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Question
Let a, b, c be positive real numbers. The following system of equations in x, y and z
(a) no solution
(b) unique solution
(c) infinitely many solutions
(d) finitely many solutions
Options
no solution
unique solution
infinitely many solutions
finitely many solutions
Solution
(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.
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