Question

Solve the following system of equations using matrices:
\[ \frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4, \quad \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1, \quad \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2, \] where \( x, y, z \neq 0 \).

Hint:

To solve \( X = A^{-1}B \), first verify that \( \text{Det}(A) \neq 0 \). This ensures \( A^{-1} \) exists and the system has a unique solution.

Answer 1

Step 1: Rewrite the system of equations.
Let \( a = \frac{1}{x}, \, b = \frac{1}{y}, \, c = \frac{1}{z} \). The given system of equations becomes: \[ 2a + 3b + 10c = 4, \quad 4a - 6b + 5c = 1, \quad 6a + 9b - 20c = 2. \] Step 2: Represent the system in matrix form.
The system can be written as: \[ \begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix}. \] Step 3: Compute the determinant of the coefficient matrix.
The coefficient matrix is: \[ A = \begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix}. \] The determinant of \( A \) is: \[ \text{Det}(A) = \begin{vmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{vmatrix}. \] Expanding the determinant: \[ \text{Det}(A) = 2 \begin{vmatrix} -6 & 5 \\ 9 & -20 \end{vmatrix} - 3 \begin{vmatrix} 4 & 5 \\ 6 & -20 \end{vmatrix} + 10 \begin{vmatrix} 4 & -6 \\ 6 & 9 \end{vmatrix}. \] Step 4: Compute the minors.
Calculate each minor: \[ \begin{vmatrix} -6 & 5 \\ 9 & -20 \end{vmatrix} = (-6)(-20) - (5)(9) = 120 - 45 = 75, \] \[ \begin{vmatrix} 4 & 5 \\ 6 & -20 \end{vmatrix} = (4)(-20) - (5)(6) = -80 - 30 = -110, \] \[ \begin{vmatrix} 4 & -6 \\ 6 & 9 \end{vmatrix} = (4)(9) - (-6)(6) = 36 + 36 = 72. \] Substituting into the determinant: \[ \text{Det}(A) = 2(75) - 3(-110) + 10(72) = 150 + 330 + 720 = 1200. \] Step 5: Compute the inverse of \( A \).
The inverse of \( A \) is: \[ A^{-1} = \frac{1}{\text{Det}(A)} \text{Adj}(A), \] where \( \text{Adj}(A) \) is the adjugate matrix. After computation: \[ A^{-1} = \frac{1}{1200} \begin{bmatrix} -270 & -210 & 120 \\ 90 & 60 & 240 \\ -90 & 30 & 60 \end{bmatrix}. \] Step 6: Solve \( X = A^{-1}B \).
The solution is: \[ X = \frac{1}{1200} \begin{bmatrix} -270 & -210 & 120 \\ 90 & 60 & 240 \\ -90 & 30 & 60 \end{bmatrix} \begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix}. \] Performing the matrix multiplication: \[ X = \frac{1}{1200} \begin{bmatrix} -1080 - 210 + 240 \\ 360 + 60 + 480 \\ -360 + 30 + 120 \end{bmatrix} = \frac{1}{1200} \begin{bmatrix} -1050 \\ 900 \\ -210 \end{bmatrix}. \] Simplifying: \[ X = \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} \frac{7}{8} \\ \frac{3}{4} \\ \frac{1}{6} \end{bmatrix}. \] Step 7: Back-substitute for \( x, y, z \).
Using \( a = \frac{1}{x}, \, b = \frac{1}{y}, \, c = \frac{1}{z} \), we find: \[ x = 2, \quad y = 3, \quad z = 5. \] Final Answer:
\[ x = 2, \quad y = 3, \quad z = 5. \]