Question

Match the statements in Column I with statements in Column II

Answer 1

A) $A=\left|\begin{array}{ccc}x+2& x+3& x+5\\ x+4& x+6& x+9\\ x+8& x+11& x+15\end{array}\right|$
Applying $R_1\to R_1-R_2,R_2\to R_2-R_3,R_3\to R_3-R_1$
$A=\left|\begin{array}{ccc}-1& -2& -3\\ -2& -3& -5\\ -3& -4& -7\end{array}\right|=-\left|\begin{array}{ccc}1& 2& 3\\ 2& 3& 5\\ 3& 4& 7\end{array}\right|=-(1(21-20)-2(14-15)+3(8-9))=-2$
B) $B=\left|\begin{array}{ccc}7& 6& x^2-13\\ 2& x^2-13& 2\\ x^2-13& 3& 7\end{array}\right|$
Applying $R_1\to R_1+R_2+R_3$
$B=\left|\begin{array}{ccc}{x}^2-4& x^2-4& x^2-4\\ 2& x^2-13& 2\\ x^2-13& 3& 7\end{array}\right|=(x^2-4)\left|\begin{array}{ccc}1& 1& 1\\ 2& x^2-13& 2\\ x^2-13& 3& 7\end{array}\right|$
Now applying $C_2\to C_2-C_1,C_3\to C_3-C_1$
$B=(x^2-4)\left|\begin{array}{ccc}1& 0& 0\\ 2& x^2-9& 0\\ x^2-13& 16-x^2& 20-x^2\end{array}\right|=(x^2-4)(x^2-9)(20-x^2)$
As $x+2$ is one of the roots the equation become
$B=(x^2-9)(20-x^2)(x-2)$
Hence sum off roots is $0$
C) $C=\left|\begin{array}{ccc}\sqrt{6}& 2i& 3+\sqrt{6}\\ \sqrt{12}& \sqrt{3}+\sqrt{8}i& 3\sqrt{2}+\sqrt{6}i\\ \sqrt{18}& \sqrt{2}+\sqrt{12}i& \sqrt{27}+2i\end{array}\right|=\sqrt{6}\left|\begin{array}{ccc}1& 2i& 3+\sqrt{6}\\ \sqrt{2}& \sqrt{3}+\sqrt{8}i& 3\sqrt{2}+\sqrt{6}i\\ \sqrt{3}& \sqrt{2}+\sqrt{12}i& \sqrt{27}+2i\end{array}\right|$
Applying $R_2\to R_2-\sqrt{2}{R}_1,R_3\to R_3-\sqrt{3}{R}_1$, we get
$C=\left|\begin{array}{ccc}1& 2i& 3+\sqrt{6}\\ 0& \sqrt{3}& \sqrt{6}i-2\sqrt{3}\\ 0& \sqrt{2}& 2i-3\sqrt{2}\end{array}\right|=\sqrt{6}(\sqrt{3}(2i-3\sqrt{2})-\sqrt{2}(\sqrt{6}i-2\sqrt{3}))=-6$
D) $f\left(\theta \right)=\left|\begin{array}{ccc}{\cos }^2\theta & \cos \theta \sin \theta & -\sin \theta \\ \cos \theta \sin \theta & {\sin }^2\theta & \cos \theta \\ \sin \theta & -\cos \theta & 0\end{array}\right|=\frac{1}{\sin \theta }\left|\begin{array}{ccc}{\cos }^2\theta \sin \theta & \cos \theta \sin \theta & -\sin \theta \\ \cos \theta {\sin }^2\theta & {\sin }^2\theta & \cos \theta \\ {\sin }^2\theta & -\cos \theta & 0\end{array}\right|$
Applying $C_1\to C_1-\cos \theta C_2$, we get
$f\left(\theta \right)=\frac{1}{\sin \theta }\left|\begin{array}{ccc}0& \cos \theta \sin \theta & -\sin \theta \\ 0& {\sin }^2\theta & \cos \theta \\ 1& -\cos \theta & 0\end{array}\right|=\frac{1}{\sin \theta }({\cos }^2\theta \sin \theta +{\sin }^3\theta )={\cos }^2\theta +{\sin }^2\theta =1\Rightarrow f\left(\frac{\pi }{3}\right)=1$