GATE AE 2025 • Q12 MCQ • 1 Mark

Engineering Mathematics

Linear Algebra

Subtopic: Eigenvalues & Eigenvectors

Question Statement:

2x + 3y + z = 0 x + y = 0 y + z = 0 The given system of equations has

Options:

  • (A) a unique solution.
  • (B) infinitely many solutions.
  • (C) no solution.
  • (D) a finite number of solutions.

Official Answer Key:

B

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2025 Q.12)** **1. Intuitive Physical Setup & Given Parameters**: We are given the system of homogeneous linear equations: $2x + 3y + z = 0$ $x + y + 0z = 0$ $0x + y + z = 0$ and asked to classify its solution set. **2. Step-by-Step Mathematical Derivation**: 1. **Matrix Representation**: We write the system as $A\mathbf{x} = \mathbf{0}$, where the coefficient matrix is: $A = \begin{bmatrix} 2 & 3 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix}$ 2. **Determinant Computation**: $\det(A) = 2(1 \cdot 1 - 0 \cdot 1) - 3(1 \cdot 1 - 0 \cdot 0) + 1(1 \cdot 1 - 1 \cdot 0)$ $\det(A) = 2(1) - 3(1) + 1(1) = 2 - 3 + 1 = 0$ 3. **Rank & Solution Dimension**: Since $\det(A) = 0$, the rank of $A$ is $\text{rank}(A) < 3$. Inspecting the $2 \times 2$ submatrix $\begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} = 1 \ne 0$, so $\text{rank}(A) = 2$. By the Rank-Nullity theorem, nullity $= n - \text{rank}(A) = 3 - 2 = 1 > 0$. A homogeneous linear system with nullity $\ge 1$ possesses non-trivial solutions, and thus has **infinitely many solutions** (forming a 1-D subspace parameterized by $t$). **3. Comprehensive Options & Distractor Analysis**: - **(A) a unique solution**: ❌ **INCORRECT**. A unique trivial solution requires $\det(A) \ne 0$. - **(B) infinitely many solutions**: ✅ **CORRECT**. Since $\det(A) = 0$ for a homogeneous system, there exists a one-parameter family of non-trivial solutions. - **(C) no solution**: ❌ **INCORRECT**. Homogeneous systems are always consistent because $\mathbf{x} = \mathbf{0}$ is always a solution. - **(D) a finite number of solutions**: ❌ **INCORRECT**. Linear systems in $\mathbb{R}$ cannot have a finite number of multiple solutions; they have either 0, 1, or $\infty$. **4. Official Master Key & Exam Takeaway**: - **Official Answer Key**: **B** - **Exam Takeaway**: A homogeneous system $Ax = 0$ is ALWAYS consistent. If $\det(A) \ne 0$, it has only the unique trivial solution $(0,0,0)$. If $\det(A) = 0$, it has infinitely many solutions.

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