====== Question 3, Exercise 2.6 ======
Solutions of Question 3 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 3(i)=====
Solve the system of linear equation by Gauss elimination method.\\
$2 x+3 y+4 z=2$\\
$2 x+y+z=5$\\
$3 x-2 y+z=-3$\\

** Solution. **
Given the system of equations:
\begin{align*}
\begin{aligned}
2x + 3y + 4z &= 2 \\
2x + y + z &= 5 \\
3x - 2y + z &= -3
\end{aligned}\end{align*}
	
The associated augmented matrix is:
\begin{align*}
A_{b} &=\quad \left[\begin{array}{cccc}
2 & 3 & 4 & 2 \\
2 & 1 & 1 & 5 \\
3 & -2 & 1 & -3
\end{array}\right]\\
& \sim \text{R}\left[\begin{array}{cccc}
2 & 3 & 4 & 2 \\
0 & -2 & -3 & 3 \\
0 & -\frac{13}{2} & -5 & -6
\end{array}\right]\quad R_2 - R_1 \text{and}\quad R_3 - \frac{3}{2}R_1\\	
& \sim \text{R}\left[\begin{array}{cccc}
2 & 3 & 4 & 2 \\
0 & -2 & -3 & 3 \\
0 & 0 & \frac{19}{4} & -\frac{63}{4}
\end{array}\right]\quad R_3 - \frac{13}{4} R_2 \end{align*}
Now \begin{align*}
\frac{19}{4}z& = -\frac{63}{4}\\
\Rightarrow z& = -\frac{63}{19}\end{align*}
From the second row, we have:
\begin{align*}
&- 2y - 3z = 3\\
\Rightarrow &-2y - 2(-\frac{63}{19}) = 3 \\
\Rightarrow &-2y = -\frac{132}{19} \\
\Rightarrow &y = \frac{66}{19}\end{align*}	
Finally, from the first row, we have:
\begin{align*}
2x + 3y + 4z &= 2 \\
2x + 3(\frac{66}{19}) + 4(-\frac{63}{19}) &= 2 \\
2x +\frac{198}{18} -\frac{252}{19} &= 2 \\
2x - \frac{54}{19} &= 2 \\
2x &= 2 + \frac{54}{19}  \\
2x &= \frac{92}{19}\\
x &= \frac{46}{19}	
\end{align*}
	
Therefore, the solution to the system is:
	
$$x = \frac{46}{19}, \quad y = \frac{66}{19}, \quad z = -\frac{63}{19}$$



=====Question 3(ii)=====
Solve the system of linear equation by Gauss elimination method.\\
$5 x-2 y+z=2$\\
$2 x+2 y+6 z=1$\\
$3 x-4 y-5 z=3$\\

** Solution. **
Given the system of equations:
\begin{align*}
5x - 2y + z &= 2 \quad \cdots (i) \\
2x + 2y + 6z &= 1 \quad \cdots (ii) \\
3x - 4y - 5z &= 3 \quad \cdots (iii)
\end{align*}
The associated augmented matrix is:
\begin{align*}	
A_b& =\quad \left[\begin{array}{cccc}
5 & -2 & 1 & 2 \\
2 & 2 & 6 & 1 \\
3 & -4 & -5 & 3
\end{array}\right]\\
&\sim \text{R}\left[\begin{array}{cccc}
2 & 2 & 6 & 1 \\
5 & -2 & 1 & 2 \\
3 & -4 & -5 & 3
\end{array}\right]\quad R_1 \sim R_2\\
&\sim \text{R}\left[\begin{array}{cccc}
2 & 2 & 6 & 1 \\
0 & -7 & -14 & \frac{1}{2} \\
0 & -7 & -14 & \frac{7}{2}
\end{array}\right]\quad R_2 - \frac{5}{2}R_1\text{and}\quad R_3 - \frac{3}{2}R_1\\
&\sim \text{R}\left[\begin{array}{cccc}
2 & 2 & 6 & 1 \\
0 & -7 & -14 & \frac{1}{2} \\
0 & 0 & 0 & 3
\end{array}\right]\quad R_3 - R_2\end{align*}
Since the last row corresponds to \(0x + 0y + 0z = 3\), which is inconsistent, the system of equations has no solution.




=====Question 3(iii)=====
Solve the system of linear equation by Gauss elimination method.\\
$2 x+z=2$\\
$2 y-z=3$\\
$x+3 y=5$\\

** Solution. **
Given the system of equations:
\begin{align*}
2x + z &= 2 \quad \cdots (i) \\
2y - z &= 3 \quad \cdots (ii) \\
x + 3y &= 5 \quad \cdots (iii)
\end{align*}
The associated augmented matrix is:
\begin{align*}
A_b &= \left[\begin{array}{cccc}
2 & 0 & 1 & 2 \\
0 & 2 & -1 & 3 \\
1 & 3 & 0 & 5
\end{array}\right]\\
& \sim \text{R}\left[\begin{array}{cccc}
1 & 3 & 0 & 5 \\
0 & 2 & -1 & 3 \\
2 & 0 & 1 & 2
\end{array}\right]\quad R_1 \sim R_3\\
& \sim \text{R}\left[\begin{array}{cccc}
1 & 3 & 0 & 5 \\
0 & 2 & -1 & 3 \\
0 & -6 & 1 & -8
\end{array}\right]\quad R_3 - 2R_1\\
& \sim \text{R}\left[\begin{array}{cccc}
1 & 3 & 0 & 5 \\
0 & 2 & -1 & 3 \\
0 & 0 & -2 & 1
\end{array}\right]\quad R_3 + 3R_2
\end{align*}
From the last row, we have:
\begin{align*}
&-2z = 1 \\
&\Rightarrow \quad z = -\frac{1}{2}\end{align*}
Put	$z = -\frac{1}{2}$ into the second row:
\begin{align*}	
&2y - (-\frac{1}{2}) = 3 \\
\Rightarrow & 2y + \frac{1}{2} = 3 \\
\Rightarrow & 2y = \frac{6}{2} - \frac{1}{2}\\
\Rightarrow & = \frac{5}{2} \\
\Rightarrow & y = \frac{5}{4}\end{align*}
Put values of $y = \frac{5}{4}$ and $z = -\frac{1}{2}$ into the first row:
\begin{align*}
&x + 3\left(\frac{5}{4}\right) = 5 \\
\Rightarrow & x + \frac{15}{4} = 5\\
\Rightarrow & x = \frac{20}{4} - \frac{15}{4} = \frac{5}{4}\end{align*}
Thus, the solution to the system is:
$$x = \frac{5}{4}, \quad y = \frac{5}{4}, \quad z = -\frac{1}{2}$$



=====Question 3(iv)=====
Solve the system of linear equation by Gauss elimination method.\\
$x+2 y+5 z=4$\\
$3 x-2 y+2 z=3$\\
$5 x-8 y-4 z=1$

** Solution. **
Do yourself.

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