Solve the following system of equations :
`{:((1)/(x)-(1)/(y)+(2)/(z)=4,,),((2)/(x)+(1)/(y)-(3)/(z)=0,,),((1)/(x)+(1)/(y)+(1)/(z)=2,x,y,z ne o):}`

Answer

To solve the given system of equations:

\[ \begin{align*} \frac{1}{x} - \frac{1}{y} + \frac{2}{z} &= 4 \quad (1) \\ \frac{2}{x} + \frac{1}{y} - \frac{3}{z} &= 0 \quad (2) \\ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} &= 2 \quad (3) \end{align*} \]

we will use the substitution method and matrix method. Let's denote:

\[ u = \frac{1}{x}, \quad v = \frac{1}{y}, \quad p = \frac{1}{z} \]

This transforms our equations into:

\[ \begin{align*} u - v + 2p &= 4 \quad (1') \\ 2u + v - 3p &= 0 \quad (2') \\ u + v + p &= 2 \quad (3') \end{align*} \]

### Step 1: Write the system in matrix form

We can express the system of equations in matrix form \(Ax = B\), where:

\[ A = \begin{pmatrix} 1 & -1 & 2 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{pmatrix}, \quad x = \begin{pmatrix} u \\ v \\ p \end{pmatrix}, \quad B = \begin{pmatrix} 4 \\ 0 \\ 2 \end{pmatrix} \]

### Step 2: Find the determinant of matrix A

To find the determinant of \(A\):

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

Calculating the minors:

\[ \begin{vmatrix} 1 & -3 \\ 1 & 1 \end{vmatrix} = (1)(1) - (-3)(1) = 1 + 3 = 4 \]

\[ \begin{vmatrix} 2 & -3 \\ 1 & 1 \end{vmatrix} = (2)(1) - (-3)(1) = 2 + 3 = 5 \]

\[ \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} = (2)(1) - (1)(1) = 2 - 1 = 1 \]

Putting it all together:

\[ \text{det}(A) = 1 \cdot 4 + 1 \cdot 5 + 2 \cdot 1 = 4 + 5 + 2 = 11 \]

### Step 3: Find the adjoint of matrix A

To find the adjoint, we need the cofactors of each element in \(A\):

\[ \text{Cofactor}(A) = \begin{pmatrix} 4 & -5 & 1 \\ 3 & -1 & -2 \\ 1 & 7 & 3 \end{pmatrix} \]

Taking the transpose gives us the adjoint:

\[ \text{adj}(A) = \begin{pmatrix} 4 & 3 & 1 \\ -5 & -1 & 7 \\ 1 & -2 & 3 \end{pmatrix} \]

### Step 4: Find the inverse of matrix A

The inverse of \(A\) is given by:

\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) = \frac{1}{11} \begin{pmatrix} 4 & 3 & 1 \\ -5 & -1 & 7 \\ 1 & -2 & 3 \end{pmatrix} \]

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

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

\[ x = \frac{1}{11} \begin{pmatrix} 4 & 3 & 1 \\ -5 & -1 & 7 \\ 1 & -2 & 3 \end{pmatrix} \begin{pmatrix} 4 \\ 0 \\ 2 \end{pmatrix} \]

Calculating the product:

\[ = \frac{1}{11} \begin{pmatrix} (4 \cdot 4) + (3 \cdot 0) + (1 \cdot 2) \\ (-5 \cdot 4) + (-1 \cdot 0) + (7 \cdot 2) \\ (1 \cdot 4) + (-2 \cdot 0) + (3 \cdot 2) \end{pmatrix} \]

Calculating each entry:

\[ = \frac{1}{11} \begin{pmatrix} 16 + 0 + 2 \\ -20 + 0 + 14 \\ 4 + 0 + 6 \end{pmatrix} = \frac{1}{11} \begin{pmatrix} 18 \\ -6 \\ 10 \end{pmatrix} \]

Thus, we have:

\[ x = \begin{pmatrix} \frac{18}{11} \\ -\frac{6}{11} \\ \frac{10}{11} \end{pmatrix} \]

### Step 6: Find values of \(x\), \(y\), and \(z\)

Since \(u = \frac{1}{x}\), \(v = \frac{1}{y}\), and \(p = \frac{1}{z}\):

\[ x = \frac{1}{u} = \frac{11}{18}, \quad y = \frac{1}{v} = -\frac{11}{6}, \quad z = \frac{1}{p} = \frac{11}{10} \]

### Final Answer

\[ x = \frac{11}{18}, \quad y = -\frac{11}{6}, \quad z = \frac{11}{10} \]