Question

If $a,h,0>0$ 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 independent of

• A. $a,b,c$
• B. $x,y,z$
• C. $a,b,c,x,y,z$
• D. none of these

Answer 1

The correct option is C $a,b,c,x,y,z$

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

Applying $C_1\to C_1-C_2-4C_3$

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

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