Question 4, Exercise 2.6

Solutions of Question 4 of Exercise 2.6 of Unit 02: Matrices and Determinants. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Solve the system of linear equation by Gauss-Jordan method.
$2 x_{1}-x_{2}-x_{3}=2$
$3 x_{1}-4 x_{2}+3 x_{3}=7$
$4 x_{1}+2 x_{2}-5 x_{3}=10$

Solution.

$$\begin{align*} 2x_1 - x_2 - x_3 &= 2, \\ 3x_1 - 4x_2 + 3x_3 &= 7, \\ 4x_1 + 2x_2 - 5x_3 &= 10, \end{align*}$$

The associative augment matrix: $$\begin{align*} A_b &= \begin{bmatrix} 2 & -1 & -1 & : & 2 \\ 3 & -4 & 3 & : & 7 \\ 4 & 2 & -5 & : & 10 \end{bmatrix}\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2} & \vert & 1 \\ 3 & -4 & 3 & \vert & 7 \\ 4 & 2 & -5 & \vert & 10 \end{bmatrix}\quad \dfrac{1}{2}\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2} & : & 1 \\ 0 & -\frac{5}{2} & \frac{9}{2} & : & 4 \\ 4 & 2 & -5 & : & 10 \end{bmatrix}\quad R_2 - 3R_1\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2} & : & 1 \\ 0 & -\frac{5}{2} & \frac{9}{2} & : & 4 \\ 0 & 4 & -3 & : & 6 \end{bmatrix}\quad R_3 - 4R_1\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2} & : & 1 \\ 0 & 1 & -\frac{9}{5} & : & -\frac{8}{5} \\ 0 & 4 & -3 & : & 6 \end{bmatrix}\quad -\frac{2}{5} R_2\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2} & : & 1 \\ 0 & 1 & -\frac{9}{5} & : & -\frac{8}{5} \\ 0 & 0 & \frac{21}{5} & : & \frac{62}{5} \end{bmatrix}\quad R_3 - 4R_2 \\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2} & : & 1 \\ 0 & 1 & -\frac{9}{5} & : & -\frac{8}{5} \\ 0 & 0 & 1 & : & \frac{62}{21} \end{bmatrix}\quad \frac{5}{21} R_3\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2} & \vert & 1 \\ 0 & 1 & 0 & \vert & \frac{26}{7} \\ 0 & 0 & 1 & \vert & \frac{62}{21} \end{bmatrix}\quad R_2 + \frac{9}{5}R_3\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{1}{2} & 0 & \vert & \frac{52}{21} \\ 0 & 1 & 0 & \vert & \frac{26}{7} \\ 0 & 0 & 1 & \vert & \frac{62}{21} \end{bmatrix}\quad R_1 + \frac{1}{2}R_3 \\ &\sim \text{R} \begin{bmatrix} 1 & 0 & 0 & \vert & \frac{13}{3} \\ 0 & 1 & 0 & \vert & \frac{26}{7} \\ 0 & 0 & 1 & \vert & \frac{62}{21} \end{bmatrix}\quad R_1 + \frac{1}{2}R_2\end{align*}$$ Thus, the solution to the system of equations is: $$\boxed{x_1 = \frac{13}{3}, \quad x_2 = \frac{26}{7}, \quad x_3 = \frac{62}{21}.}$$

Solve the system of linear equation by Gauss-Jordan method.
$2 x_{1}-3 x_{2}+7 x_{3}=1$
$4 x_{1}+5 x_{2}-3 x_{3}=4$
$10 x_{1}-4 x_{2}+18 x_{3}=7$

Solution.

$$\begin{align*} 2x_1 - 3x_2 + 7x_3 &= 1, \\ 4x_1 + 5x_2 - 3x_3 &= 4, \\ 10x_1 - 4x_2 + 18x_3 &= 7. \end{align*}$$

The associated augmented matrix is: $$\begin{align*} A_b &= \begin{bmatrix} 2 & -3 & 7 & : & 1 \\ 4 & 5 & -3 & : & 4 \\ 10 & -4 & 18 & : & 7 \end{bmatrix}\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{3}{2} & \frac{7}{2} & : & \frac{1}{2} \\ 4 & 5 & -3 & : & 4 \\ 10 & -4 & 18 & : & 7 \end{bmatrix}\quad \text{(Divide } R_1 \text{ by 2)}\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{3}{2} & \frac{7}{2} & : & \frac{1}{2} \\ 0 & 11 & -17 & : & 2 \\ 10 & -4 & 18 & : & 7 \end{bmatrix}\quad \text{(} R_2 \text{ - 4}R_1)\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{3}{2} & \frac{7}{2} & : & \frac{1}{2} \\ 0 & 11 & -17 & : & 2 0 & 11 & -17 & : & 2 \end{bmatrix}\quad \text{(} R_3 \text{ - 10}R_1)\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{3}{2} & \frac{7}{2} & : & \frac{1}{2} \\ 0 & 1 & -\frac{17}{11} & : & \frac{2}{11} \\ 0 & 0 & 0 & : & 0 \end{bmatrix}\quad \text{(Divide } R_2 \text{ by 11 and } R_3 - R_2). \end{align*}$$ There is no value of $x$. Then

$x_3 = 0$.

From the second row: $$\begin{align*} x_2 & = \frac{2}{11} \end{align*}$$ From the first row: $$\begin{align*} &x_1 - \frac{3}{2}x_2 + \frac{7}{2}x_3 = \frac{1}{2}\\ \Rightarrow &x_1 = \frac{1}{2} + \frac{3}{2}\frac{2}{11}\\ \Rightarrow & x_1= \frac{1}{2} + \frac{3}{11}\\ \Rightarrow & x_1= \frac{17}{22}\end{align*}$$ Thus, the general solution is: $$\boxed{x_1 = \frac{22}{11}, \quad x_2 = \frac{2}{11}, x_3=0}$$

Solve the system of linear equation by Gauss-Jordan method.
$x_{1}+x_{2}+x_{3}=3$
$2 x_{1}-3 x_{2}+2 x_{3}=7$
$4 x_{1}+2 x_{2}-5 x_{3}=10$

Solution. Do yourself.

Solve the system of linear equation by Gauss-Jordan method.
$2 x_{1}-7 x_{2}+10 x_{3}=1$
$x_{1}+2 x_{2}-4 x_{3}=8$
$2 x_{1}-11 x_{2}+13 x_{3}=7$

Solution. Do yourself.

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