Solve the following system of equations :
`{:((1)/(x)-(1)/(y)+(2)/(z)=7),((3)/(x)+(4)/(y)-(5)/(z)=-5),((2)/(x)-(1)/(y)+(3)/(z)=12):}`

Answer

To solve the system of equations given by:

1. \(\frac{1}{x} - \frac{1}{y} + \frac{2}{z} = 7\)
2. \(\frac{3}{x} + \frac{4}{y} - \frac{5}{z} = -5\)
3. \(\frac{2}{x} - \frac{1}{y} + \frac{3}{z} = 12\)

we will use the matrix method. Let's denote:
- \(u = \frac{1}{x}\)
- \(v = \frac{1}{y}\)
- \(p = \frac{1}{z}\)

Thus, we can rewrite the equations as:

1. \(u - v + 2p = 7\)
2. \(3u + 4v - 5p = -5\)
3. \(2u - v + 3p = 12\)

### Step 1: Form the matrix equation \(AX = B\)

The matrix \(A\), the vector \(X\), and the vector \(B\) can be defined as follows:

\[ A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{bmatrix}, \quad X = \begin{bmatrix} u \\ v \\ p \end{bmatrix}, \quad B = \begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix} \]

### Step 2: Find the inverse of matrix \(A\)

To find \(X\), we need to calculate \(X = A^{-1}B\). First, we need to find the determinant of \(A\).

\[ \text{det}(A) = 1 \cdot (4 \cdot 3 - (-5)(-1)) - (-1)(3 \cdot 3 - (-5)(2)) + 2(3 \cdot (-1) - 4 \cdot 2) \]

Calculating each term:

1. \(1 \cdot (12 - 5) = 7\)
2. \(-(-1)(9 - 10) = 1\)
3. \(2(-3 - 8) = 2 \cdot (-11) = -22\)

So,

\[ \text{det}(A) = 7 + 1 - 22 = -14 \]

### Step 3: Calculate the adjoint of \(A\)

Next, we need to find the cofactor matrix and then the adjoint (transpose of the cofactor matrix).

Calculating the cofactors:

\[ \text{Cofactor}(A) = \begin{bmatrix} \text{det}\begin{bmatrix} 4 & -5 \\ -1 & 3 \end{bmatrix} & -\text{det}\begin{bmatrix} 3 & -5 \\ 2 & 3 \end{bmatrix} & \text{det}\begin{bmatrix} 3 & 4 \\ 2 & -1 \end{bmatrix} \\ -\text{det}\begin{bmatrix} -1 & 2 \\ -1 & 3 \end{bmatrix} & \text{det}\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix} & -\text{det}\begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \\ \text{det}\begin{bmatrix} -1 & 2 \\ 4 & -5 \end{bmatrix} & -\text{det}\begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix} & \text{det}\begin{bmatrix} 1 & -1 \\ 3 & 4 \end{bmatrix} \end{bmatrix} \]

Calculating each determinant:

1. \(4 \cdot 3 - (-5)(-1) = 12 - 5 = 7\)
2. \(3 \cdot 3 - (-5)(2) = 9 + 10 = 19\)
3. \(3 \cdot (-1) - 4 \cdot 2 = -3 - 8 = -11\)
4. \(-(-1)(3) - 2(-1) = 3 + 2 = 5\)
5. \(1 \cdot 3 - 2 \cdot 2 = 3 - 4 = -1\)
6. \(1 \cdot (-1) - (-1)(2) = -1 + 2 = 1\)
7. \(-1 \cdot (-5) - 2 \cdot 4 = 5 - 8 = -3\)
8. \(1 \cdot (-5) - 2 \cdot 3 = -5 - 6 = -11\)
9. \(1 \cdot 4 - (-1)(-1) = 4 - 1 = 3\)

Thus, the cofactor matrix is:

\[ \text{Cofactor}(A) = \begin{bmatrix} 7 & -19 & -11 \\ 5 & -1 & 1 \\ -3 & 11 & 3 \end{bmatrix} \]

Taking the transpose gives us the adjoint:

\[ \text{adj}(A) = \begin{bmatrix} 7 & 5 & -3 \\ -19 & -1 & 11 \\ -11 & 1 & 3 \end{bmatrix} \]

### Step 4: Calculate \(A^{-1}\)

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

\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) = \frac{1}{-14} \begin{bmatrix} 7 & 5 & -3 \\ -19 & -1 & 11 \\ -11 & 1 & 3 \end{bmatrix} \]

### Step 5: Multiply \(A^{-1}\) by \(B\)

Now we will compute \(X = A^{-1}B\):

\[ X = \frac{1}{-14} \begin{bmatrix} 7 & 5 & -3 \\ -19 & -1 & 11 \\ -11 & 1 & 3 \end{bmatrix} \begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix} \]

Calculating the multiplication:

1. First row: \(7 \cdot 7 + 5 \cdot (-5) + (-3) \cdot 12 = 49 - 25 - 36 = -12\)
2. Second row: \(-19 \cdot 7 + (-1) \cdot (-5) + 11 \cdot 12 = -133 + 5 + 132 = 4\)
3. Third row: \(-11 \cdot 7 + 1 \cdot (-5) + 3 \cdot 12 = -77 - 5 + 36 = -46\)

Thus,

\[ X = \frac{1}{-14} \begin{bmatrix} -12 \\ 4 \\ -46 \end{bmatrix} = \begin{bmatrix} \frac{12}{14} \\ \frac{-4}{14} \\ \frac{46}{14} \end{bmatrix} = \begin{bmatrix} \frac{6}{7} \\ -\frac{2}{7} \\ \frac{23}{7} \end{bmatrix} \]

### Step 6: Solve for \(x\), \(y\), and \(z\)

Recalling that:

- \(u = \frac{1}{x}\)
- \(v = \frac{1}{y}\)
- \(p = \frac{1}{z}\)

We have:

1. \(u = \frac{6}{7} \Rightarrow x = \frac{1}{u} = \frac{7}{6}\)
2. \(v = -\frac{2}{7} \Rightarrow y = \frac{1}{v} = -\frac{7}{2}\) (not valid as \(y\) cannot be negative)
3. \(p = \frac{23}{7} \Rightarrow z = \frac{1}{p} = \frac{7}{23}\)

### Final Result

The valid solutions for \(x\) and \(z\) are:

\[ x = \frac{7}{6}, \quad y = -\frac{7}{2} \text{ (not valid)}, \quad z = \frac{7}{23} \]