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

$\begin{array}{1 1}(a)\;No \;solution&(b)\;unique\;solution\\(c)\;infinitely\;many\;solutions&(d)\;finitely\;many\;solutions\end{array}$

Let $\large\frac{x^2}{a^2}=$$X \large\frac{y^2}{b^2}$$=Y$
$\large\frac{z^2}{c^2}$$=Z Then the given system of equations becomes X+Y-Z=1 X-Y+Z=1 -X+Y+Z=1 This is the new system of equations. For New system we have, D=\begin{vmatrix}1&1&-1\\1&-1&1\\-1&1&1\end{vmatrix} \;\;=1(-1-1)-1(1+1)-1(1-1) \;\;\;=-4\neq 0 New system of equations has unique solution. D_1=\begin{vmatrix}1 &1&-1\\1&-1&1\\1&1&1\end{vmatrix} \;\;\;=1(-1-1)-1(1-1)-1(1+1) \;\;\;=-4 D_2=\begin{vmatrix}1 &1&-1\\1&-1&1\\1&1&1\end{vmatrix} \;\;\;=1(-1-1)-1(1-1)-1(1+1) \;\;\;=-4 D_3=\begin{vmatrix}1 &1&-1\\1&-1&1\\1&1&1\end{vmatrix} \;\;\;=1(-1-1)-1(1+1)-1(1-1) \;\;\;=-4 Now X=\large\frac{D_1}{D}=\frac{-4}{-4}=$$1$
$Y=\large\frac{D_2}{D}=\frac{-4}{-4}=$$1 Z=\large\frac{D_3}{D}=\frac{-4}{-4}=$$1$
$\Rightarrow x=\pm a,y=\pm b,z=\pm c$
Hence (b) is the correct answer.
answered Nov 21, 2013
edited Mar 19, 2014