The solution of the system of equations
\(\displaystyle \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 4,\quad \frac{2}{x} - \frac{1}{y} + \frac{1}{z} = 8,\quad \frac{1}{x} + \frac{2}{y} - \frac{1}{z} = -3\)
where \(x, y, z \ne 0\), is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle x = \frac{1}{2},\ y = -1,\ z = \frac{1}{3}\)
Let \(\displaystyle u = \frac{1}{x},\ v = \frac{1}{y},\ w = \frac{1}{z}\). The system becomes linear in u, v, w:
\(u + v + w = 4\)
\(2u - v + w = 8\)
\(u + 2v - w = -3\)
\(\displaystyle \begin{vmatrix}1 & 1 & 1\\ 2 & -1 & 1\\ 1 & 2 & -1\end{vmatrix} = 1(1 - 2) - 1(-2 - 1) + 1(4 + 1) = 7\)
The determinant is not zero, so there is a unique solution.
Adding the first and third equations:
\(2u + 3v = 1\)
Adding the second and third equations:
\(3u + v = 5\), so \(v = 5 - 3u\)
\(2u + 15 - 9u = 1\)
\(u = 2,\ v = -1\)
\(w = 4 - 2 + 1 = 3\)
\(\displaystyle x = \frac{1}{2},\ y = -1,\ z = \frac{1}{3}\)
Why the other options are incorrect
Option (1) \(x = 2,\ y = -1,\ z = 3\) — Incorrect. These are the values of u, v, w; x, y, z are their reciprocals.
Option (3) \(\displaystyle x = \frac{1}{2},\ y = 1,\ z = \frac{1}{3}\) — Incorrect. The first equation gives 2 + 1 + 3 = 6, not 4.
Option (4) \(\displaystyle x = -\frac{1}{2},\ y = -1,\ z = \frac{1}{3}\) — Incorrect. The first equation gives −2 − 1 + 3 = 0, not 4.