Solve the following equations, using inverse of a matrix :
`{:(x+y+z=6),(y+3z=11),(x-2y+z=0):}`

Answer

To solve the system of equations using the inverse of a matrix, we can follow these steps:

### Step 1: Write the equations in matrix form
We have the following equations:
1. \( x + y + z = 6 \)
2. \( y + 3z = 11 \)
3. \( x - 2y + z = 0 \)

We can express these equations in the form \( AX = B \), where:
- \( A \) is the coefficient matrix,
- \( X \) is the variable matrix,
- \( B \) is the constant matrix.

The coefficient matrix \( A \) and the constant matrix \( B \) are as follows:

\[ A = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 6 \\ 11 \\ 0 \end{pmatrix} \]

### Step 2: Find the inverse of matrix \( A \)
To find the inverse of matrix \( A \), we first need to calculate its determinant.

#### Determinant of \( A \):
\[ \text{det}(A) = 1 \cdot (1 \cdot 1 - 3 \cdot (-2)) - 1 \cdot (0 \cdot 1 - 3 \cdot 1) + 1 \cdot (0 \cdot (-2) - 1 \cdot 1) \]
Calculating this step-by-step:
- \( 1 \cdot (1 + 6) = 7 \)
- \( -1 \cdot (-3) = 3 \)
- \( +1 \cdot (-1) = -1 \)

So,
\[ \text{det}(A) = 7 + 3 - 1 = 9 \]

#### Adjoint of \( A \):
Next, we calculate the adjoint of \( A \) by finding the cofactor matrix and then taking its transpose.

The cofactor matrix \( C \) is calculated as follows:
\[ C = \begin{pmatrix} \text{C}_{11} & \text{C}_{12} & \text{C}_{13} \\ \text{C}_{21} & \text{C}_{22} & \text{C}_{23} \\ \text{C}_{31} & \text{C}_{32} & \text{C}_{33} \end{pmatrix} \]

Calculating each cofactor:
- \( \text{C}_{11} = 1 \cdot 1 - 3 \cdot (-2) = 7 \)
- \( \text{C}_{12} = -(0 \cdot 1 - 3 \cdot 1) = 3 \)
- \( \text{C}_{13} = 0 \cdot (-2) - 1 \cdot 1 = -1 \)

- \( \text{C}_{21} = -(1 \cdot 1 - 3 \cdot 1) = -(-2) = 2 \)
- \( \text{C}_{22} = 1 \cdot 1 - 1 \cdot 1 = 0 \)
- \( \text{C}_{23} = -(1 \cdot (-2) - 1 \cdot 1) = 1 \)

- \( \text{C}_{31} = 1 \cdot (-2) - 1 \cdot 1 = -3 \)
- \( \text{C}_{32} = -(1 \cdot 1 - 1 \cdot 1) = 0 \)
- \( \text{C}_{33} = 1 \cdot 1 - 0 \cdot 1 = 1 \)

Thus, the cofactor matrix \( C \) is:
\[ C = \begin{pmatrix} 7 & 3 & -1 \\ 2 & 0 & 1 \\ -3 & 0 & 1 \end{pmatrix} \]

Now, taking the transpose of \( C \) gives us the adjoint \( \text{adj}(A) \):
\[ \text{adj}(A) = \begin{pmatrix} 7 & 2 & -3 \\ 3 & 0 & 0 \\ -1 & 1 & 1 \end{pmatrix} \]

#### Inverse of \( A \):
Now we can find the inverse of \( A \):
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) = \frac{1}{9} \begin{pmatrix} 7 & 2 & -3 \\ 3 & 0 & 0 \\ -1 & 1 & 1 \end{pmatrix} \]

### Step 3: Solve for \( X \)
Now we can find \( X \) using the formula \( X = A^{-1}B \):
\[ X = \frac{1}{9} \begin{pmatrix} 7 & 2 & -3 \\ 3 & 0 & 0 \\ -1 & 1 & 1 \end{pmatrix} \begin{pmatrix} 6 \\ 11 \\ 0 \end{pmatrix} \]

Calculating the product:
1. First row: \( 7 \cdot 6 + 2 \cdot 11 + (-3) \cdot 0 = 42 + 22 + 0 = 64 \)
2. Second row: \( 3 \cdot 6 + 0 \cdot 11 + 0 \cdot 0 = 18 + 0 + 0 = 18 \)
3. Third row: \( -1 \cdot 6 + 1 \cdot 11 + 1 \cdot 0 = -6 + 11 + 0 = 5 \)

Thus,
\[ X = \frac{1}{9} \begin{pmatrix} 64 \\ 18 \\ 5 \end{pmatrix} = \begin{pmatrix} \frac{64}{9} \\ 2 \\ \frac{5}{9} \end{pmatrix} \]

### Step 4: Final Values
The final values of \( x, y, z \) are:
- \( x = \frac{64}{9} \)
- \( y = 2 \)
- \( z = \frac{5}{9} \)