For theta Belonging to 0, 4Pi the linear equations with no solutions

The number of $\theta \in (0,4\pi )$ for which the system of linear equations

$\begin{gathered} 3(\sin 3\theta )x - y + z = 2 \hfill \\ 3(\cos 2\theta )x + 4y + 3z = 3 \hfill \\ 6x + 7y + 7z = 9 \hfill \\ \end{gathered}$

has no solution, is

(A) 6
(B) 7
(C) 8
(D) 9

Solution:

Tip for solving this question:

By converting linear equations into matrix form, find the determinant of the matrix. Also, check out What is Determinant of Matrix in detail here.
Then using properties of linear equations, we find the number for no solutions.

Step 1:

Matrix Form of Linear Equations:

$\left[ {\begin{array}{*{20}{c}} {3(\sin 3\theta )}&{ - 1}&1 \\ {3(\cos 2\theta )}&4&3 \\ 6&7&7 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2 \\ 3 \\ 9 \end{array}} \right]$

Now, the determinant of a matrix is

$\begin{gathered} \Delta = \left[ {\begin{array}{*{20}{c}} {3(\sin 3\theta )}&{ - 1}&1 \\ {3(\cos 2\theta )}&4&3 \\ 6&7&7 \end{array}} \right] \hfill \\ = 3(\sin 3\theta )(28 - 21) + 1(21\cos 2\theta - 18) + 1(21\cos 2\theta - 24) \hfill \\ = 21\sin 3\theta + 42\cos 2\theta - 42 \hfill \\ \end{gathered}$

Step 2:

We have given that system of linear equations has no solution

For no solution, = 0

$\begin{gathered} \Rightarrow 21\sin 3\theta + 42\cos 2\theta - 42 = 0 \hfill \\ \Rightarrow 21\sin 3\theta + 42\cos 2\theta = 42 \hfill \\ \Rightarrow 21(\sin 3\theta + 2\cos 2\theta ) = 42 \hfill \\ \Rightarrow \sin 3\theta + 2\cos 2\theta = 2 \hfill \\ \Rightarrow \sin 3\theta = 2(1 - \cos 2\theta ) \hfill \\ \Rightarrow \sin 3\theta = 2 \times 2{\sin ^2}\theta \{ \because 1 - \cos 2\theta = 2{\sin ^2}\theta \} \hfill \\ \Rightarrow 3\sin \theta - 4{\sin ^3}\theta = 4{\sin ^2}\theta \{ \because \sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta \} \hfill \\ \Rightarrow - 4{\sin ^3}\theta - 4{\sin ^2}\theta + 3\sin \theta = 0 \hfill \\ \Rightarrow 4{\sin ^3}\theta + 4{\sin ^2}\theta - 3\sin \theta = 0 \hfill \\ \Rightarrow \sin \theta (4{\sin ^2}\theta + 4\sin \theta - 3) = 0 \hfill \\ \Rightarrow \sin \theta = 0or4{\sin ^2}\theta + 4\sin \theta - 3 = 0 \hfill \\ \Rightarrow \sin \theta = 0or4{\sin ^2}\theta + 4\sin \theta + 1 = 4 \hfill \\ \Rightarrow \sin \theta = 0or{(2\sin \theta + 1)^2} = 4 \hfill \\ \end{gathered}$

$\text{Now, for } \sin \theta = 0$

$\theta \in \left\{ {\pi ,2\pi ,3\pi } \right\}\because \theta \in (0,4\pi ) ...... (1)$

$\begin{gathered}\text{For } {(2\sin \theta + 1)^2} = 4 \hfill \\ 2\sin \theta + 1 = 2\hspace{1mm}\&\hspace{1mm} 2\sin \theta + 1 = - 2 \hfill \\ \sin \theta = \frac{1}{2} \hspace{1mm} \& \hspace{1mm} \sin \theta = \frac{{ - 3}}{2} \hfill \\ \sin \theta = \frac{1}{2} \Rightarrow \theta \in \left\{ {\frac{\pi }{6},\frac{{5\pi }}{6},\frac{{13\pi }}{6},\frac{{17\pi }}{6}} \right\}\because \theta \in (0,4\pi )......(2) \hfill \end{gathered}$

and $\sin \theta = \frac{{ - 3}}{2}$

No solution is there for $\theta \in (0,4\pi )$

Therefore, from (1) and (2)

$\text{For } \theta \in (0,4\pi)\text{ we get }\theta = \pi ,2\pi ,3\pi ,\frac{\pi }{6},\frac{{5\pi }}{6},\frac{{13\pi }}{6},\frac{{17\pi }}{6}$

Number of solutions = 7

Final Answer:

Option (B) is correct.

Solution:

Leave a Reply Cancel reply