If `A= [[1,1,1],[1,1,1],[1,1,1]]` then

Options

A. `A^(3) = 9 A `
B. `A^(3) = 27 A `
C. `A + A = A^(2)`
D. `A^(-1)` does not exist

Explanation

To solve the problem involving the matrix \( A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} \), we will perform the following steps:

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

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

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

To perform the multiplication, we calculate each element of the resulting matrix:

- The element at position (1,1):
\[ 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 = 3 \]

- The element at position (1,2):
\[ 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 = 3 \]

- The element at position (1,3):
\[ 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 = 3 \]

Repeating this for all rows, we find:

\[ A^2 = \begin{bmatrix} 3 & 3 & 3 \\ 3 & 3 & 3 \\ 3 & 3 & 3 \end{bmatrix} \]

### Step 2: Express \( A^2 \) in terms of \( A \)

We can see that:

\[ A^2 = 3A \]

### Step 3: Calculate \( A^3 \)

Next, we calculate \( A^3 \) using the result from \( A^2 \):

\[ A^3 = A^2 \cdot A = (3A) \cdot A = 3A^2 \]

Substituting \( A^2 = 3A \):

\[ A^3 = 3(3A) = 9A \]

### Step 4: Analyze the options

Now, we can analyze the options given in the problem:

1. \( A^2 = 3A \) (True)
2. \( A^3 = 9A \) (True)
3. The inverse of \( A \) does not exist.

To confirm that the inverse does not exist, we calculate the determinant of \( A \):

### Step 5: Calculate the determinant of \( A \)

The determinant of a matrix with two identical rows (or columns) is zero. Since all rows of \( A \) are identical, we have:

\[ \text{det}(A) = 0 \]

Since the determinant is zero, the inverse of \( A \) does not exist.

### Conclusion

The correct conclusions are:
- \( A^2 = 3A \)
- \( A^3 = 9A \)
- The inverse of \( A \) does not exist.