The system of linear equations (sin theta) x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For a system of homogeneous linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.Let $D = \begin

Vedclass · Accepted Answer

no value of $	heta$

Vedclass · Answer

(A) For a system of homogeneous linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.Let $D = \begin{vmatrix} \sin 	heta & 1 & -2 \ 2 & -1 & \cos 	heta \ -3 & \sec 	heta & 3 \end{vmatrix} = 0$.Expanding along the first row:$D = \sin 	heta (-3 - \cos 	heta \sec 	heta) - 1(6 + 3 \cos 	heta) - 2(2 \sec 	heta - 3) = 0$.Since $\cos 	heta \sec 	heta = 1$,we have:$D = \sin 	heta (-3 - 1) - 6 - 3 \cos 	heta - 4 \sec 	heta + 6 = 0$.$-4 \sin 	heta - 3 \cos 	heta - 4 \sec 	heta = 0$.Multiply by $\cos 	heta$ (given $	heta 
eq (2n+1)\frac{\pi}{2}$,so $\cos 	heta 
eq 0$):$-4 \sin 	heta \cos 	heta - 3 \cos^2 	heta - 4 = 0$.$-2 \sin(2	heta) - 3 \cos^2 	heta - 4 = 0$.Using $\cos^2 	heta = \frac{1 + \cos(2	heta)}{2}$:$-2 \sin(2	heta) - \frac{3}{2} - \frac{3}{2} \cos(2	heta) - 4 = 0$.$-4 \sin(2	heta) - 3 \cos(2	heta) = 11$.Since the maximum value of $a \sin x + b \cos x$ is $\sqrt{a^2 + b^2} = \sqrt{(-4)^2 + (-3)^2} = 5$,and $5 Thus,the system has no non-trivial solution for any $	heta$.

The system of linear equations $(\sin \theta) x + y - 2z = 0$,$2x - y + (\cos \theta) z = 0$,and $-3x + (\sec \theta) y + 3z = 0$,where $\theta \neq (2n + 1) \frac{\pi}{2}$,has a non-trivial solution for:

Similar Questions

Consider the following system of equations: $\alpha x + 2y + z = 1$; $2\alpha x + 3y + z = 1$; $3x + \alpha y + 2z = \beta$. For some $\alpha, \beta \in \mathbb{R}$. Which of the following is $NOT$ correct?

If the system of equations
$x+y+az=b$
$2x+5y+2z=6$
$x+2y+3z=3$
has infinitely many solutions,then $2a+3b$ is equal to $...........$.

If $A=\begin{bmatrix} 1 & 5 & 3 \\ 2 & 4 & 0 \\ 3 & -1 & -5 \end{bmatrix}$,$B=\begin{bmatrix} -1 \\ -2 \\ 4 \end{bmatrix}$ and $[x \ y \ z] A^{T}=B^{T}$,then $x+y+z=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module