Matrices And Determinants Ques 88
The number of values of $\theta \in(0, \pi)$ for which the system of linear equations
$ \begin{aligned} & x+3 y+7 z=0, \\ & -x+4 y+7 z=0, \\ & (\sin 3 \theta) x+(\cos 2 \theta) y+2 z=0 \end{aligned} $
has a non-trivial solution, is
(2019 Main, 10 Jan II)
(a) two
(b) three
(c) four
(d) one
Show Answer
Answer:
Correct Answer: 88.(a)
Solution:
Formula:
System of equations with 3 variables:
- We know that,
the system of linear equations
$ \begin{aligned} & a_{1} x+b_{1} y+c_{1} z=0 \\ & a_{2} x+b_{2} y+c_{2} z=0 \\ & a_{3} x+b_{3} y+c_{3} z=0 \end{aligned} $
has a non-trivial solution, if
$ \left|\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right|=0 $
Now, if the given system of linear equations
$ \begin{aligned} x+3 y+7 z & =0 \\ -x+4 y+7 z & =0 \end{aligned} $
and $(\sin 3 \theta) x+(\cos 2 \theta) y+2 z=0$
has non-trivial solution, then
$ \left|\begin{array}{ccc} 1 & 3 & 7 \\ -1 & 4 & 7 \\ \sin 3 \theta & \cos 2 \theta & 2 \end{array}\right|=0 $
$\Rightarrow 1(8-7 \cos 2 \theta)-3(-2-7 \sin 3 \theta)$ $ +7(-\cos 2 \theta-4 \sin 3 \theta)=0 $
$\Rightarrow 8-7 \cos 2 \theta+6+21 \sin 3 \theta$ $-7 \cos 2 \theta-28 \sin 3 \theta=0$
$\Rightarrow-7 \sin 3 \theta-14 \cos 2 \theta+14=0$
$\Rightarrow-7\left(3 \sin \theta-4 \sin ^{3} \theta\right)-14\left(1-2 \sin ^{2} \theta\right)+14=0$
$ \left[\because \sin 3 A=3 \sin A-4 \sin ^{3} A\right. \text { and } $
$\left.\cos 2 A=1-2 \sin ^{2} A\right]$
$\Rightarrow 28 \sin ^{3} \theta+28 \sin ^{2} \theta-21 \sin \theta-14+14=0$
$\Rightarrow 7 \sin \theta\left[4 \sin ^{2} \theta+4 \sin \theta-3\right]=0$
$\Rightarrow \sin \theta\left[4 \sin ^{2} \theta+6 \sin \theta-2 \sin \theta-3\right]=0$
$\Rightarrow \sin \theta[2 \sin \theta(2 \sin \theta+3)-1(2 \sin \theta+3)]=0$
$\Rightarrow(\sin \theta)(2 \sin \theta-1)(2 \sin \theta+3)=0$
Now, either $\sin \theta=0$ or $\frac{1}{2}$
$ \because \sin \theta \neq-\frac{3}{2} \text { as }-1 \leq \sin \theta \leq 1 $
In given interval $(0, \pi)$,
$ \begin{aligned} \sin \theta & =\frac{1}{2} \\ \Rightarrow \quad \theta & =\frac{\pi}{6}, \frac{5 \pi}{6} \quad[\because \sin \theta \neq 0, \theta \in(0, \pi)] \end{aligned} $
Hence, 2 solutions in $(0, \pi)$
BETA
