 4: Combine the results to find \( A^2 - 4A - 5I \)

Now we combine the results:

\[ A^2 - 4A - 5I = \begin{pmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{pmatrix} - \begin{pmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{pmatrix} - \begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix} \]

Calculating each element:

- First row:
\[ 9 - 4 - 5 = 0, \quad 8 - 8 - 0 = 0, \quad 8 - 8 - 0 = 0 \]

- Second row:
\[ 8 - 8 - 0 = 0, \quad 9 - 4 - 5 = 0, \quad 8 - 8 - 0 = 0 \]

- Third row:
\[ 8 - 8 - 0 = 0, \quad 8 - 8 - 0 = 0, \quad 9 - 4 - 5 = 0 \]

Thus, we have:

\[ A^2 - 4A - 5I = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \]

### Final Answer:
\[ A^2 - 4A - 5I = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \]