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GATE CS 1996 Q1.7 — Linear Algebra: Matrices, Determinants, System of Linear Equations

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Let Ax=bAx = b be a system of linear equations where AA is an m×nm \times n matrix and bb is a m×1m \times 1 column vector and XX is an n×1n \times1 column vector of unknowns. Which of the following is false?

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Answer: C

To find the false statement, let us analyze each option individually:

  • Option A (True): According to the Rouché–Capelli theorem, a system of linear equations Ax=bAx = b is consistent (has at least one solution) if and only if the rank of the coefficient matrix AA is equal to the rank of the augmented matrix [A∣b][A \mid b].
  • Option B (True): For a homogeneous system Ax=0Ax = 0 where m<nm < n (fewer equations than variables), the rank of AA is at most mm. The number of free variables (nullity) is given by: Nullity=n−rank(A)≥n−m>0\text{Nullity} = n - \text{rank}(A) \ge n - m > 0 Since there is at least one free variable, the system must have infinitely many solutions.
  • Option C (False): If m=nm = n and b≠0b \neq 0, the system Ax=bAx = b has a unique solution only if AA is non-singular (i.e., rank(A)=n\text{rank}(A) = n). If AA is singular (rank(A)<n\text{rank}(A) < n), the system will either have infinitely many solutions or no solution at all.
    Counterexample: Let A=(1111)A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} and b=(12)b = \begin{pmatrix} 1 \\ 2 \end{pmatrix}. Here, m=n=2m = n = 2 and b≠0b \neq 0, but the system has no solution because rank(A)=1\text{rank}(A) = 1 while rank([A∣b])=2\text{rank}([A \mid b]) = 2.
  • Option D (True): For a homogeneous system Ax=0Ax = 0 with m=nm = n and rank(A)=n\text{rank}(A) = n, the matrix AA is invertible. Multiplying both sides by A−1A^{-1} gives the unique trivial solution x=0x = 0.

Thus, statement C is false.

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