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

Options

A. `[{:(-6,0,0),(0,-6,0),(0,0,-6):}]`
B. `[{:(0,(1)/(6),-(1)/(6)),((1)/(3),(1)/(6),(1)/(2)),((1)/(2),(1)/(3),(1)/(6)):}]`
C. `[{:(3,0,0),(0,3,0),(0,0,3):}]`
D. `[{:(1,0,0),(0,1,0),(0,0,1):}]`

Answer

To solve the problem, we need to compute the expression \( (A \cdot \text{adj}(A) \cdot A^{-1}) \cdot A \) where \( A \) is the given matrix:

\[ A = \begin{pmatrix} 0 & 1 & -1 \\ 2 & 1 & 3 \\ 3 & 2 & 1 \end{pmatrix} \]

### Step 1: Calculate the Determinant of \( A \)

First, we need to find the determinant of matrix \( A \).

\[ \text{det}(A) = 0 \cdot \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & 3 \\ 3 & 1 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} \]

Calculating the minors:

1. \( \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} = (1 \cdot 1) - (3 \cdot 2) = 1 - 6 = -5 \)
2. \( \begin{vmatrix} 2 & 3 \\ 3 & 1 \end{vmatrix} = (2 \cdot 1) - (3 \cdot 3) = 2 - 9 = -7 \)
3. \( \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} = (2 \cdot 2) - (1 \cdot 3) = 4 - 3 = 1 \)

Now substituting back into the determinant formula:

\[ \text{det}(A) = 0 \cdot (-5) - 1 \cdot (-7) + 1 \cdot (1) = 0 + 7 + 1 = 8 \]

### Step 2: Calculate the Adjoint of \( A \)

The adjoint of \( A \), denoted as \( \text{adj}(A) \), is the transpose of the cofactor matrix. We need to find the cofactors of each element in \( A \).

Calculating the cofactors:

1. \( C_{11} = \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} = -5 \)
2. \( C_{12} = -\begin{vmatrix} 2 & 3 \\ 3 & 1 \end{vmatrix} = 7 \)
3. \( C_{13} = \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} = 1 \)

4. \( C_{21} = -\begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} = 1 \)
5. \( C_{22} = \begin{vmatrix} 0 & -1 \\ 3 & 1 \end{vmatrix} = 3 \)
6. \( C_{23} = -\begin{vmatrix} 0 & 1 \\ 3 & 2 \end{vmatrix} = -3 \)

7. \( C_{31} = \begin{vmatrix} 1 & -1 \\ 1 & 3 \end{vmatrix} = 4 \)
8. \( C_{32} = -\begin{vmatrix} 0 & -1 \\ 2 & 3 \end{vmatrix} = 2 \)
9. \( C_{33} = \begin{vmatrix} 0 & 1 \\ 2 & 1 \end{vmatrix} = -2 \)

Now, the cofactor matrix is:

\[ \text{Cof}(A) = \begin{pmatrix} -5 & 7 & 1 \\ 1 & 3 & -3 \\ 4 & 2 & -2 \end{pmatrix} \]

Taking the transpose gives us the adjoint:

\[ \text{adj}(A) = \begin{pmatrix} -5 & 1 & 4 \\ 7 & 3 & 2 \\ 1 & -3 & -2 \end{pmatrix} \]

### Step 3: Calculate the Inverse of \( A \)

The inverse of \( A \) can be calculated using the formula:

\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]

Substituting the determinant and adjoint:

\[ A^{-1} = \frac{1}{8} \begin{pmatrix} -5 & 1 & 4 \\ 7 & 3 & 2 \\ 1 & -3 & -2 \end{pmatrix} \]

### Step 4: Calculate \( A \cdot \text{adj}(A) \cdot A^{-1} \)

Now we compute \( A \cdot \text{adj}(A) \):

\[ A \cdot \text{adj}(A) = A \cdot \begin{pmatrix} -5 & 1 & 4 \\ 7 & 3 & 2 \\ 1 & -3 & -2 \end{pmatrix} \]

Calculating this product will give us a matrix that we can then multiply by \( A^{-1} \).

### Step 5: Final Calculation

Finally, we multiply the result from the previous step by \( A \) to get the final result.

\[ (A \cdot \text{adj}(A) \cdot A^{-1}) \cdot A = \text{det}(A) \cdot I \cdot A = 8 \cdot A \]

Thus, the final result is:

\[ 8A = 8 \begin{pmatrix} 0 & 1 & -1 \\ 2 & 1 & 3 \\ 3 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 8 & -8 \\ 16 & 8 & 24 \\ 24 & 16 & 8 \end{pmatrix} \]