Question

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

(a) no solution
(b) unique solution
(c) infinitely many solutions
(d) finitely many solutions

Answer 1

$$\begin{gathered} \left(\mathrm{b}\right)\mathrm{unique}\mathrm{solution} \\ \text{The given system of equations can be written in matrix form as follows:} \\ \left[\begin{array}{ccc}\frac{1}{a^2}& \frac{1}{b^2}& \frac{-1}{c^2}\\ \frac{1}{a^2}& \frac{-1}{b^2}& \frac{1}{c^2}\\ \frac{-1}{a^2}& \frac{1}{b^2}& \frac{1}{c^2}\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right] \\ \text{Here,} \\ \mathit{\text{A}}\text{=}\left[\begin{array}{ccc}\frac{1}{a^2}& \frac{1}{b^2}& \frac{-1}{c^2}\\ \frac{1}{a^2}& \frac{-1}{b^2}& \frac{1}{c^2}\\ \frac{-1}{a^2}& \frac{1}{b^2}& \frac{1}{c^2}\end{array}\right]\text{,}\mathit{\text{X}}\text{=}\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\text{and}\mathit{\text{B}}=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right] \\ \mathrm{Now}, \\ \left|A\right|=\left|\begin{array}{ccc}\frac{1}{a^2}& \frac{1}{b^2}& \frac{-1}{c^2}\\ \frac{1}{a^2}& \frac{-1}{b^2}& \frac{1}{c^2}\\ \frac{-1}{a^2}& \frac{1}{b^2}& \frac{1}{c^2}\end{array}\right| \\ =\frac{1}{a^2{b}^2{c}^2}\left|\begin{array}{ccc}1& 1& -1\\ 1& -1& 1\\ -1& 1& 1\end{array}\right| \\ =\frac{1}{a^2{b}^2{c}^2}\times 1\left(-1-1\right)-1\left(1+1\right)-1\left(1-1\right) \\ =\frac{1}{a^2{b}^2{c}^2}\times \left(-2-2\right) \\ =\frac{-4}{a^2{b}^2{c}^2} \\ \Rightarrow \left|A\right|\text{≠ 0} \\ \text{So, the given system of equations has a unique solution.} \end{gathered}$$