Solve the following equations, using inverse of a matrix :
`{:(5x-y+z=4),(3x+2y-5z=2),(x+3y-2z=5):}`

Answer

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

### Step 1: Write the system of equations in matrix form
The given equations are:
1. \(5x - y + z = 4\)
2. \(3x + 2y - 5z = 2\)
3. \(x + 3y - 2z = 5\)

We can express these equations in the form of \(AX = B\), where:
\[ A = \begin{bmatrix} 5 & -1 & 1 \\ 3 & 2 & -5 \\ 1 & 3 & -2 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 4 \\ 2 \\ 5 \end{bmatrix} \]

### Step 2: Find the determinant of matrix \(A\)
To find the inverse of matrix \(A\), we first need to calculate its determinant (\(|A|\)):
\[ |A| = 5 \begin{vmatrix} 2 & -5 \\ 3 & -2 \end{vmatrix} - (-1) \begin{vmatrix} 3 & -5 \\ 1 & -2 \end{vmatrix} + 1 \begin{vmatrix} 3 & 2 \\ 1 & 3 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \(\begin{vmatrix} 2 & -5 \\ 3 & -2 \end{vmatrix} = (2)(-2) - (3)(-5) = -4 + 15 = 11\)
2. \(\begin{vmatrix} 3 & -5 \\ 1 & -2 \end{vmatrix} = (3)(-2) - (1)(-5) = -6 + 5 = -1\)
3. \(\begin{vmatrix} 3 & 2 \\ 1 & 3 \end{vmatrix} = (3)(3) - (1)(2) = 9 - 2 = 7\)

Now substituting back into the determinant formula:
\[ |A| = 5(11) + 1(-1) + 1(7) = 55 - 1 + 7 = 61 \]

### Step 3: Find the adjoint of matrix \(A\)
The adjoint of \(A\) is the transpose of the cofactor matrix. We will calculate the cofactor matrix first.

Cofactors:
1. \(C_{11} = \begin{vmatrix} 2 & -5 \\ 3 & -2 \end{vmatrix} = 11\)
2. \(C_{12} = -\begin{vmatrix} 3 & -5 \\ 1 & -2 \end{vmatrix} = 1\)
3. \(C_{13} = \begin{vmatrix} 3 & 2 \\ 1 & 3 \end{vmatrix} = 7\)
4. \(C_{21} = -\begin{vmatrix} -1 & 1 \\ 3 & -2 \end{vmatrix} = -(-1)(-2) - (1)(3) = -2 - 3 = -5\)
5. \(C_{22} = \begin{vmatrix} 5 & 1 \\ 1 & -2 \end{vmatrix} = (5)(-2) - (1)(1) = -10 - 1 = -11\)
6. \(C_{23} = -\begin{vmatrix} 5 & -1 \\ 1 & 3 \end{vmatrix} = -((5)(3) - (-1)(1)) = -15 + 1 = -14\)
7. \(C_{31} = \begin{vmatrix} -1 & 1 \\ 2 & -5 \end{vmatrix} = (-1)(-5) - (1)(2) = 5 - 2 = 3\)
8. \(C_{32} = -\begin{vmatrix} 5 & 1 \\ 3 & -5 \end{vmatrix} = -((5)(-5) - (1)(3)) = -(-25 - 3) = 28\)
9. \(C_{33} = \begin{vmatrix} 5 & -1 \\ 3 & 2 \end{vmatrix} = (5)(2) - (-1)(3) = 10 + 3 = 13\)

Thus, the cofactor matrix is:
\[ C = \begin{bmatrix} 11 & 1 & 7 \\ -5 & -11 & -14 \\ 3 & 28 & 13 \end{bmatrix} \]

The adjoint of \(A\) is the transpose of the cofactor matrix:
\[ \text{adj}(A) = \begin{bmatrix} 11 & -5 & 3 \\ 1 & -11 & 28 \\ 7 & -14 & 13 \end{bmatrix} \]

### Step 4: Find the inverse of matrix \(A\)
The inverse of matrix \(A\) is given by:
\[ A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A) = \frac{1}{61} \begin{bmatrix} 11 & -5 & 3 \\ 1 & -11 & 28 \\ 7 & -14 & 13 \end{bmatrix} \]

### Step 5: Multiply \(A^{-1}\) by \(B\) to find \(X\)
Now we will multiply \(A^{-1}\) by \(B\):
\[ X = A^{-1}B = \frac{1}{61} \begin{bmatrix} 11 & -5 & 3 \\ 1 & -11 & 28 \\ 7 & -14 & 13 \end{bmatrix} \begin{bmatrix} 4 \\ 2 \\ 5 \end{bmatrix} \]

Calculating the product:
1. First row: \(11 \cdot 4 + (-5) \cdot 2 + 3 \cdot 5 = 44 - 10 + 15 = 49\)
2. Second row: \(1 \cdot 4 + (-11) \cdot 2 + 28 \cdot 5 = 4 - 22 + 140 = 122\)
3. Third row: \(7 \cdot 4 + (-14) \cdot 2 + 13 \cdot 5 = 28 - 28 + 65 = 65\)

Thus,
\[ X = \frac{1}{61} \begin{bmatrix} 49 \\ 122 \\ 65 \end{bmatrix} \]

### Step 6: Final values of \(x\), \(y\), and \(z\)
Now we can find \(x\), \(y\), and \(z\):
\[ x = \frac{49}{61}, \quad y = \frac{122}{61}, \quad z = \frac{65}{61} \]

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