Question

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

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

Answer 1

The correct option is B

unique solution

Explanation for the correct option:

The given equations,

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

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

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

To find the number of solution,

Taking determinant of coefficients of above system of matrices

$$\begin{gathered} ∆=\left|\begin{array}{ccc}\frac{1}{a^2}& \frac{1}{b^2}& \frac{-1}{c^2}\\ \frac{1}{a^2}& \frac{-1}{b^2}& \frac{1}{c^2}\\ \frac{-1}{a^2}& \frac{1}{b^2}& \frac{1}{c^2}\end{array}\right| \\ =\frac{1}{a^2{b}^2{c}^2}\left|\begin{array}{ccc}1& 1& -1\\ 1& -1& 1\\ -1& 1& 1\end{array}\right| \\ =\frac{1}{a^2{b}^2{c}^2}\left|\begin{array}{ccc}1& 1& -1\\ 1& -1& 1\\ 0& 0& 2\end{array}\right|\left[R_3\to R_2+R_3\right] \\ =\frac{1}{a^2{b}^2{c}^2}\left[1\left(-2-0\right)-1(2-0)-1(0-0)\right] \\ =\frac{-4}{a^2{b}^2{c}^2}\ne 0 \end{gathered}$$

Since determinant of coefficient's are non zero,

Therefore it has unique solution.

Hence, the correct option is (B).