Solve the following equations, using inverse of a matrix :
`{:(3x-y+z=5),(2x-2y+3z=7),(x+y-z=-1):}`

Answer

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

### Step 1: Write the equations in matrix form
The given equations are:
1. \(3x - y + z = 5\)
2. \(2x - 2y + 3z = 7\)
3. \(x + y - z = -1\)

We can express this in matrix form as:
\[ A \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} = B \]
where
\[ A = \begin{pmatrix} 3 & -1 & 1 \\ 2 & -2 & 3 \\ 1 & 1 & -1 \end{pmatrix}, \quad B = \begin{pmatrix} 5 \\ 7 \\ -1 \end{pmatrix} \]

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

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

Now substituting back into the determinant formula:
\[ \text{det}(A) = 3(-1) + 1(-5) + 1(4) = -3 - 5 + 4 = -4 \]

### Step 3: Find the adjoint of matrix \(A\)
The adjoint of \(A\) is the transpose of the cofactor matrix. We calculate the cofactors for each element of \(A\):

1. \(C_{11} = \begin{vmatrix} -2 & 3 \\ 1 & -1 \end{vmatrix} = -1\)
2. \(C_{12} = -\begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix} = 5\)
3. \(C_{13} = \begin{vmatrix} 2 & -2 \\ 1 & 1 \end{vmatrix} = 4\)

4. \(C_{21} = -\begin{vmatrix} -1 & 1 \\ 1 & -1 \end{vmatrix} = 0\)
5. \(C_{22} = \begin{vmatrix} 3 & 1 \\ 1 & -1 \end{vmatrix} = -4\)
6. \(C_{23} = -\begin{vmatrix} 3 & -1 \\ 1 & 1 \end{vmatrix} = -2\)

7. \(C_{31} = \begin{vmatrix} -1 & 1 \\ -2 & 3 \end{vmatrix} = 1\)
8. \(C_{32} = -\begin{vmatrix} 3 & 1 \\ 2 & 3 \end{vmatrix} = -7\)
9. \(C_{33} = \begin{vmatrix} 3 & -1 \\ 2 & -2 \end{vmatrix} = -4\)

Thus, the cofactor matrix \(C\) is:
\[ C = \begin{pmatrix} -1 & 5 & 4 \\ 0 & -4 & -2 \\ 1 & -7 & -4 \end{pmatrix} \]

Now, taking the transpose to find the adjoint:
\[ \text{adj}(A) = C^T = \begin{pmatrix} -1 & 0 & 1 \\ 5 & -4 & -7 \\ 4 & -2 & -4 \end{pmatrix} \]

### Step 4: Calculate the inverse of matrix \(A\)
The inverse of \(A\) is given by:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) = -\frac{1}{4} \begin{pmatrix} -1 & 0 & 1 \\ 5 & -4 & -7 \\ 4 & -2 & -4 \end{pmatrix} \]

Thus,
\[ A^{-1} = \begin{pmatrix} \frac{1}{4} & 0 & -\frac{1}{4} \\ -\frac{5}{4} & 1 & \frac{7}{4} \\ -1 & \frac{1}{2} & 1 \end{pmatrix} \]

### Step 5: Solve for \(X\)
Now we can find \(X\) using:
\[ X = A^{-1}B \]
where \(B = \begin{pmatrix} 5 \\ 7 \\ -1 \end{pmatrix}\).

Calculating \(X\):
\[ X = \begin{pmatrix} \frac{1}{4} & 0 & -\frac{1}{4} \\ -\frac{5}{4} & 1 & \frac{7}{4} \\ -1 & \frac{1}{2} & 1 \end{pmatrix} \begin{pmatrix} 5 \\ 7 \\ -1 \end{pmatrix} \]

Calculating the first row:
\[ x = \frac{1}{4}(5) + 0(7) - \frac{1}{4}(-1) = \frac{5}{4} + \frac{1}{4} = \frac{6}{4} = \frac{3}{2} \]

Calculating the second row:
\[ y = -\frac{5}{4}(5) + 1(7) + \frac{7}{4}(-1) = -\frac{25}{4} + 7 - \frac{7}{4} = -\frac{25}{4} + \frac{28}{4} - \frac{7}{4} = -\frac{4}{4} = -1 \]

Calculating the third row:
\[ z = -1(5) + \frac{1}{2}(7) + 1(-1) = -5 + \frac{7}{2} - 1 = -5 - 1 + \frac{7}{2} = -6 + \frac{7}{2} = -\frac{12}{2} + \frac{7}{2} = -\frac{5}{2} \]

Thus, the solution is:
\[ X = \begin{pmatrix} \frac{3}{2} \\ -1 \\ -\frac{5}{2} \end{pmatrix} \]

### Final Answer:
The solution to the system of equations is:
\[ x = \frac{3}{2}, \quad y = -1, \quad z = -\frac{5}{2} \]

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