====== 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.
=====Question 2(i)=====
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$=\left[ \begin{array}{c}
- \dfrac{10}{11} x_{3} \\
\dfrac{11}{13}x_3 \\
x_3
\end{array} \right]$
=====Question 2(ii)=====
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*}
====Go to ====
[[math-11-nbf:sol:unit02:ex2-6-p1|< Question 1]]
[[math-11-nbf:sol:unit02:ex2-6-p3|Question 3 >]]