Question

Let $a,b$ and $c$ be positive real numbers. The following system of equations in $x,y$ and $z$
$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1,\frac{x^2}{a^2}-\frac{y^2}{b^2}+\frac{z^2}{c^2}=1,-\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ has

• A. no solution
• B. unique solution
• C. infinitely many solutions
• D. finitely many solutions

Answer 1

The correct option is D finitely many solutions

Let $\frac{x^2}{a^2}=X,\frac{y^2}{b^2}=Y,\frac{z^2}{c^2}=Z$

Then the given system of the equations becomes
$X+Y-Z=1$
$X-Y+Z=1$
$-X+Y+Z=1$

The coefficient matrix $A=\left[\begin{array}{ccc}1& 1& -1\\ 1& -1& 1\\ -1& 1& 1\end{array}\right]$

$|A|=-4\ne 0$
Hence, the transformed system has a unique solution.
On solving, we get $X=Y=Z=1$
Hence, $x=\pm a,y=\pm b,z=\pm c$