Question

Match the statements in Column I with the statements in Column II

Answer 1

A) $A=\left|\begin{array}{ccc}\frac{1}{c}& \frac{1}{c}& -\frac{(a+b)}{c^2}\\ -\frac{(a+b)}{a^2}& \frac{1}{a}& \frac{1}{a}\\ -\frac{b(b+c)}{a^{2}c}& \frac{(a+2b+c)}{ac}& -\frac{b(b+c)}{{ac}^2}\end{array}\right|$
Multiplying $C_1$ by $a$, $C_2$ by $b$ and $C_3$ by $c$, we get

$A=\frac{1}{abc}\left|\begin{array}{ccc}\frac{a}{c}& \frac{b}{c}& -\frac{(a+b)}{c}\\ -\frac{(a+b)}{a}& \frac{b}{a}& \frac{c}{a}\\ -\frac{b(b+c)}{ac}& \frac{b(a+2b+c)}{ac}& -\frac{b(b+c)}{ac}\end{array}\right|$

Now applying $C_1\to C_1+C_2+C_3$, we get
$A=\frac{1}{abc}\left|\begin{array}{ccc}0& \frac{b}{c}& -\frac{(a+b)}{c}\\ 0& \frac{b}{a}& \frac{c}{a}\\ 0& \frac{b(a+2b+c)}{ac}& -\frac{b(b+c)}{ac}\end{array}\right|=0$
B) $B=\left|\begin{array}{ccc}\sin a\cos b& \sin a\sin b& \cos a\\ \cos a\cos b& \cos a\sin b& -\sin a\\ -\sin a\sin b& \sin a\cos b& 0\end{array}\right|$

Applying $C_1\to C_1-\cot b{C}_2$, we get

$B=\left|\begin{array}{ccc}0& \sin a\sin b& \cos a\\ 0& \cos a\sin b& -\sin a\\ -\frac{\sin a}{\sin b}& \sin a\cos b& 0\end{array}\right|=-\frac{\sin a}{\sin b}[-\sin b{\sin }^{2}a-{\cos }^{2}a\sin b]=\sin a$
C) $C=\left|\begin{array}{ccc}\frac{1}{\sin a\cos b}& \frac{1}{\sin a\sin b}& \frac{1}{\cos a}\\ \frac{-\cos a}{{\sin }^{2}a\cos b}& \frac{-\cos a}{{\sin }^{2}a\sin b}& \frac{\sin a}{{\cos }^{2}a}\\ \frac{\sin b}{\sin a{\cos }^{2}b}& \frac{-\cos b}{\sin a{\sin }^{2}b}& 0\end{array}\right|$

Taking out $\frac{1}{\sin a\cos b}$ from $C_1$, $\frac{1}{\sin a\sin b}$ from $C_2$ and $\frac{1}{\cos a}$ from $C_3$

$C=\frac{1}{{\sin }^{2}a\cos a\sin b\cos b}\left|\begin{array}{ccc}1& 1& 1\\ -\cot a& -\cot a& \tan a\\ \tan b& -\cot b& 0\end{array}\right|$

Applying $C_1\to C_1-C_2$, we get
$C=\frac{\tan a+\cot a}{\sin b\cos b}=\frac{1}{\sin b\cos b\sin a\cos a}$
D) $\Delta =\left|\begin{array}{ccc}{a}^2& b\sin A& c\sin A\\ b\sin A& 1& \cos A\\ c\sin A& \cos A& 1\end{array}\right|$

Using $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k$

$\Delta =\left|\begin{array}{ccc}{a}^2& bak& cak\\ bak& 1& \cos A\\ cak& \cos A& 1\end{array}\right|=a^2\left|\begin{array}{ccc}1& bk& ck\\ bk& 1& \cos A\\ ck& \cos A& 1\end{array}\right|=a^2\left|\begin{array}{ccc}1& \sin B& \sin C\\ \sin B& 1& \cos A\\ \sin C& \cos A& 1\end{array}\right|$

Applying $C_2\to C_2-\sin B{C}_1,C_3\to C_3-\sin C{C}_1$, we get

$\Delta =a^2\left|\begin{array}{ccc}1& 0& 0\\ \sin B& 1-{\sin }^{2}B& \cos A-\sin B\sin C\\ \sin C& \cos A-\sin B\sin C& 1-{\sin }^{2}C\end{array}\right|=a^2[{\cos }^{2}B{\cos }^{2}C-{(\cos A-\sin B\sin C)}^2]$

As $A+B+C=\pi$

$\cos A=\cos (\pi -(B+C))=-\cos (B+C)=-(\cos B\cos C-\sin B\sin C)\Rightarrow \cos A-\sin B\sin C=\cos B\cos C$

$\Rightarrow \Delta =a^2[{\cos }^{2}B{\cos }^{2}C-{\left(\cos B\cos C\right)}^2]=0$