If `{:A=[(0,1,-1),(2,1,3),(3,2,1)]:}` then `(A(adj A)A^(-1))A=`

Options

A. `2{:[(3,0,0),(0,3,0),(0,0,3)]:}`
B. `{:[(-6,0,0),(0,-6,0),(0,0,-6)]:}`
C. `{:[(0,1//6,-1//6),(2//6,1//6,3//6),(3//6,2//6,1//6)]:}`
D. none of these

Explanation

To solve the problem, we need to evaluate the expression \( (A \cdot \text{adj} A \cdot A^{-1}) A \) where \( A = \begin{pmatrix} 0 & 1 & -1 \\ 2 & 1 & 3 \\ 3 & 2 & 1 \end{pmatrix} \).

### Step-by-Step Solution:

1. **Identify the Matrix A**:
\[ A = \begin{pmatrix} 0 & 1 & -1 \\ 2 & 1 & 3 \\ 3 & 2 & 1 \end{pmatrix} \]

**Hint**: Make sure to write down the matrix correctly as it will be used in further calculations.

2. **Calculate the Determinant of A**:
The determinant of a 3x3 matrix can be calculated using the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix:
\[ \text{det}(A) = 0(1 \cdot 1 - 3 \cdot 2) - 1(2 \cdot 1 - 3 \cdot 3) + (-1)(2 \cdot 2 - 1 \cdot 3) \]
Simplifying:
\[ = 0 - 1(2 - 9) - 1(4 - 3) = 0 + 7 - 1 = 6 \]

**Hint**: Remember that the determinant is a scalar value that helps in finding the adjoint and inverse of the matrix.

3. **Use the Property of Matrices**:
We know that:
\[ A \cdot \text{adj} A = \text{det}(A) \cdot I \]
where \( I \) is the identity matrix. Therefore:
\[ A \cdot \text{adj} A = 6I \]

**Hint**: This property is crucial as it simplifies our calculations significantly.

4. **Calculate \( A^{-1} \cdot A \)**:
We know that:
\[ A^{-1} \cdot A = I \]

**Hint**: The inverse of a matrix multiplied by the matrix itself always results in the identity matrix.

5. **Combine the Results**:
Now substituting back into our original expression:
\[ (A \cdot \text{adj} A \cdot A^{-1}) A = (6I \cdot I) A = 6I \cdot A \]
This simplifies to:
\[ 6A \]

**Hint**: Multiplying by the identity matrix does not change the matrix.

6. **Calculate \( 6A \)**:
Now, we multiply the matrix \( A \) by 6:
\[ 6A = 6 \begin{pmatrix} 0 & 1 & -1 \\ 2 & 1 & 3 \\ 3 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 6 & -6 \\ 12 & 6 & 18 \\ 18 & 12 & 6 \end{pmatrix} \]

**Hint**: Ensure to multiply each element of the matrix by 6 correctly.

### Final Result:
Thus, the final result is:
\[ (A \cdot \text{adj} A \cdot A^{-1}) A = \begin{pmatrix} 0 & 6 & -6 \\ 12 & 6 & 18 \\ 18 & 12 & 6 \end{pmatrix} \]