If `A=[{:(1,1,1),(1,0,2),(3,1,1):}]`, find `A^(-1)`. Hence, solve the system of equations :
`x+y+z=6`, `x+2z=7`, `3x+y+z=12`

Answer

To find the inverse of the matrix \( A \) and solve the given system of equations, we will follow these steps:

### Step 1: Write the matrix \( A \)
The matrix \( A \) is given as:
\[ A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1 \end{pmatrix} \]

### Step 2: Calculate the determinant of \( A \)
The determinant of a \( 3 \times 3 \) matrix can be calculated using the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix:
\[ \text{det}(A) = 1(0 \cdot 1 - 2 \cdot 1) - 1(1 \cdot 1 - 2 \cdot 3) + 1(1 \cdot 1 - 0 \cdot 3) \]
Calculating this step-by-step:
\[ = 1(0 - 2) - 1(1 - 6) + 1(1 - 0) \]
\[ = -2 + 5 + 1 = 4 \]

### Step 3: Calculate the cofactor matrix of \( A \)
The cofactor matrix is obtained by calculating the determinant of the \( 2 \times 2 \) matrices formed by removing one row and one column for each element, and applying the checkerboard pattern of signs.

The cofactor matrix \( C \) is:
\[ C = \begin{pmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{pmatrix} \]
Calculating each cofactor:
- \( C_{11} = \text{det} \begin{pmatrix} 0 & 2 \\ 1 & 1 \end{pmatrix} = (0 \cdot 1 - 2 \cdot 1) = -2 \)
- \( C_{12} = -\text{det} \begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix} = - (1 \cdot 1 - 2 \cdot 3) = -(-5) = 5 \)
- \( C_{13} = \text{det} \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} = (1 \cdot 1 - 0 \cdot 3) = 1 \)
- \( C_{21} = -\text{det} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = - (1 \cdot 1 - 1 \cdot 1) = 0 \)
- \( C_{22} = \text{det} \begin{pmatrix} 1 & 1 \\ 3 & 1 \end{pmatrix} = (1 \cdot 1 - 1 \cdot 3) = -2 \)
- \( C_{23} = -\text{det} \begin{pmatrix} 1 & 1 \\ 3 & 1 \end{pmatrix} = - (1 \cdot 1 - 1 \cdot 3) = 2 \)
- \( C_{31} = \text{det} \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} = (1 \cdot 2 - 1 \cdot 0) = 2 \)
- \( C_{32} = -\text{det} \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} = - (1 \cdot 2 - 1 \cdot 1) = -1 \)
- \( C_{33} = \text{det} \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = (1 \cdot 0 - 1 \cdot 1) = -1 \)

Thus, the cofactor matrix \( C \) is:
\[ C = \begin{pmatrix} -2 & 5 & 1 \\ 0 & -2 & 2 \\ 2 & -1 & -1 \end{pmatrix} \]

### Step 4: Calculate the adjoint of \( A \)
The adjoint of \( A \) is the transpose of the cofactor matrix \( C \):
\[ \text{adj}(A) = C^T = \begin{pmatrix} -2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1 \end{pmatrix} \]

### Step 5: Calculate the inverse of \( A \)
The inverse of \( A \) is given by:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]
Substituting the values:
\[ A^{-1} = \frac{1}{4} \begin{pmatrix} -2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1 \end{pmatrix} = \begin{pmatrix} -\frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{4} & -\frac{1}{2} & -\frac{1}{4} \\ \frac{1}{4} & \frac{1}{2} & -\frac{1}{4} \end{pmatrix} \]

### Step 6: Solve the system of equations
The system of equations can be represented in matrix form as \( Ax = b \), where:
\[ b = \begin{pmatrix} 6 \\ 7 \\ 12 \end{pmatrix} \]
To find \( x \), we use:
\[ x = A^{-1}b \]
Calculating:
\[ x = \begin{pmatrix} -\frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{4} & -\frac{1}{2} & -\frac{1}{4} \\ \frac{1}{4} & \frac{1}{2} & -\frac{1}{4} \end{pmatrix} \begin{pmatrix} 6 \\ 7 \\ 12 \end{pmatrix} \]
Calculating each component:
1. First row:
\[ -\frac{1}{2} \cdot 6 + 0 \cdot 7 + \frac{1}{2} \cdot 12 = -3 + 0 + 6 = 3 \]
2. Second row:
\[ \frac{5}{4} \cdot 6 - \frac{1}{2} \cdot 7 - \frac{1}{4} \cdot 12 = \frac{30}{4} - \frac{14}{4} - \frac{12}{4} = \frac{30 - 14 - 12}{4} = \frac{4}{4} = 1 \]
3. Third row:
\[ \frac{1}{4} \cdot 6 + \frac{1}{2} \cdot 7 - \frac{1}{4} \cdot 12 = \frac{6}{4} + \frac{7}{2} - \frac{12}{4} = \frac{6 + 14 - 12}{4} = \frac{8}{4} = 2 \]

Thus, we have:
\[ x = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} \]

### Final Answer
The solution to the system of equations is:
\[ x = 3, \quad y = 1, \quad z = 2 \]