Question

If $a,b,c>0$ and $x,y,z\in R$, then the determinant $\left|\begin{array}{ccc}{(a^x+a^{-x})}^2& {(a^x-a^{-x})}^2& 1\\ {(b^y+b^{-y})}^2& {(b^y-b^{-y})}^2& 1\\ {(c^z+c^{-z})}^2& {(c^z-c^{-z})}^2& 1\end{array}\right|$ is equal to :

• A. $a^z{b}^y{c}^z$
• B. $a^{-x}{b}^{-y}{c}^{-z}$
• C. $0$
• D. $a^{2x}{b}^{2y}{c}^{2z}$

Answer 1

The correct option is C $0$

$\left|\begin{array}{ccc}(a^x+a^{-x}{)}^2& (a^x-a^{-x}{)}^2& 1\\ (b^y+b^{-y}{)}^2& (b^y-b^{-y}{)}^2& 1\\ (c^z+c^{-z}{)}^2& (c^z-c^{-z}{)}^2& 1\end{array}\right|$

applying $C_1\to C_1-C_2$ gives

$\left|\begin{array}{ccc}4a^x{a}^{-x}& (a^x-a^{-x}{)}^2& 1\\ 4b^y{b}^{-y}& (b^y-b^{-y}{)}^2& 1\\ 4c^z{c}^{-z}& (c^z-c^{-z}{)}^2& 1\end{array}\right|$

$\Rightarrow \left|\begin{array}{ccc}4& (a^x-a^{-x}{)}^2& 1\\ 4& (b^y-b^{-y}{)}^2& 1\\ 4& (c^z-c^{-z}{)}^2& 1\end{array}\right|=0$ (as $C_1$ and $C_3$ are proportional)

$\therefore \left|\begin{array}{ccc}(a^x+a^{-x}{)}^2& (a^x-a^{-x}{)}^2& 1\\ (b^y+b^{-y}{)}^2& (b^y-b^{-y}{)}^2& 1\\ (c^z+c^{-z}{)}^2& (c^z-c^{-z}{)}^2& 1\end{array}\right|=0$

Hence, option C.