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.

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:

$$x = \frac{46}{19}, \quad y = \frac{66}{19}, \quad z = -\frac{63}{19}$$

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.

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: $$x = \frac{5}{4}, \quad y = \frac{5}{4}, \quad z = -\frac{1}{2}$$

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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