The system of equations lambda x - y + (costh | 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 $AX = 0$ to have a non-trivial solution,the determinant of the coefficient matrix must be zero,i.e.,$

Vedclass · Accepted Answer

for no value of $\lambda$ and $	heta$

Vedclass · Answer

(A) For a system of homogeneous linear equations $AX = 0$ to have a non-trivial solution,the determinant of the coefficient matrix must be zero,i.e.,$|A| = 0$.The coefficient matrix is:$A = \begin{bmatrix} \lambda & -1 & \cos	heta \ 3 & 1 & 2 \ \cos	heta & 1 & 2 \end{bmatrix}$Calculating the determinant $|A|$:$|A| = \lambda(1 	imes 2 - 1 	imes 2) - (-1)(3 	imes 2 - \cos	heta 	imes 2) + \cos	heta(3 	imes 1 - \cos	heta 	imes 1)$$|A| = \lambda(0) + (6 - 2\cos	heta) + \cos	heta(3 - \cos	heta)$$|A| = 6 - 2\cos	heta + 3\cos	heta - \cos^2	heta$$|A| = 6 + \cos	heta - \cos^2	heta$For a non-trivial solution,$|A| = 0$:$-\cos^2	heta + \cos	heta + 6 = 0$$\cos^2	heta - \cos	heta - 6 = 0$Let $t = \cos	heta$. Then $t^2 - t - 6 = 0$.$(t - 3)(t + 2) = 0$So,$\cos	heta = 3$ or $\cos	heta = -2$.Since the range of $\cos	heta$ is $[-1, 1]$,neither $\cos	heta = 3$ nor $\cos	heta = -2$ is possible for any real $	heta$.Therefore,$|A|$ is never zero for any $	heta \in (0, 2\pi)$.Thus,the system has only the trivial solution $(x=0, y=0, z=0)$ for all values of $\lambda$ and $	heta$.

$List-I$	$List-II$
$(I)$ If $\frac{q}{r}=10$,then the system of linear equations has	$(P)$ $x=0, y=\frac{10}{9}, z=-\frac{1}{9}$ as a solution
$(II)$ If $\frac{p}{r} \neq 100$,then the system of linear equations has	$(Q)$ $x=\frac{10}{9}, y=-\frac{1}{9}, z=0$ as a solution
$(III)$ If $\frac{p}{q} \neq 10$,then the system of linear equations has	$(R)$ infinitely many solutions
$(IV)$ If $\frac{p}{q}=10$,then the system of linear equations has	$(S)$ no solution
	$(T)$ at least one solution

The system of equations $\lambda x - y + (\cos\theta) z = 0$,$3x + y + 2z = 0$,and $(\cos\theta) x + y + 2z = 0$ for $0 < \theta < 2\pi$ has non-trivial solution$(s)$:

Similar Questions

If the system of equations $x + ay = 0,$ $az + y = 0$ and $ax + z = 0$ has infinite solutions,then the value of $a$ is

If $\left[\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}2 \\ 4\end{array}\right]$,then the values of $x$ and $y$ respectively are

Consider the system of equations: $ax + by + cz = 2$, $bx + cy + az = 2$, $cx + ay + bz = 2$, where $a, b, c$ are real numbers such that $a + b + c = 0$. Then, the system

If the system of linear equations
$2x + y - z = 3$
$x - y - z = \alpha$
$3x + 3y + \beta z = 3$
has infinitely many solutions,then $\alpha + \beta - \alpha \beta$ is equal to .... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module