Question 1, Exercise 2.6

Solutions of Question 1 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 homogeneous linear equation for non-trivial solution if exists
$ 2 x_{1}-3 x_{2}+4 x_{3}=0$
$x_{1}-2 x_{2}+3 x_{3}=0$
$4 x_{1}+x_{2}-6 x_{3}=0$

Solution. \begin{align*} &2 x_{1}-3 x_{2}+4 x_{3}=0\cdots (i)\\ &x_{1}-2 x_{2}+3 x_{3}=0\cdots (ii)\\ &4 x_{1}+x_{2}-6 x_{3}=0\cdots (iii)\\ \end{align*} For system of equation, \begin{align*} A &= \left[ \begin{array}{ccc} 2 & -3 & 4 \\ 1 & -2 & 3 \\ 4 & 1 & -6 \end{array} \right]\quad \left[ \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right]\\ |A|&=2(9)+3(-18)+4(9)\\ &=18-54+36=0 \end{align*} So the system has non-trivial solution.

\text{By}\quad(i)-2(ii), we have \begin{align*} &\begin{array}{cccc} 2x_1&-3 x_{2}&+4 x_{3}&=0\\ \mathop+\limits_{-}2x_1&\mathop-\limits_{+}4x_2&\mathop+\limits_{-}6x_3&=0 \\ \hline &x_2&-2x_3 &=0\\ \end{array} \\ \implies &x_2=2x_3\\ \end{align*} Put the value of $x_3$ in (iii), we have \begin{align*} &4 x_{1}+2x_{3}-6 x_{3}=0\\ &4 x_{1}-4 x_{3}=0\\ &4 x_{1}=4 x_{3}\\ &x_1=x_3 \end{align*} With the different values of $x_3$, there are infinite solutions.

Hence solution is;

\begin{align*} \left[ \begin{array}{c} x_3 \\ 2x_3 \\ x_3 \end{array} \right] \end{align*}

Solve the system of homogeneous linear equation for non-trivial solution if exists
$2 x_{1}-3 x_{2}+4 x_{3}=0$
$x_{1}+x_{2}+x_{3}=0$
$x_{1}-4 x_{2}+3 x_{3}=0$

Solution. \begin{align*} &2x_1 - 3x_2 + 4x_3 = 0 \quad \text{(i)}\\ &x_1 + x_2 + x_3 = 0 \quad \text{(ii)}\\ &x_1 - 4x_2 + 3x_3 = 0 \quad \text{(iii)} \end{align*} For the system of equations, we have: \begin{align*} A &= \left[ \begin{array}{ccc} 2 & -3 & 4 \\ 1 & 1 & 1 \\ 1 & -4 & 3 \end{array} \right], \quad \left[ \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right] \end{align*}

The determinant of matrix $A$ is: \begin{align*} |A| &= 2(7) + 3(2) + 4(-5)\\ & = 14 + 6 - 20 = 0 \end{align*} Since $|A| = 0,$ the system has a non-trivial solution.

By $(i)- 2(ii)$, we get: \begin{align*} &\begin{array}{cccc} &2x_1 &- 3x_2& + 4x_3 = 0\\ \mathop+\limits_{-}2x_1&\mathop+\limits_{-}2x_2&\mathop+\limits_{-}2x_3&=0 \\ \hline &-5x_2&+2x_3 &=0\\ \end{array} \\ \implies & x_2 = \frac{2}{5}x_3 \end{align*} Put the value of $x_2$ in (iii), we have \begin{align*} &x_1 - 4\left(\frac{2}{5}x_3\right) + 3x_3 = 0 \\ &x_1 - \frac{8}{5}x_3 + 3x_3 = 0 \\ &x_1 + \frac{7}{5}x_3 = 0 \\ &x_1 = -\frac{7}{5}x_3 \end{align*} Therefore, the solution set is: \begin{align*} &\left[ \begin{array}{c} x_3 \\ -\frac{7}{5}x_3 \\ \frac{2}{5}x_3 \end{array} \right] \end{align*} With different values of $x_3$, the system has infinitely many non-trivial solutions.

Solve the system of homogeneous linear equation for non-trivial solution if exists
$x_{1}+x_{2}-3 x_{3}=0$
$3 x_{1}-2 x_{2}+x_{3}=0$
$4 x_{1}-x_{2}-2 x_{3}=0$

Solution.

Do yourself.

Solve the system of homogeneous linear equation for non-trivial solution if exists
$5 x_{1}+6 x_{2}-7 x_{3}=0$
$2 x_{1}-x_{2}+x_{3}=0$
$x_{1}+2 x_{2}+2 x_{3}=0$

Solution.

Do yourself.

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