Question

Solve the following system of equations by matrix method:

\[\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4, \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1, \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2; x, y, z \neq 0\]

 

Answer 1

\[\text{ Let }\frac{1}{x}\text{ be }a,\frac{1}{y}\text{ be } b \text{ and }\frac{1}{z}\text{ be }c.\]
Here,
\[A = \begin{bmatrix}2 & 3 & 10 \\ 4 & - 6 & 5 \\ 6 & 9 & - 20\end{bmatrix}\]
\[\left| A \right| = \begin{vmatrix}2 & 3 & 10 \\ 4 & - 6 & 5 \\ 6 & 9 & - 20\end{vmatrix}\]
\[ = 2\left( 120 - 45 \right) - 3\left( - 80 - 30 \right) + 10(36 + 36)\]
\[ = 150 + 330 + 720\]
\[ = 1200\]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 6 & 5 \\ 9 & - 20\end{vmatrix} = 75, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}4 & 5 \\ 6 & - 20\end{vmatrix} = 110, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}4 & - 6 \\ 6 & 9\end{vmatrix} = 72\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}3 & 10 \\ 9 & - 20\end{vmatrix} = 150, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}2 & 10 \\ 6 & - 20\end{vmatrix} = - 100, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}2 & 3 \\ 6 & 9\end{vmatrix} = 0\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}3 & 10 \\ - 6 & 5\end{vmatrix} = 75, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}2 & 10 \\ 4 & 5\end{vmatrix} = 30, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}2 & 3 \\ 4 & - 6\end{vmatrix} = - 24\]
\[adj A = \begin{bmatrix}75 & 110 & 72 \\ 150 & - 100 & 0 \\ 75 & 30 & - 24\end{bmatrix}^T \]
\[ = \begin{bmatrix}75 & 150 & 75 \\ 110 & - 100 & 30 \\ 72 & 0 & - 24\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{1200} \begin{bmatrix}75 & 150 & 75 \\ 110 & - 100 & 30 \\ 72 & 0 & - 24\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \begin{bmatrix}a \\ b \\ c\end{bmatrix} = \frac{1}{1200} \begin{bmatrix}75 & 150 & 75 \\ 110 & - 100 & 30 \\ 72 & 0 & - 24\end{bmatrix}\begin{bmatrix}4 \\ 1 \\ 2\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}a \\ b \\ c\end{bmatrix} = \frac{1}{1200}\begin{bmatrix}300 + 150 + 150 \\ 440 - 100 + 60 \\ 288 - 48\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}a \\ b \\ c\end{bmatrix} = \frac{1}{1200}\begin{bmatrix}600 \\ 400 \\ 240\end{bmatrix}\]
\[ \Rightarrow x = \frac{1}{a} = \frac{1200}{600}, y = \frac{1}{b} = \frac{1200}{400}\text{ and }z = \frac{1}{c} = \frac{1200}{240}\]
\[ \therefore x = 2, y = 3\text{ and }z = 5\]