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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Find the value of $\theta$ satisfying $\begin{bmatrix} 1 & 1 & \sin3\theta \\ -4 & 3 & \cos2\theta \\ 7 & -7 & -2 \end{bmatrix}\;=\;0$

$\begin{array}{1 1}(A)\frac{\pi}{6} \\ (B)\frac{\pi}{3} \\ (C)\frac{\pi}{4} \\ (D)\frac{\pi}{2}\end{array} $

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  • If $A=\begin{vmatrix}a_{11} & a_{21} & a_{31}\\a_{12} & a_{22} & a_{32}\\a_{13} & a_{23} & a_{33}\end{vmatrix}$
  • $|A|=a_{11}(a_22\times a_{33}-a_{32}\times a_{23})-a_{21}(a_{12}\times a_{33}-a_{32}\times a_{13})+a_{31}(a_{12}\times a_{23}-a_{22}\times a_{13})$
Let $\Delta=\begin{vmatrix}1 & 1 & sin\theta\\-4 & 3 & cos 2\theta\\7 & -7& -2\end{vmatrix}$
Expanding along $R_1$ we get,
$\Delta=1(-6+7cos 2\theta)-1(8-7cos2\theta)+sin3\theta(28-21)$
$\quad=-6+7cos2\theta-8+7cos 2\theta+7sin 3\theta$
$\quad=14cos 2\theta+7sin 3\theta-14=0.$
(Because given $\Delta=0.$)
Taking 7 as the common factor we get,
$7(2cos 2\theta+sin 3\theta-2)=0.$
$\Rightarrow 2cos 2\theta+sin3\theta-2=0.$
We know $cos 2\theta=1-2sin^2\theta$ and $sin 3\theta=3sin\theta-4sin^3\theta$
Therefore $2(1-2sin^2\theta)+(3sin\theta-4sin^3\theta)-2=0.$
On simplifying we get,
Taking -sin$\theta$ as the common factor,
Now evaluating for sin$\theta$ we get,
$\Rightarrow 2sin\theta(2sin\theta+3)-1(2sin\theta+3)=0.$
$\Rightarrow (2sin\theta-1)(2sin\theta+3)=0.$
$\Rightarrow either\;2sin\theta-1=0$
$\Rightarrow sin\theta=\frac{1}{2}$
$\Rightarrow \theta=\frac{\pi}{6}$
or $2sin\theta+3=1.$
$\Rightarrow \theta=sin^{-1}\big(-\frac{3}{2}\big)$
But the value of $sin\theta$ range from -1 to +1.Hence cannot take the value $sin^{-1}\big(\frac{3}{2}\big)$
Hence the principle solutions are
$\theta=n\pi$ or $\theta=n\pi+(-1)^n\frac{\pi}{6}$
answered Mar 16, 2013 by vijayalakshmi_ramakrishnans

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