Let A be a `3xx3` matrix such that A `[(1,2,3),(0,2,3),(0,1,1)]=[(0,0,1),(1,0,0),(0,1,0)]`, then `A^(-1)` is :

Options

A. `[(3,1,2),(3,0,2),(1,0,1)]`
B. `[(3,2,1),(3,2,0),(1,1,0)]`
C. `[(0,1,3),(0,2,3),(2,1,1)]`
D. `[(1,2,3),(0,1,1),(0,2,3)]`

Explanation

To find the inverse of the matrix \( A \) given that \( A \cdot B = I \), where \( B \) is the matrix \( \begin{pmatrix} 1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 1 & 1 \end{pmatrix} \) and \( I \) is the identity matrix \( \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \), we can follow these steps:

### Step 1: Set up the equation
We have the equation:
\[ A \cdot B = I \]
This can be rearranged to find \( A \):
\[ A = I \cdot B^{-1} \]

### Step 2: Find the inverse of matrix \( B \)
To find \( A^{-1} \), we need to compute \( B^{-1} \). We can use the formula for the inverse of a \( 3 \times 3 \) matrix, which is given by:
\[ B^{-1} = \frac{1}{\text{det}(B)} \cdot \text{adj}(B) \]
where \( \text{det}(B) \) is the determinant of \( B \) and \( \text{adj}(B) \) is the adjugate of \( B \).

### Step 3: Calculate the determinant of \( B \)
The determinant of matrix \( B \) is calculated as follows:
\[ B = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 1 & 1 \end{pmatrix} \]
Using the determinant formula for a \( 3 \times 3 \) matrix:
\[ \text{det}(B) = 1 \cdot (2 \cdot 1 - 3 \cdot 1) - 2 \cdot (0 \cdot 1 - 3 \cdot 0) + 3 \cdot (0 \cdot 1 - 2 \cdot 0) = 1 \cdot (-1) = -1 \]

### Step 4: Calculate the adjugate of \( B \)
To find the adjugate, we need to calculate the cofactor matrix and then transpose it. The cofactor matrix is calculated as follows:
\[ \text{Cofactor}(B) = \begin{pmatrix} \text{det}(M_{11}) & -\text{det}(M_{12}) & \text{det}(M_{13}) \\ -\text{det}(M_{21}) & \text{det}(M_{22}) & -\text{det}(M_{23}) \\ \text{det}(M_{31}) & -\text{det}(M_{32}) & \text{det}(M_{33}) \end{pmatrix} \]
where \( M_{ij} \) is the minor matrix obtained by deleting the \( i \)-th row and \( j \)-th column.

Calculating the cofactors:
- \( \text{det}(M_{11}) = \text{det} \begin{pmatrix} 2 & 3 \\ 1 & 1 \end{pmatrix} = 2 - 3 = -1 \)
- \( \text{det}(M_{12}) = \text{det} \begin{pmatrix} 0 & 3 \\ 0 & 1 \end{pmatrix} = 0 \)
- \( \text{det}(M_{13}) = \text{det} \begin{pmatrix} 0 & 2 \\ 0 & 1 \end{pmatrix} = 0 \)
- \( \text{det}(M_{21}) = \text{det} \begin{pmatrix} 2 & 3 \\ 1 & 1 \end{pmatrix} = -1 \)
- \( \text{det}(M_{22}) = \text{det} \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} = 1 \)
- \( \text{det}(M_{23}) = \text{det} \begin{pmatrix} 1 & 2 \\ 0 & 0 \end{pmatrix} = 0 \)
- \( \text{det}(M_{31}) = \text{det} \begin{pmatrix} 2 & 3 \\ 2 & 3 \end{pmatrix} = 0 \)
- \( \text{det}(M_{32}) = \text{det} \begin{pmatrix} 1 & 3 \\ 0 & 2 \end{pmatrix} = 2 \)
- \( \text{det}(M_{33}) = \text{det} \begin{pmatrix} 1 & 2 \\ 0 & 2 \end{pmatrix} = 2 \)

Thus, the cofactor matrix is:
\[ \text{Cofactor}(B) = \begin{pmatrix} -1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & -2 & 2 \end{pmatrix} \]
Transposing gives us the adjugate:
\[ \text{adj}(B) = \begin{pmatrix} -1 & 1 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & 2 \end{pmatrix} \]

### Step 5: Calculate \( B^{-1} \)
Now we can find \( B^{-1} \):
\[ B^{-1} = \frac{1}{-1} \cdot \begin{pmatrix} -1 & 1 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & 2 \end{pmatrix} = \begin{pmatrix} 1 & -1 & 0 \\ 0 & -1 & 2 \\ 0 & 0 & -2 \end{pmatrix} \]

### Step 6: Find \( A^{-1} \)
Since \( A = B^{-1} \), we have:
\[ A^{-1} = \begin{pmatrix} 1 & -1 & 0 \\ 0 & -1 & 2 \\ 0 & 0 & -2 \end{pmatrix} \]

### Final Answer
Thus, the inverse of matrix \( A \) is:
\[ A^{-1} = \begin{pmatrix} 1 & -1 & 0 \\ 0 & -1 & 2 \\ 0 & 0 & -2 \end{pmatrix} \]