Solve the following equations, using inverse of a matrix :
`{:(3x-2y+3z=8),(2x+y-z=1),(4x-3y+2z=4):}`

Answer

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

### Step 1: Write the equations in matrix form
The given equations are:
1. \(3x - 2y + 3z = 8\)
2. \(2x + y - z = 1\)
3. \(4x - 3y + 2z = 4\)

We can express this in the matrix form \(AX = B\), where:
\[ A = \begin{pmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 8 \\ 1 \\ 4 \end{pmatrix} \]

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

#### Step 2.1: Calculate the determinant of \(A\)
The determinant of a \(3 \times 3\) matrix is calculated as follows:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix \(A\):
\[ \text{det}(A) = 3(1 \cdot 2 - (-1) \cdot (-3)) - (-2)(2 \cdot 2 - (-1) \cdot 4) + 3(2 \cdot (-3) - 1 \cdot 4) \]
Calculating each term:
- First term: \(3(2 - 3) = 3(-1) = -3\)
- Second term: \(-2(4 - 4) = -2(0) = 0\)
- Third term: \(3(-6 - 4) = 3(-10) = -30\)

So,
\[ \text{det}(A) = -3 + 0 - 30 = -33 \]

#### Step 2.2: Find the adjoint of \(A\)
The adjoint of a matrix is the transpose of the cofactor matrix. We will calculate the cofactors for each element of \(A\).

Calculating cofactors:
- \(C_{11} = \text{det}\begin{pmatrix} 1 & -1 \\ -3 & 2 \end{pmatrix} = (1)(2) - (-1)(-3) = 2 - 3 = -1\)
- \(C_{12} = -\text{det}\begin{pmatrix} 2 & -1 \\ 4 & 2 \end{pmatrix} = -(2 \cdot 2 - (-1)(4)) = -(4 + 4) = -8\)
- \(C_{13} = \text{det}\begin{pmatrix} 2 & 1 \\ 4 & -3 \end{pmatrix} = (2)(-3) - (1)(4) = -6 - 4 = -10\)

Continuing this process for all elements, we find the cofactor matrix \(C\):
\[ C = \begin{pmatrix} -1 & -8 & -10 \\ -1 & 10 & 6 \\ -1 & -8 & 7 \end{pmatrix} \]

Now, take the transpose to find the adjoint:
\[ \text{adj}(A) = C^T = \begin{pmatrix} -1 & -1 & -1 \\ -8 & 10 & -8 \\ -10 & 6 & 7 \end{pmatrix} \]

#### Step 2.3: Calculate \(A^{-1}\)
Using the formula \(A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A)\):
\[ A^{-1} = \frac{1}{-33} \begin{pmatrix} -1 & -1 & -1 \\ -8 & 10 & -8 \\ -10 & 6 & 7 \end{pmatrix} = \begin{pmatrix} \frac{1}{33} & \frac{1}{33} & \frac{1}{33} \\ \frac{8}{33} & -\frac{10}{33} & \frac{8}{33} \\ \frac{10}{33} & -\frac{6}{33} & -\frac{7}{33} \end{pmatrix} \]

### Step 3: Solve for \(X\)
Now, we can find \(X\) using \(X = A^{-1}B\):
\[ X = \begin{pmatrix} \frac{1}{33} & \frac{1}{33} & \frac{1}{33} \\ \frac{8}{33} & -\frac{10}{33} & \frac{8}{33} \\ \frac{10}{33} & -\frac{6}{33} & -\frac{7}{33} \end{pmatrix} \begin{pmatrix} 8 \\ 1 \\ 4 \end{pmatrix} \]

Calculating the multiplication:
1. First row: \(\frac{1}{33}(8) + \frac{1}{33}(1) + \frac{1}{33}(4) = \frac{8 + 1 + 4}{33} = \frac{13}{33}\)
2. Second row: \(\frac{8}{33}(8) - \frac{10}{33}(1) + \frac{8}{33}(4) = \frac{64 - 10 + 32}{33} = \frac{86}{33}\)
3. Third row: \(\frac{10}{33}(8) - \frac{6}{33}(1) - \frac{7}{33}(4) = \frac{80 - 6 - 28}{33} = \frac{46}{33}\)

Thus, we have:
\[ X = \begin{pmatrix} \frac{13}{33} \\ \frac{86}{33} \\ \frac{46}{33} \end{pmatrix} \]

### Final Solution
The values of \(x\), \(y\), and \(z\) are:
\[ x = \frac{13}{33}, \quad y = \frac{86}{33}, \quad z = \frac{46}{33} \]