Question 2, Exercise 2.6

Solutions of Question 2 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.

Find the value of $\lambda$ for which the system of homogeneous linear equation may have non-trivial solution. Also solve the system for value of $\lambda$.
$2 x_{1}-\lambda x_{2}+x_{3}=0$
$2 x_{1}+3 x_{2}-x_{3}=0$
$3 x_{1}-2 x_{2}+4 x_{3}=0$

Solution. \begin{align*} &2 x_{1}-\lambda x_{2}+x_{3}=0 \cdots(i)\\ &2 x_{1}+3 x_{2}-x_{3}=0\cdots(ii)\\ &3 x_{1}-2 x_{2}+4 x_{3}=0\cdots(iii)\\ \end{align*} Homogenous system has non-trivial solution, if \begin{align*} &\left| \begin{array}{ccc} 2 & -\lambda & 1 \\ 2 & 3 & -1 \\ 3 & -2 & 4 \end{array} \right|=0\\ &2(12-2)+\lambda(8+3)+1(-4-9)=0\\ &20+11\lambda-13=0\\ &\lambda =-\frac{7}{11}\\ \end{align*} The system becomes

\begin{align*} &2 x_{1}+ \frac{7}{11}x_{2}+x_{3}=0 \cdots(iv)\\ &2 x_{1}+3 x_{2}-x_{3}=0\cdots(v)\\ &3 x_{1}-2 x_{2}+4 x_{3}=0\cdots(vi)\\ \end{align*} (iv)-(v), we have
\begin{align*} &\begin{array}{cccc} 2x_1&+\frac{7}{11} x_{2}&+ x_{3}&=0\\ \mathop+\limits_{-}2x_1&\mathop+\limits_{-}3x_2&\mathop-\limits_{+}x_3&=0 \\ \hline &-\frac{26}{11}x_2&+2x_3 &=0\\ \end{array} \\ \implies &x_2=\frac{11}{13}x_3\\ \end{align*} Put the value of $x_2$ in (vi), we have \begin{align*} &3 x_{1}-2(\frac{11}{13})x_{3}+4 x_{3}=0\\ &3 x_{1}+ \frac{30}{11} x_{3}=0\\ & x_{1}=- \frac{10}{11} x_{3}\\ \end{align*} Hence S.S$=\left[ \begin{array}{c} - \dfrac{10}{11} x_{3}
\dfrac{11}{13}x_3
x_3 \end{array} \right]$

Find the value of $\lambda$ for which the system of homogeneous linear equation may have non-trivial solution. Also solve the system for value of $\lambda$.
$x_{1}-4 x_{2}+3 x_{3}=0$
$2 x_{1}+\lambda x_{2}+x_{3}=0$
$x_{1}-2 x_{2}+\lambda x_{3}=0$

Solution. \begin{align*} &x_{1}-4 x_{2}+3 x_{3}=0 \quad \text{(i)}\\ &2 x_{1}+\lambda x_{2}+x_{3}=0 \quad \text{(ii)}\\ &x_{1}-2 x_{2}+\lambda x_{3}=0 \quad \text{(iii)} \end{align*} For the system to have a non-trivial solution, the determinant must be zero: \begin{align*} &\left| \begin{array}{ccc} 1 & -4 & 3 \\ 2 & \lambda & 1 \\ 1 & -2 & \lambda \end{array} \right| = 0\\ &1\left(\lambda^2 + 2\right) + 4\left(2\lambda - 1\right) + 3\left(-4 -\lambda \right) = 0\\ &\lambda^2 + 2 + 8\lambda - 4 - 12-3\lambda = 0\\ &\lambda^2 + 5\lambda - 14 = 0\\ &\lambda = \frac{-5 \pm \sqrt{25 + 56}}{2} \\ &\lambda = \frac{-5 \pm \sqrt{81}}{2} \\ &\lambda = \frac{-5 \pm 9}{2} \\ &\lambda = 2 \quad \lambda =-7 \end{align*} If $\lambda=2$, put the value of $\lambda$ we have \begin{align*} x_{1} - 4x_{2} + 3x_{3} &= 0 \quad \text{(i)} \\ 2x_{1} + 2x_{2} + x_{3} &= 0 \quad \text{(ii)} \\ x_{1} - 2x_{2} + 2x_{3} &= 0 \quad \text{(iii)} \end{align*} By using (i), we have \begin{align*} x_{1} &= 4x_{2} - 3x_{3} \quad \text{(iv)} \end{align*} Put the value if $x_1$ in (ii), we have \begin{align*} 2(4x_{2} - 3x_{3}) + 2x_{2} + x_{3} &= 0 \\ 8x_{2} - 6x_{3} + 2x_{2} + x_{3} &= 0 \\ 10x_{2} - 5x_{3} &= 0 \\ x_{2} &= \frac{1}{2}x_{3} \end{align*} Put the value of $x_2$ in $x_1$ \begin{align*} x_{1} &= 4(\frac{1}{2}x_{3}) - 3x_{3}\\ &=-x_3 \end{align*} Henc solution set is: \begin{align*} \left[ \begin{array}{c} -x_3 \\ \frac{1}{2}x_3 \\ x_3 \end{array} \right] \end{align*} If $\lambda =-7$ \begin{align*} x_{1} - 4x_{2} + 3x_{3} &= 0 \quad \text{(1)} \\ 2x_{1} - 7x_{2} + x_{3} &= 0 \quad \text{(2)} \\ x_{1} - 2x_{2} - 7x_{3} &= 0 \quad \text{(3)} \end{align*} From equation (1), we have \begin{align*} x_{1} &= 4x_{2} - 3x_{3} \end{align*} Put the value of $x_1$ in (2) \begin{align*} 2(4x_{2} - 3x_{3}) - 7x_{2} + x_{3} &= 0 \\ 8x_{2} - 6x_{3} - 7x_{2} + x_{3} &= 0 \\ x_{2} - 5x_{3} &= 0 \\ x_{2} &= 5x_{3} \end{align*} Put the value of $x_2$ in $x_1$, we have \begin{align*} x_{1} &= 4x_{2} - 3x_{3} &=20x_3-3x_3\\ &=17x_3\end{align*} Hence solution set is \begin{align*} \left[ \begin{array}{c} 17x_3 \\ 5x_3 \\ x_3 \end{array} \right] \end{align*}

< Question 1 Question 3 >