Statement -1: The system of linear equationsx | vedclass

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

Question

(B) For a system of linear equations $AX = 0$ to have a non-trivial solution,the determinant of the coefficient matrix must be zero,i.e.,$|A| = 0$.For

Vedclass · Accepted Answer

Statement $-1$ is true,Statement $-2$ is true,Statement $-2$ is a correct explanation for Statement $-1$.

Vedclass · Answer

(B) For a system of linear equations $AX = 0$ to have a non-trivial solution,the determinant of the coefficient matrix must be zero,i.e.,$|A| = 0$.For Statement $-1$,the determinant is:$\Delta_1 = \left| \begin{matrix} 1 & \sin \alpha & \cos \alpha \ 1 & \cos \alpha & \sin \alpha \ 1 & -\sin \alpha & -\cos \alpha \end{matrix} ight|$Applying $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$:$\Delta_1 = \left| \begin{matrix} 1 & \sin \alpha & \cos \alpha \ 0 & \cos \alpha - \sin \alpha & \sin \alpha - \cos \alpha \ 0 & -2\sin \alpha & -2\cos \alpha \end{matrix} ight|$$= 1 \cdot [(\cos \alpha - \sin \alpha)(-2\cos \alpha) - (\sin \alpha - \cos \alpha)(-2\sin \alpha)]$$= -2\cos^2 \alpha + 2\sin \alpha \cos \alpha + 2\sin^2 \alpha - 2\sin \alpha \cos \alpha = 2(\sin^2 \alpha - \cos^2 \alpha) = -2\cos(2\alpha)$.Setting $\Delta_1 = 0$,we get $\cos(2\alpha) = 0$. In $(0, \frac{\pi}{2})$,$2\alpha \in (0, \pi)$,so $2\alpha = \frac{\pi}{2} \Rightarrow \alpha = \frac{\pi}{4}$. Thus,Statement $-1$ is true.For Statement $-2$,the determinant is:$\Delta_2 = \left| \begin{matrix} \cos \alpha & \sin \alpha & \cos \alpha \ \sin \alpha & \cos \alpha & \sin \alpha \ \cos \alpha & -\sin \alpha & -\cos \alpha \end{matrix} ight|$Applying $C_3 	o C_3 - C_1$:$\Delta_2 = \left| \begin{matrix} \cos \alpha & \sin \alpha & 0 \ \sin \alpha & \cos \alpha & 0 \ \cos \alpha & -\sin \alpha & -2\cos \alpha \end{matrix} ight|$$= -2\cos \alpha (\cos^2 \alpha - \sin^2 \alpha) = -2\cos \alpha \cos(2\alpha) = 0$.This implies $\cos \alpha = 0$ or $\cos(2\alpha) = 0$. In $(0, \frac{\pi}{2})$,$\cos \alpha 
eq 0$ and $\cos(2\alpha) = 0$ only at $\alpha = \frac{\pi}{4}$. Thus,Statement $-2$ is true and explains Statement $-1$.

Similar Questions

The solution of the equation $\begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$ is $(x, y, z) = $

If the system of linear equations
$7x + 11y + \alpha z = 13$
$5x + 4y + 7z = \beta$
$175x + 194y + 57z = 361$
has infinitely many solutions,then $\alpha + \beta + 2$ is equal to

If $\begin{bmatrix} \alpha & \beta & \gamma \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & -5 \\ 1 & 2 & 5 \end{bmatrix} = \begin{bmatrix} 3 & 5 & 2 \end{bmatrix}$, then $\alpha^3 + \beta^3 + \gamma^3 = $

Solve the system of linear equations using the matrix method: $2x + 3y + 3z = 5$,$x - 2y + z = -4$,$3x - y - 2z = 3$.

If the system of equations $x + y + z = 5$, $x + 2y + 3z = 9$, $x + 3y + \lambda z = \mu$ has infinitely many solutions, then the value of $\lambda + \mu$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module