Question

Let $a,b,c$ be the positive numbers. The following system of equations in $x.y$ and $z$ has

$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$

$-\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$

• 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 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,

$D=\left|\begin{array}{ccc}1& 1& -1\\ 1& -1& 1\\ -1& 1& 1\end{array}\right|$

$\Rightarrow D=1(-1-1)-1(1-(-1\left)\right)+(-1)(1-1)=-2-2+0=-4\ne 0$

New system of equations has unique solution.

$D_1=\left|\begin{array}{ccc}1& 1& -1\\ 1& -1& 1\\ 1& 1& 1\end{array}\right|$

$\Rightarrow D=1(-1-1)-1(1-1)+(-1)(1-(-1\left)\right)=-2-0-2=-4\ne 0$

$D_2=\left|\begin{array}{ccc}1& 1& -1\\ 1& -1& 1\\ 1& 1& 1\end{array}\right|$

$\Rightarrow D=1(-1-1)-1(1-1)+(-1)(1-(-1\left)\right)=-2-0-2=-4\ne 0$

$D_3=\left|\begin{array}{ccc}1& 1& -1\\ 1& -1& 1\\ 1& 1& 1\end{array}\right|$

$\Rightarrow D=1(-1-1)-1(1-1)+(-1)(1-(-1\left)\right)=-2-0-2=-4\ne 0$

Now,

$X=\frac{D_1}{D}=\frac{-4}{-4}=1$

$\therefore \frac{x^2}{a^2}=1\Rightarrow x=\pm a$

$Y=\frac{D_2}{D}=\frac{-4}{-4}=1$

$\therefore \frac{y^2}{b^2}=1\Rightarrow y=\pm b$

$Z=\frac{D_3}{D}=\frac{-4}{-4}=1$

$\therefore \frac{z^2}{c^2}=1\Rightarrow z=\pm c$

Since, $x,y,z$, all have more than $1$ values, hence, the given system of equations have finitely many solutions.