Let `A =[[1,2,2],[2,1,2],[2,2,1]]`, then

Options

A. `A^(2) - 4A-5I_(3)=O`
B. `A^(-1) = 1/5 (A-4I_(3))`
C. `A^(3)` is not invertible
D. `A^(2)` is invertible

Explanation

To solve the problem step by step, we will analyze the matrix \( A \) and check the given options.

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

### Step 1: Calculate \( A^2 \)
To find \( A^2 \), we multiply matrix \( A \) by itself.

\[ A^2 = A \cdot A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \]

Calculating the elements 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 \( A^2 - 4A - 5I_3 \)
Next, we need to calculate \( A^2 - 4A - 5I_3 \).

Where \( I_3 \) is the identity matrix of order 3:
\[ I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]

Calculating \( 4A \):
\[ 4A = 4 \cdot \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix} \]

Calculating \( 5I_3 \):
\[ 5I_3 = 5 \cdot \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} \]

Now, we can compute \( A^2 - 4A - 5I_3 \):
\[ A^2 - 4A - 5I_3 = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} - \begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix} - \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} \]

Calculating the result:
\[ = \begin{bmatrix} 9 - 4 - 5 & 8 - 8 - 0 & 8 - 8 - 0 \\ 8 - 8 - 0 & 9 - 4 - 5 & 8 - 8 - 0 \\ 8 - 8 - 0 & 8 - 8 - 0 & 9 - 4 - 5 \end{bmatrix} \]
\[ = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \]

### Step 3: Conclusion
Since \( A^2 - 4A - 5I_3 = 0 \), we can conclude that the first option is correct.

### Step 4: Finding \( A^{-1} \)
Using the equation derived from the previous step:
\[ A^2 - 4A + 5I_3 = 0 \implies A^2 - 4A = -5I_3 \implies A(A - 4I_3) = -5I_3 \]
Thus,
\[ A^{-1} = -\frac{1}{5}(A - 4I_3) \]

### Step 5: Determinant of \( A \)
To check if \( A^3 \) is invertible, we find the determinant of \( A \):
\[ \text{det}(A) = 1(1 \cdot 1 - 2 \cdot 2) - 2(2 \cdot 1 - 2 \cdot 2) + 2(2 \cdot 2 - 1 \cdot 2) \]
\[ = 1(1 - 4) - 2(2 - 4) + 2(4 - 2) \]
\[ = 1(-3) - 2(-2) + 2(2) = -3 + 4 + 4 = 5 \]

Since \( \text{det}(A) \neq 0 \), \( A \) is invertible, and consequently \( A^3 \) is also invertible.

### Step 6: Determinant of \( A^2 \)
\[ \text{det}(A^2) = (\text{det}(A))^2 = 5^2 = 25 \neq 0 \]
Thus, \( A^2 \) is also invertible.

### Final Answer
The correct options are:
1. \( A^2 - 4A - 5I_3 = 0 \)
2. \( A^{-1} = \frac{1}{5}(A - 4I_3) \)
3. \( A^3 \) is invertible.
4. \( A^2 \) is invertible.