Question

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

\[\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1, \frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1, - \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \text { has }\]
(a) no solution
(b) unique solution
(c) infinitely many solutions
(d) finitely many solutions

विकल्प

• no solution

• unique solution

• infinitely many solutions

•  finitely many solutions

Answer 1

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