The system of equations (sintheta) x + 2z = 0 | vedclass

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(B) The system of equations can be represented in matrix form $AX = B$,where $A = \begin{bmatrix} \sin	heta & 0 & 2 \ \cos	heta & \sin	heta & 0 \

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a unique solution which is a function of $a$ and $	heta$

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(B) The system of equations can be represented in matrix form $AX = B$,where $A = \begin{bmatrix} \sin	heta & 0 & 2 \ \cos	heta & \sin	heta & 0 \ 0 & \cos	heta & 2 \end{bmatrix}$.To check for a unique solution,we calculate the determinant $D = |A|$.$D = \sin	heta (\sin	heta \cdot 2 - 0 \cdot \cos	heta) - 0 + 2 (\cos	heta \cdot \cos	heta - 0 \cdot \sin	heta)$$D = 2\sin^2	heta + 2\cos^2	heta = 2(\sin^2	heta + \cos^2	heta) = 2(1) = 2$.Since $D = 2 
eq 0$,the system has a unique solution.Using Cramer's Rule,$x = \frac{D_x}{D}$,$y = \frac{D_y}{D}$,and $z = \frac{D_z}{D}$.$D_x = \begin{vmatrix} 0 & 0 & 2 \ 0 & \sin	heta & 0 \ a & \cos	heta & 2 \end{vmatrix} = 2(0 - a\sin	heta) = -2a\sin	heta$.$D_y = \begin{vmatrix} \sin	heta & 0 & 2 \ \cos	heta & 0 & 0 \ 0 & a & 2 \end{vmatrix} = -0 + 0 - 2(a\cos	heta) = -2a\cos	heta$.$D_z = \begin{vmatrix} \sin	heta & 0 & 0 \ \cos	heta & \sin	heta & 0 \ 0 & \cos	heta & a \end{vmatrix} = \sin	heta(a\sin	heta - 0) = a\sin^2	heta$.Thus,$x = -a\sin	heta$,$y = -a\cos	heta$,and $z = \frac{a\sin^2	heta}{2}$.Since the solution depends on both $a$ and $	heta$,the correct option is $B$.

The system of equations $(\sin\theta) x + 2z = 0$,$(\cos\theta) x + (\sin\theta) y = 0$,and $(\cos\theta) y + 2z = a$ has

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