With the help of metrices the solution of the equations `3x+y+2z=3,2x-3y-z= 3 , x+2y+z=4 ` are given by

Options

A. `x=1, y=2,z=-1`
B. `x=-1, y=2,z=1`
C. `x=1, y=-2,z=-1`
D. `x=-1, y=-2,z=1`

Answer

To solve the system of equations using matrices, we will follow these steps:

### Given Equations:
1. \(3x + y + 2z = 3\)
2. \(2x - 3y - z = 3\)
3. \(x + 2y + z = 4\)

### Step 1: Represent the equations in matrix form
We can represent the equations in the form \(A \mathbf{x} = \mathbf{B}\), where:

\[ A = \begin{pmatrix} 3 & 1 & 2 \\ 2 & -3 & -1 \\ 1 & 2 & 1 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad \mathbf{B} = \begin{pmatrix} 3 \\ 3 \\ 4 \end{pmatrix} \]

### Step 2: Find the inverse of matrix \(A\)
To find \(\mathbf{x}\), we need to compute \(A^{-1}\). The formula for the inverse of a matrix is:

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

#### Step 2.1: Calculate the determinant of \(A\)
The determinant of \(A\) can be calculated as follows:

\[ \text{det}(A) = 3 \cdot (-3) \cdot 1 + 1 \cdot (-1) \cdot 2 + 2 \cdot 2 \cdot 2 - (2 \cdot (-3) \cdot 2 + (-1) \cdot 1 \cdot 1 + 3 \cdot (-1) \cdot 1) \]

Calculating each term:

- \(3 \cdot (-3) \cdot 1 = -9\)
- \(1 \cdot (-1) \cdot 2 = -2\)
- \(2 \cdot 2 \cdot 2 = 8\)
- \(2 \cdot (-3) \cdot 2 = -12\)
- \((-1) \cdot 1 \cdot 1 = -1\)
- \(3 \cdot (-1) \cdot 1 = -3\)

Now, substituting these values into the determinant formula:

\[ \text{det}(A) = -9 - 2 + 8 + 12 - 1 + 3 = 11 \]

#### Step 2.2: Calculate the adjoint of \(A\)
The adjoint of \(A\) is the transpose of the cofactor matrix. We will compute the cofactor matrix first.

Calculating the cofactors:

\[ C = \begin{pmatrix} (-3)(1) - (2)(-1) & -(2)(1) - (3)(2) & (2)(-3) - (1)(2) \\ -(1)(1) - (2)(-1) & (3)(1) - (2)(2) & -(3)(2) - (1)(1) \\ (1)(-1) - (3)(-3) & -(3)(-3) - (2)(2) & (3)(2) - (2)(1) \end{pmatrix} \]

Calculating each cofactor:

- \(C_{11} = -3 + 2 = -1\)
- \(C_{12} = -2 - 6 = -8\)
- \(C_{13} = -6 - 2 = -8\)
- \(C_{21} = -1 + 2 = 1\)
- \(C_{22} = 3 - 4 = -1\)
- \(C_{23} = -6 - 1 = -7\)
- \(C_{31} = -1 + 9 = 8\)
- \(C_{32} = 9 - 4 = 5\)
- \(C_{33} = 6 - 2 = 4\)

Thus, the cofactor matrix is:

\[ C = \begin{pmatrix} -1 & -8 & -8 \\ 1 & -1 & -7 \\ 8 & 5 & 4 \end{pmatrix} \]

Now, taking the transpose gives us the adjoint:

\[ \text{adj}(A) = \begin{pmatrix} -1 & 1 & 8 \\ -8 & -1 & 5 \\ -8 & -7 & 4 \end{pmatrix} \]

#### Step 2.3: Calculate \(A^{-1}\)

Now we can find \(A^{-1}\):

\[ A^{-1} = \frac{1}{11} \begin{pmatrix} -1 & 1 & 8 \\ -8 & -1 & 5 \\ -8 & -7 & 4 \end{pmatrix} \]

### Step 3: Calculate \(\mathbf{x} = A^{-1} \mathbf{B}\)
Now we can find \(\mathbf{x}\):

\[ \mathbf{x} = A^{-1} \mathbf{B} = \frac{1}{11} \begin{pmatrix} -1 & 1 & 8 \\ -8 & -1 & 5 \\ -8 & -7 & 4 \end{pmatrix} \begin{pmatrix} 3 \\ 3 \\ 4 \end{pmatrix} \]

Calculating the product:

1. First row: \(-1 \cdot 3 + 1 \cdot 3 + 8 \cdot 4 = -3 + 3 + 32 = 32\)
2. Second row: \(-8 \cdot 3 - 1 \cdot 3 + 5 \cdot 4 = -24 - 3 + 20 = -7\)
3. Third row: \(-8 \cdot 3 - 7 \cdot 3 + 4 \cdot 4 = -24 - 21 + 16 = -29\)

Thus, we have:

\[ \mathbf{x} = \frac{1}{11} \begin{pmatrix} 32 \\ -7 \\ -29 \end{pmatrix} = \begin{pmatrix} \frac{32}{11} \\ -\frac{7}{11} \\ -\frac{29}{11} \end{pmatrix} \]

### Final Step: Result
The solutions are:

\[ x = \frac{32}{11}, \quad y = -\frac{7}{11}, \quad z = -\frac{29}{11} \]