The solution of the system of equations
\(\displaystyle\frac{2}{x} + \frac{4}{y} + \frac{1}{z} = 3\)
\(\displaystyle\frac{1}{x} - \frac{2}{y} + \frac{3}{z} = -3\)
\(\displaystyle\frac{3}{x} + \frac{2}{y} - \frac{1}{z} = 5\)
where \(x, y, z \ne 0\), is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(x = 1,\ y = 2,\ z = -1\)
Let \(\displaystyle u = \frac{1}{x}\), \(\displaystyle v = \frac{1}{y}\), \(\displaystyle w = \frac{1}{z}\). The substitution turns the system into a linear one, AU = B:
\( \displaystyle A = \begin{bmatrix}2 & 4 & 1\\ 1 & -2 & 3\\ 3 & 2 & -1\end{bmatrix},\quad U = \begin{bmatrix}u\\ v\\ w\end{bmatrix},\quad B = \begin{bmatrix}3\\ -3\\ 5\end{bmatrix} \)
The cofactors of A are
\( C_{11} = 2 - 6 = -4,\ C_{12} = -(-1 - 9) = 10,\ C_{13} = 2 + 6 = 8 \)
\( C_{21} = -(-4 - 2) = 6,\ C_{22} = -2 - 3 = -5,\ C_{23} = -(4 - 12) = 8 \)
\( C_{31} = 12 + 2 = 14,\ C_{32} = -(6 - 1) = -5,\ C_{33} = -4 - 4 = -8 \)
Expanding along the first row:
\( |A| = 2(-4) + 4(10) + 1(8) = 40 \ne 0 \)
The adjoint is the transpose of the cofactor matrix, and \(\displaystyle U = \frac{1}{|A|}(\mathrm{adj}\,A)B\):
\( \displaystyle U = \frac{1}{40}\begin{bmatrix}-4 & 6 & 14\\ 10 & -5 & -5\\ 8 & 8 & -8\end{bmatrix}\begin{bmatrix}3\\ -3\\ 5\end{bmatrix} \)
\( \displaystyle = \frac{1}{40}\begin{bmatrix}-12 - 18 + 70\\ 30 + 15 - 25\\ 24 - 24 - 40\end{bmatrix} \)
\( \displaystyle = \frac{1}{40}\begin{bmatrix}40\\ 20\\ -40\end{bmatrix} = \begin{bmatrix}1\\ \frac{1}{2}\\ -1\end{bmatrix} \)
So \(u = 1\), \(\displaystyle v = \frac{1}{2}\), \(w = -1\). Taking reciprocals:
\( x = 1,\ y = 2,\ z = -1 \)
Why the other options are incorrect
Option (2) \(x = 1,\ y = \frac{1}{2},\ z = -1\) — Incorrect. These are the values of u, v, w; y is the reciprocal of \(v = \frac{1}{2}\), so y = 2. In the first equation it gives \(2 + 8 - 1 = 9\), not 3.
Option (3) \(x = -1,\ y = 2,\ z = 1\) — Incorrect. In the first equation it gives \(-2 + 2 + 1 = 1\), not 3.
Option (4) \(x = 2,\ y = 1,\ z = -1\) — Incorrect. In the first equation it gives \(1 + 4 - 1 = 4\), not 3.