====== 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. 

=====Question 4(i)=====
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:
$$\boxed{x_1 = \frac{13}{3}, \quad x_2 = \frac{26}{7}, \quad x_3 = \frac{62}{21}.}$$




=====Question 4(ii)=====
 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:
$$\boxed{x_1 = \frac{22}{11}, \quad x_2 = \frac{2}{11}, x_3=0}$$



=====Question 4(iii)=====
 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.


=====Question 4(iv)=====
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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