====== 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. =====Question 1(i)===== 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*} =====Question 1(ii)===== 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. =====Question 1(iii)===== 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. =====Question 1(iv)===== 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. ====Go to ==== [[math-11-nbf:sol:unit02:ex2-6-p2|Question 2 >]]