Question

If $\left|\begin{array}{ccc}x& \sin \mathrm{\theta }& \cos \mathrm{\theta }\\ -\sin \mathrm{\theta }& -x& 1\\ \cos \mathrm{\theta }& 1& x\end{array}\right|=8$, write the value of x.

Answer 1

$\left|\begin{array}{ccc}x& \sin \mathrm{\theta }& \cos \mathrm{\theta }\\ -\sin \mathrm{\theta }& -x& 1\\ \cos \mathrm{\theta }& 1& x\end{array}\right|=8$
Expanding along R₁, we get
$$\begin{gathered} x\left(-x^2-1\right)-\sin \theta \left(-x\sin \theta -\cos \theta \right)+\cos \theta \left(-\sin \theta +x\cos \theta \right)=8 \\ \Rightarrow -x^3-x+x{\sin }^2\theta +\sin \theta \cos \theta -\sin \theta \cos \theta +x{\cos }^2\theta =8 \\ \Rightarrow -x^3-x+x\left({\sin }^2\theta +{\cos }^2\theta \right)=8 \\ \Rightarrow -x^3-x+x=8 \\ \Rightarrow x^3+8=0 \\ \Rightarrow \left(x+2\right)\left(x^2-2x+4\right)=0 \\ \Rightarrow x+2=0\left[\because x^2-2x+4>0\forall x\right] \\ \Rightarrow x=-2 \end{gathered}$$