Solve the following system of linear equations using matrix method :
`1/(x)+1/(y)+1/(z)=9`
`2/(x)+5/(y)+7/(z)=52`
`2/(x)+1/(y)-1/(z)=0`

Answer

To solve the given system of linear equations using the matrix method, we first need to rewrite the equations in a more manageable form. The equations provided are:

1. \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 9 \)
2. \( \frac{2}{x} + \frac{5}{y} + \frac{7}{z} = 52 \)
3. \( \frac{2}{x} + \frac{1}{y} - \frac{1}{z} = 0 \)

### Step 1: Substitute Variables
Let:
- \( a = \frac{1}{x} \)
- \( b = \frac{1}{y} \)
- \( c = \frac{1}{z} \)

Now, we can rewrite the equations as:
1. \( a + b + c = 9 \)
2. \( 2a + 5b + 7c = 52 \)
3. \( 2a + b - c = 0 \)

### Step 2: Write in Matrix Form
We can express the system of equations in matrix form \( Ax = B \), where:

\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 5 & 7 \\ 2 & 1 & -1 \end{bmatrix}, \quad x = \begin{bmatrix} a \\ b \\ c \end{bmatrix}, \quad B = \begin{bmatrix} 9 \\ 52 \\ 0 \end{bmatrix} \]

### Step 3: Calculate the Determinant of Matrix A
To find the inverse of matrix \( A \), we first need to calculate its determinant \( |A| \):

\[ |A| = 1 \cdot \begin{vmatrix} 5 & 7 \\ 1 & -1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & 7 \\ 2 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 5 \\ 2 & 1 \end{vmatrix} \]

Calculating the minors:
1. \( \begin{vmatrix} 5 & 7 \\ 1 & -1 \end{vmatrix} = (5)(-1) - (7)(1) = -5 - 7 = -12 \)

2. \( \begin{vmatrix} 2 & 7 \\ 2 & -1 \end{vmatrix} = (2)(-1) - (7)(2) = -2 - 14 = -16 \)

3. \( \begin{vmatrix} 2 & 5 \\ 2 & 1 \end{vmatrix} = (2)(1) - (5)(2) = 2 - 10 = -8 \)

Putting it all together:
\[ |A| = 1(-12) - 1(-16) + 1(-8) = -12 + 16 - 8 = -4 \]

### Step 4: Calculate the Inverse of Matrix A
The inverse of \( A \) can be calculated using the adjoint method. First, we find the matrix of cofactors and then the adjoint.

The cofactor matrix \( C \) is calculated as follows:
\[ C = \begin{bmatrix} -12 & 16 & -8 \\ 8 & -4 & -2 \\ -3 & -4 & 3 \end{bmatrix} \]

The adjoint \( \text{adj}(A) \) is the transpose of the cofactor matrix:
\[ \text{adj}(A) = \begin{bmatrix} -12 & 8 & -3 \\ 16 & -4 & -4 \\ -8 & -2 & 3 \end{bmatrix} \]

Now, the inverse \( A^{-1} \) is given by:
\[ A^{-1} = \frac{1}{|A|} \text{adj}(A) = -\frac{1}{4} \begin{bmatrix} -12 & 8 & -3 \\ 16 & -4 & -4 \\ -8 & -2 & 3 \end{bmatrix} \]

### Step 5: Multiply \( A^{-1} \) by \( B \)
Now we can find \( x \) by calculating \( A^{-1}B \):
\[ x = A^{-1}B = -\frac{1}{4} \begin{bmatrix} -12 & 8 & -3 \\ 16 & -4 & -4 \\ -8 & -2 & 3 \end{bmatrix} \begin{bmatrix} 9 \\ 52 \\ 0 \end{bmatrix} \]

Calculating each element:
1. First row: \( -\frac{1}{4}((-12)(9) + (8)(52) + (-3)(0)) = -\frac{1}{4}(-108 + 416) = -\frac{1}{4}(308) = -77 \)
2. Second row: \( -\frac{1}{4}((16)(9) + (-4)(52) + (-4)(0)) = -\frac{1}{4}(144 - 208) = -\frac{1}{4}(-64) = 16 \)
3. Third row: \( -\frac{1}{4}((-8)(9) + (-2)(52) + (3)(0)) = -\frac{1}{4}(-72 - 104) = -\frac{1}{4}(-176) = 44 \)

Thus, we have:
\[ x = \begin{bmatrix} 77 \\ 16 \\ 44 \end{bmatrix} \]

### Step 6: Back Substitute to Find \( x, y, z \)
Since \( a = 77, b = 16, c = 44 \):
- \( x = \frac{1}{a} = \frac{1}{77} \)
- \( y = \frac{1}{b} = \frac{1}{16} \)
- \( z = \frac{1}{c} = \frac{1}{44} \)

### Final Solution
Thus, the solution to the system of equations is:
\[ x = \frac{1}{77}, \quad y = \frac{1}{16}, \quad z = \frac{1}{44} \]