Match List-I with List-II
| List-I (System of equations) | List-II (Solutions) |
|---|---|
| (A) \(x + y + z = 3,\ x - y + z = 1,\ x + y - z = 1\) | (I) No solution |
| (B) \(x + y + z = 3,\ 2x + 2y + 2z = 6,\ x - y = 0\) | (II) Infinitely many solutions |
| (C) \(x + y + z = 3,\ x + y + z = 5,\ x - y + z = 1\) | (III) Unique solution \(x = y = z = 1\) |
| (D) \(x + 2y + 3z = 0,\ 2x + y + z = 0,\ x - y + 2z = 0\) | (IV) Only the solution \(x = y = z = 0\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
Each system has the form AX = B. If \(|A| \ne 0\), there is a unique solution. If \(|A| = 0\), the system has either no solution or infinitely many.
(A) The coefficient determinant is
\(\displaystyle \begin{vmatrix}1 & 1 & 1\\ 1 & -1 & 1\\ 1 & 1 & -1\end{vmatrix} = 1(1 - 1) - 1(-1 - 1) + 1(1 + 1) = 4 \ne 0\)
So the solution is unique, and x = y = z = 1 satisfies all three equations (3, 1, 1) → (III).
(B) The second equation is twice the first, so only two different equations remain for three unknowns. For example, x = y = t, z = 3 − 2t works for every t → (II).
(C) The first two equations say \(x + y + z = 3\) and \(x + y + z = 5\) together, which is impossible → (I). This system is inconsistent.
(D) The system is homogeneous, and
\(\displaystyle \begin{vmatrix}1 & 2 & 3\\ 2 & 1 & 1\\ 1 & -1 & 2\end{vmatrix} = 1(2 + 1) - 2(4 - 1) + 3(-2 - 1) = -12 \ne 0\)
So the zero solution is the only one → (IV).