PYQs / GATE EC / 2020 / Set 1 / Q36 GATE EC 2020 Set 1 — Question 36 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +2 / -0.67 Medium Systems of Linear Equations Linear Algebra Engineering Mathematics Existence & Uniqueness
Engineering Mathematics → Linear Algebra → Existence & Uniqueness
Last updated 5 September 2026
Question Consider the following system of linear equations.
x 1 + 2 x 2 = b 1 ; 2 x 1 + 4 x 2 = b 2 ; 3 x 1 + 7 x 2 = b 3 ; 3 x 1 + 9 x 2 = b 4 x_1 + 2x_2 = b_1; \quad 2x_1 + 4x_2 = b_2; \quad 3x_1 + 7x_2 = b_3; \quad 3x_1 + 9x_2 = b_4 x 1 + 2 x 2 = b 1 ; 2 x 1 + 4 x 2 = b 2 ; 3 x 1 + 7 x 2 = b 3 ; 3 x 1 + 9 x 2 = b 4 Which one of the following conditions ensures that a solution exists for the above system?
Correct answer (A) b₂ = 2b₁ and 6b₁ - 3b₃ + b₄ = 0
Solution For a solution to exist, the augmented matrix
[ A ∣ B ] [A|B] [ A ∣ B ] must have
rank ( A ) = rank ( A ∣ B ) \text{rank}(A) = \text{rank}(A|B) rank ( A ) = rank ( A ∣ B ) .
[ A ∣ B ] = [ 1 2 ∣ b 1 2 4 ∣ b 2 3 7 ∣ b 3 3 9 ∣ b 4 ] [A|B] = \begin{bmatrix} 1 & 2 & | & b_1 \\ 2 & 4 & | & b_2 \\ 3 & 7 & | & b_3 \\ 3 & 9 & | & b_4 \end{bmatrix} [ A ∣ B ] = 1 2 3 3 2 4 7 9 ∣ ∣ ∣ ∣ b 1 b 2 b 3 b 4 Perform row operations:
1. R 2 → R 2 − 2 R 1 ⟹ [ 1 2 ∣ b 1 0 0 ∣ b 2 − 2 b 1 3 7 ∣ b 3 3 9 ∣ b 4 ] R_2 \to R_2 - 2R_1 \implies \begin{bmatrix} 1 & 2 & | & b_1 \\ 0 & 0 & | & b_2 - 2b_1 \\ 3 & 7 & | & b_3 \\ 3 & 9 & | & b_4 \end{bmatrix} R 2 → R 2 − 2 R 1 ⟹ 1 0 3 3 2 0 7 9 ∣ ∣ ∣ ∣ b 1 b 2 − 2 b 1 b 3 b 4 b 2 − 2 b 1 = 0 ⟹ b 2 = 2 b 1 b_2 - 2b_1 = 0 \implies b_2 = 2b_1 b 2 − 2 b 1 = 0 ⟹ b 2 = 2 b 1 2. R 3 → R 3 − 3 R 1 ⟹ [ 1 2 ∣ b 1 0 0 ∣ 0 0 1 ∣ b 3 − 3 b 1 3 9 ∣ b 4 ] R_3 \to R_3 - 3R_1 \implies \begin{bmatrix} 1 & 2 & | & b_1 \\ 0 & 0 & | & 0 \\ 0 & 1 & | & b_3 - 3b_1 \\ 3 & 9 & | & b_4 \end{bmatrix} R 3 → R 3 − 3 R 1 ⟹ 1 0 0 3 2 0 1 9 ∣ ∣ ∣ ∣ b 1 0 b 3 − 3 b 1 b 4 3. R 4 → R 4 − 3 R 1 ⟹ [ 1 2 ∣ b 1 0 0 ∣ 0 0 1 ∣ b 3 − 3 b 1 0 3 ∣ b 4 − 3 b 1 ] R_4 \to R_4 - 3R_1 \implies \begin{bmatrix} 1 & 2 & | & b_1 \\ 0 & 0 & | & 0 \\ 0 & 1 & | & b_3 - 3b_1 \\ 0 & 3 & | & b_4 - 3b_1 \end{bmatrix} R 4 → R 4 − 3 R 1 ⟹ 1 0 0 0 2 0 1 3 ∣ ∣ ∣ ∣ b 1 0 b 3 − 3 b 1 b 4 − 3 b 1 4. R 4 → R 4 − 3 R 3 ⟹ [ 1 2 ∣ b 1 0 0 ∣ 0 0 1 ∣ b 3 − 3 b 1 0 0 ∣ ( b 4 − 3 b 1 ) − 3 ( b 3 − 3 b 1 ) ] R_4 \to R_4 - 3R_3 \implies \begin{bmatrix} 1 & 2 & | & b_1 \\ 0 & 0 & | & 0 \\ 0 & 1 & | & b_3 - 3b_1 \\ 0 & 0 & | & (b_4 - 3b_1) - 3(b_3 - 3b_1) \end{bmatrix} R 4 → R 4 − 3 R 3 ⟹ 1 0 0 0 2 0 1 0 ∣ ∣ ∣ ∣ b 1 0 b 3 − 3 b 1 ( b 4 − 3 b 1 ) − 3 ( b 3 − 3 b 1 ) The last row gives
b 4 − 3 b 1 − 3 b 3 + 9 b 1 = 6 b 1 − 3 b 3 + b 4 = 0 b_4 - 3b_1 - 3b_3 + 9b_1 = 6b_1 - 3b_3 + b_4 = 0 b 4 − 3 b 1 − 3 b 3 + 9 b 1 = 6 b 1 − 3 b 3 + b 4 = 0 .
Thus, the conditions are
b 2 = 2 b 1 b_2 = 2b_1 b 2 = 2 b 1 and
6 b 1 − 3 b 3 + b 4 = 0 6b_1 - 3b_3 + b_4 = 0 6 b 1 − 3 b 3 + b 4 = 0 .
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Correct answer (A) b₂ = 2b₁ and 6b₁ - 3b₃ + b₄ = 0
Solution For a solution to exist, the augmented matrix
[ A ∣ B ] [A|B] [ A ∣ B ] must have
rank ( A ) = rank ( A ∣ B ) \text{rank}(A) = \text{rank}(A|B) rank ( A ) = rank ( A ∣ B ) .
[ A ∣ B ] = [ 1 2 ∣ b 1 2 4 ∣ b 2 3 7 ∣ b 3 3 9 ∣ b 4 ] [A|B] = \begin{bmatrix} 1 & 2 & | & b_1 \\ 2 & 4 & | & b_2 \\ 3 & 7 & | & b_3 \\ 3 & 9 & | & b_4 \end{bmatrix} [ A ∣ B ] = 1 2 3 3 2 4 7 9 ∣ ∣ ∣ ∣ b 1 b 2 b 3 b 4 Perform row operations:
1. R 2 → R 2 − 2 R 1 ⟹ [ 1 2 ∣ b 1 0 0 ∣ b 2 − 2 b 1 3 7 ∣ b 3 3 9 ∣ b 4 ] R_2 \to R_2 - 2R_1 \implies \begin{bmatrix} 1 & 2 & | & b_1 \\ 0 & 0 & | & b_2 - 2b_1 \\ 3 & 7 & | & b_3 \\ 3 & 9 & | & b_4 \end{bmatrix} R 2 → R 2 − 2 R 1 ⟹ 1 0 3 3 2 0 7 9 ∣ ∣ ∣ ∣ b 1 b 2 − 2 b 1 b 3 b 4 b 2 − 2 b 1 = 0 ⟹ b 2 = 2 b 1 b_2 - 2b_1 = 0 \implies b_2 = 2b_1 b 2 − 2 b 1 = 0 ⟹ b 2 = 2 b 1 2. R 3 → R 3 − 3 R 1 ⟹ [ 1 2 ∣ b 1 0 0 ∣ 0 0 1 ∣ b 3 − 3 b 1 3 9 ∣ b 4 ] R_3 \to R_3 - 3R_1 \implies \begin{bmatrix} 1 & 2 & | & b_1 \\ 0 & 0 & | & 0 \\ 0 & 1 & | & b_3 - 3b_1 \\ 3 & 9 & | & b_4 \end{bmatrix} R 3 → R 3 − 3 R 1 ⟹ 1 0 0 3 2 0 1 9 ∣ ∣ ∣ ∣ b 1 0 b 3 − 3 b 1 b 4 3. R 4 → R 4 − 3 R 1 ⟹ [ 1 2 ∣ b 1 0 0 ∣ 0 0 1 ∣ b 3 − 3 b 1 0 3 ∣ b 4 − 3 b 1 ] R_4 \to R_4 - 3R_1 \implies \begin{bmatrix} 1 & 2 & | & b_1 \\ 0 & 0 & | & 0 \\ 0 & 1 & | & b_3 - 3b_1 \\ 0 & 3 & | & b_4 - 3b_1 \end{bmatrix} R 4 → R 4 − 3 R 1 ⟹ 1 0 0 0 2 0 1 3 ∣ ∣ ∣ ∣ b 1 0 b 3 − 3 b 1 b 4 − 3 b 1 4. R 4 → R 4 − 3 R 3 ⟹ [ 1 2 ∣ b 1 0 0 ∣ 0 0 1 ∣ b 3 − 3 b 1 0 0 ∣ ( b 4 − 3 b 1 ) − 3 ( b 3 − 3 b 1 ) ] R_4 \to R_4 - 3R_3 \implies \begin{bmatrix} 1 & 2 & | & b_1 \\ 0 & 0 & | & 0 \\ 0 & 1 & | & b_3 - 3b_1 \\ 0 & 0 & | & (b_4 - 3b_1) - 3(b_3 - 3b_1) \end{bmatrix} R 4 → R 4 − 3 R 3 ⟹ 1 0 0 0 2 0 1 0 ∣ ∣ ∣ ∣ b 1 0 b 3 − 3 b 1 ( b 4 − 3 b 1 ) − 3 ( b 3 − 3 b 1 ) The last row gives
b 4 − 3 b 1 − 3 b 3 + 9 b 1 = 6 b 1 − 3 b 3 + b 4 = 0 b_4 - 3b_1 - 3b_3 + 9b_1 = 6b_1 - 3b_3 + b_4 = 0 b 4 − 3 b 1 − 3 b 3 + 9 b 1 = 6 b 1 − 3 b 3 + b 4 = 0 .
Thus, the conditions are
b 2 = 2 b 1 b_2 = 2b_1 b 2 = 2 b 1 and
6 b 1 − 3 b 3 + b 4 = 0 6b_1 - 3b_3 + b_4 = 0 6 b 1 − 3 b 3 + b 4 = 0 .
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