If the system of equationsx+(sqrt{2} sin alph | vedclass

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

Question

(C) For a system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.$\Delta = \begin{vmatrix} 1

Vedclass · Accepted Answer

$\frac{5 \pi}{24}$

Vedclass · Answer

(C) For a system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.$\Delta = \begin{vmatrix} 1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \alpha \ 1 & \cos \alpha & \sin \alpha \ 1 & \sin \alpha & -\cos \alpha \end{vmatrix} = 0$Expanding along the first row:$1(-\cos^2 \alpha - \sin^2 \alpha) - \sqrt{2} \sin \alpha(-\cos \alpha - \sin \alpha) + \sqrt{2} \cos \alpha(\sin \alpha - \cos \alpha) = 0$$-1 + \sqrt{2} \sin \alpha \cos \alpha + \sqrt{2} \sin^2 \alpha + \sqrt{2} \sin \alpha \cos \alpha - \sqrt{2} \cos^2 \alpha = 0$$-1 + 2\sqrt{2} \sin \alpha \cos \alpha - \sqrt{2}(\cos^2 \alpha - \sin^2 \alpha) = 0$$-1 + \sqrt{2} \sin 2\alpha - \sqrt{2} \cos 2\alpha = 0$$\sqrt{2}(\sin 2\alpha - \cos 2\alpha) = 1$$\sin 2\alpha - \cos 2\alpha = \frac{1}{\sqrt{2}}$$\frac{1}{\sqrt{2}} \sin 2\alpha - \frac{1}{\sqrt{2}} \cos 2\alpha = \frac{1}{2}$$\sin(2\alpha - \frac{\pi}{4}) = \frac{1}{2}$Since $\alpha \in (0, \frac{\pi}{2})$,then $2\alpha \in (0, \pi)$,so $2\alpha - \frac{\pi}{4} \in (-\frac{\pi}{4}, \frac{3\pi}{4})$.Thus,$2\alpha - \frac{\pi}{4} = \frac{\pi}{6}$ or $2\alpha - \frac{\pi}{4} = \frac{5\pi}{6}$.Case $1$: $2\alpha = \frac{\pi}{6} + \frac{\pi}{4} = \frac{5\pi}{12} \Rightarrow \alpha = \frac{5\pi}{24}$.Case $2$: $2\alpha = \frac{5\pi}{6} + \frac{\pi}{4} = \frac{13\pi}{12} \Rightarrow \alpha = \frac{13\pi}{24}$ (Outside the range).Therefore,$\alpha = \frac{5\pi}{24}$.

If the system of equations
$x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0$
$x+(\cos \alpha) y+(\sin \alpha) z=0$
$x+(\sin \alpha) y-(\cos \alpha) z=0$
has a non-trivial solution,then $\alpha \in \left(0, \frac{\pi}{2}\right)$ is equal to :

Similar Questions

$\left| \begin{matrix} 0 & a & -b \\ -a & 0 & c \\ b & -c & 0 \end{matrix} \right| = $

The set of all values of $t \in R$,for which the matrix $\left[\begin{array}{ccc}e^t & e^{-t}(\sin t-2 \cos t) & e^{-t}(-2 \sin t-\cos t) \\e^t & e^{-t}(2 \sin t+\cos t) & e^{-t}(\sin t-2 \cos t) \\e^t & e^{-t} \cos t & e^{-t} \sin t \end{array}\right]$ is invertible.

If $k = p + q + r$,then the value of $\left|\begin{array}{ccc} k+r & p & q \\ r & k+p & q \\ r & p & k+q \end{array}\right|$ is equal to:

For how many value$(s)$ of $x$ in the closed interval $[-4, -1]$ is the matrix $\begin{bmatrix} 3 & -1+x & 2 \\ 3 & -1 & x+2 \\ x+3 & -1 & 2 \end{bmatrix}$ singular?

Prove that the determinant $\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|$ is independent of $\theta$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

If the system of equations$x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0$$x+(\cos \alpha) y+(\sin \alpha) z=0$$x+(\sin \alpha) y-(\cos \alpha) z=0$has a non-trivial solution,then $\alpha \in \left(0, \frac{\pi}{2}\right)$ is equal to :