English

Solve the Following System of Equations by Matrix Method: 2 X − 3 Y + 3 Z = 10 1 X + 1 Y + 1 Z = 10 3 X − 1 Y + 2 Z = 13

Advertisements
Advertisements

Question

Solve the following system of equations by matrix method:
\[\frac{2}{x} - \frac{3}{y} + \frac{3}{z} = 10\]
\[\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 10\]
\[\frac{3}{x} - \frac{1}{y} + \frac{2}{z} = 13\]

Advertisements

Solution

\[\text{ Let }\frac{1}{x}\text{ be }a,\frac{1}{y}\text{ be }b \text{ and}\frac{1}{z}\text{ be }c.\]
Here,
\[A = \begin{bmatrix}2 & - 3 & 3 \\ 1 & 1 & 1 \\ 3 & - 1 & 2\end{bmatrix}\]
\[\left| A \right| = \begin{vmatrix}2 & - 3 & 3 \\ 1 & 1 & 1 \\ 3 & - 1 & 2\end{vmatrix}\]
\[ = 2\left( 2 + 1 \right) + 3\left( 2 - 3 \right) + 3( - 1 - 3)\]
\[ = 6 - 3 - 12\]
\[ = - 9\]
\[ {\text{ Let }C}_{ij} {\text{be the cofactors of the elements a}}_{ij}\text{ in }A\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}1 & 1 \\ - 1 & 2\end{vmatrix} = 3, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}1 & 1 \\ 3 & 2\end{vmatrix} = 1 , C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}1 & 1 \\ 3 & - 1\end{vmatrix} = - 4\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 3 & 3 \\ - 1 & 2\end{vmatrix} = 3 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}2 & 3 \\ 3 & 2\end{vmatrix} = - 5, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}2 & - 3 \\ 3 & - 1\end{vmatrix} = - 7\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 3 & 3 \\ 1 & 1\end{vmatrix} = - 6 , C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}2 & 3 \\ 1 & 1\end{vmatrix} = 1, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}2 & - 3 \\ 1 & 1\end{vmatrix} = 5\]
\[adj A = \begin{bmatrix}3 & 1 & - 4 \\ 3 & - 5 & - 7 \\ - 6 & 1 & 5\end{bmatrix}^T \]
\[ = \begin{bmatrix}3 & 3 & - 6 \\ 1 & - 5 & 1 \\ - 4 & - 7 & 5\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{- 9}\begin{bmatrix}3 & 3 & - 6 \\ 1 & - 5 & 1 \\ - 4 & - 7 & 5\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \begin{bmatrix}a \\ b \\ c\end{bmatrix} = \frac{1}{- 9}\begin{bmatrix}3 & 3 & - 6 \\ 1 & - 5 & 1 \\ - 4 & - 7 & 5\end{bmatrix}\begin{bmatrix}10 \\ 10 \\ 13\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}a \\ b \\ c\end{bmatrix} = \frac{1}{- 9}\begin{bmatrix}30 + 30 - 78 \\ 10 - 50 + 13 \\ - 40 - 70 + 65\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}a \\ b \\ c\end{bmatrix} = \frac{1}{- 9}\begin{bmatrix}- 18 \\ - 27 \\ - 45\end{bmatrix}\]
\[ \Rightarrow x = \frac{1}{a} = \frac{- 9}{- 18}, y = \frac{1}{b} = \frac{- 9}{- 27}\text{ and }z = \frac{1}{c} = \frac{- 9}{- 45}\]
\[ \therefore x = \frac{1}{a} = \frac{1}{2}, y = \frac{1}{b} = \frac{1}{3}\text{ and }z = \frac{1}{c} = \frac{1}{5}\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 7: Solution of Simultaneous Linear Equations - Exercise 8.1 [Page 14]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 7 Solution of Simultaneous Linear Equations
Exercise 8.1 | Q 2.05 | Page 14
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×