Subtopic: Systems of Linear Equations & Rank Theorem
Question Statement:
The following system of equations:
$2x + 3y + z = 5$
$x + y + z = 3$
$3x + 4y + 2z = 8$
(A) has no solution.
(B) has a unique solution.
(C) has three solutions.
(D) has infinitely many solutions.
Options:
(A) has no solution.
(B) has a unique solution.
(C) has three solutions.
(D) has infinitely many solutions.
Official Answer Key:
D
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2019 Q.36 / AE Q.26)**
**1. Intuitive Physical Setup & Given Parameters**:
- Augmented matrix $[A|B]$ for the linear system:
$\begin{bmatrix} 2 & 3 & 1 & | & 5 \\ 1 & 1 & 1 & | & 3 \\ 3 & 4 & 2 & | & 8 \end{bmatrix}$
**2. Step-by-Step Mathematical Derivation**:
1. **Row Operations on Augmented Matrix**:
- Notice that Equation 3 is the exact sum of Equation 1 and Equation 2:
$\text{Row}_1 + \text{Row}_2 = (2+1)x + (3+1)y + (1+1)z = 3x + 4y + 2z$
$5 + 3 = 8$
- Performing $\text{Row}_3 \leftarrow \text{Row}_3 - (\text{Row}_1 + \text{Row}_2)$:
$\begin{bmatrix} 2 & 3 & 1 & | & 5 \\ 1 & 1 & 1 & | & 3 \\ 0 & 0 & 0 & | & 0 \end{bmatrix}$
2. **Rank Comparison**:
- $\text{Rank}(A) = 2$
- $\text{Rank}([A|B]) = 2$
- Number of unknowns: $n = 3$ ($x, y, z$).
3. **Rouché-Capelli Theorem**:
- Since $\text{Rank}(A) = \text{Rank}([A|B]) = 2 < n = 3$, the system is consistent and possesses $n - r = 3 - 2 = 1$ free variable.
- Therefore, the system has **infinitely many solutions**.
**3. Comprehensive Options & Distractor Analysis**:
- **(A)** (no solution): ❌ Incorrect. The system is consistent.
- **(B)** (unique solution): ❌ Incorrect. Requires $\text{Rank}(A) = 3$.
- **(C)** (three solutions): ❌ Incorrect. Linear systems cannot have finite solutions other than 0 or 1.
- **(D)** (infinitely many solutions): ✅ **CORRECT**. Directly established by $\text{Rank}(A) = \text{Rank}([A|B]) < 3$.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **D**
- **Exam Rule**: Linear equations can only have 0, 1, or $\infty$ solutions—never 2 or 3.
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