Find the inverse of the matrix :
`A=[{:(0,1,2),(0,1,1),(1,0,2):}]`

Answer

To find the inverse of the matrix \( A = \begin{pmatrix} 0 & 1 & 2 \\ 0 & 1 & 1 \\ 1 & 0 & 2 \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate the Determinant of Matrix A
The determinant of a \( 3 \times 3 \) matrix \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \) can be calculated using the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix:
\[ A = \begin{pmatrix} 0 & 1 & 2 \\ 0 & 1 & 1 \\ 1 & 0 & 2 \end{pmatrix} \]
We can expand the determinant along the first row:
\[ \text{det}(A) = 0 \cdot (1 \cdot 2 - 1 \cdot 0) - 1 \cdot (0 \cdot 2 - 1 \cdot 1) + 2 \cdot (0 \cdot 0 - 1 \cdot 1) \]
Calculating this gives:
\[ = 0 - 1 \cdot (-1) + 2 \cdot (-1) = 1 - 2 = -1 \]

### Step 2: Calculate the Matrix of Cofactors
Next, we need to find the matrix of cofactors. The cofactor \( C_{ij} \) is given by \( (-1)^{i+j} \) times the determinant of the submatrix obtained by deleting the \( i^{th} \) row and \( j^{th} \) column.

Calculating the cofactors:
- \( C_{11} = \text{det}\begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} = (1 \cdot 2 - 1 \cdot 0) = 2 \)
- \( C_{12} = -\text{det}\begin{pmatrix} 0 & 1 \\ 1 & 2 \end{pmatrix} = -(0 \cdot 2 - 1 \cdot 1) = 1 \)
- \( C_{13} = \text{det}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = (0 \cdot 0 - 1 \cdot 1) = -1 \)

- \( C_{21} = -\text{det}\begin{pmatrix} 1 & 2 \\ 0 & 2 \end{pmatrix} = -(1 \cdot 2 - 2 \cdot 0) = -2 \)
- \( C_{22} = \text{det}\begin{pmatrix} 0 & 2 \\ 1 & 2 \end{pmatrix} = (0 \cdot 2 - 2 \cdot 1) = -2 \)
- \( C_{23} = -\text{det}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = -(0 \cdot 0 - 1 \cdot 1) = 1 \)

- \( C_{31} = \text{det}\begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix} = (1 \cdot 1 - 2 \cdot 1) = -1 \)
- \( C_{32} = -\text{det}\begin{pmatrix} 0 & 2 \\ 0 & 1 \end{pmatrix} = -(0 \cdot 1 - 2 \cdot 0) = 0 \)
- \( C_{33} = \text{det}\begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix} = (0 \cdot 1 - 1 \cdot 0) = 0 \)

The cofactor matrix is:
\[ C = \begin{pmatrix} 2 & 1 & -1 \\ -2 & -2 & 1 \\ -1 & 0 & 0 \end{pmatrix} \]

### Step 3: Transpose the Cofactor Matrix to Get the Adjoint
The adjoint of matrix \( A \) is the transpose of the cofactor matrix:
\[ \text{adj}(A) = C^T = \begin{pmatrix} 2 & -2 & -1 \\ 1 & -2 & 0 \\ -1 & 1 & 0 \end{pmatrix} \]

### Step 4: Calculate the Inverse of Matrix A
The inverse of matrix \( A \) is given by:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]
Since \( \text{det}(A) = -1 \):
\[ A^{-1} = -1 \cdot \begin{pmatrix} 2 & -2 & -1 \\ 1 & -2 & 0 \\ -1 & 1 & 0 \end{pmatrix} = \begin{pmatrix} -2 & 2 & 1 \\ -1 & 2 & 0 \\ 1 & -1 & 0 \end{pmatrix} \]

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