Question

If $a,b,c>0$ and $x,y,zϵR$, 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. $a^x{b}^y{c}^x$
• B. $a^{-x}{b}^{-y}{c}^{-z}$
• C. $a^{2x}{b}^{2y}{c}^{2z}$
• D. zero

Answer 1

Answer: (D)

zero

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

$C_1\to C_1-C_2$

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

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

$C_1$ and $C_3$ are same

$\Rightarrow △=0$