GATE 2020 EC Q.26: solution and answer

GATE 2020 · ECQ.26 · 2 marks · MCQ

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}

Which one of the following conditions ensures that a solution exists for the above system?

(A)
$b_{2} = 2 b_{1}$ and $6 b_{1} - 3 b_{3} + b_{4} = 0$
(B)
$b_{3} = 2 b_{1}$ and $6 b_{1} - 3 b_{3} + b_{4} = 0$
(C)
$b_{2} = 2 b_{1}$ and $3 b_{1} - 6 b_{3} + b_{4} = 0$
(D)
$b_{3} = 2 b_{1}$ and $3 b_{1} - 6 b_{3} + b_{4} = 0$

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Answer and solution

ANSWER

The system has a solution if and only if $b_{2} = 2 b_{1}$ and $6 b_{1} - 3 b_{3} + b_{4} = 0$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2020 EC (IIT Delhi)

A solution exists exactly when the system is consistent, i.e. $rank (A) = rank ([A ∣ b])$ . The coefficient matrix is $A = 12332479$ . Rows 1 and 3 are independent because $1 \cdot 7 - 2 \cdot 3 = 1 \neq = 0$ , and $A$ has only 2 columns, so $rank (A) = 2$ . With 4 equations and rank 2, there are $4 - 2 = 2$ independent conditions on $b$ .
First condition: the left side of equation 2 is exactly twice the left side of equation 1, $2 x_{1} + 4 x_{2} = 2 (x_{1} + 2 x_{2})$ . For consistency the right sides must satisfy the same relation: $b_{2} = 2 b_{1}$ .
Second condition: solve equations 1 and 3 for the unique $(x_{1}, x_{2})$ . Subtracting $3 \times$ (eq. 1) from eq. 3 gives $x_{2} = b_{3} - 3 b_{1}$ . Then $x_{1} = b_{1} - 2 x_{2} = b_{1} - 2 (b_{3} - 3 b_{1}) = 7 b_{1} - 2 b_{3}$ .
Substitute into equation 4: $3 x_{1} + 9 x_{2} = 3 (7 b_{1} - 2 b_{3}) + 9 (b_{3} - 3 b_{1}) = 21 b_{1} - 6 b_{3} + 9 b_{3} - 27 b_{1} = - 6 b_{1} + 3 b_{3}$ Consistency requires this to equal $b_{4}$ , so $- 6 b_{1} + 3 b_{3} = b_{4}$ , i.e. $6 b_{1} - 3 b_{3} + b_{4} = 0$ .
Cross-check with the row dependency: $6 (1, 2) - 3 (3, 7) + 1 (3, 9) = (6 - 9 + 3, 12 - 21 + 9) = (0, 0)$ . The same combination applied to the right-hand side has to vanish, which again gives $6 b_{1} - 3 b_{3} + b_{4} = 0$ .
The two conditions are also sufficient. If they hold, the $(x_{1}, x_{2})$ found from equations 1 and 3 satisfies equation 2 (because $2 x_{1} + 4 x_{2} = 2 b_{1} = b_{2}$ ) and equation 4 (step 4), so a solution exists.
Both conditions, $b_{2} = 2 b_{1}$ and $6 b_{1} - 3 b_{3} + b_{4} = 0$ , appear together only in option (A). Options (B) and (D) drop the required $b_{2} = 2 b_{1}$ . Option (C) replaces the second condition with $3 b_{1} - 6 b_{3} + b_{4} = 0$ , which does not guarantee a solution: for example $b = (1, 2, 0, - 3)$ satisfies (C), but $6 b_{1} - 3 b_{3} + b_{4} = 3 \neq = 0$ , so that system has no solution.

KEY IDEA

A system $A x = b$ is consistent exactly when $b$ satisfies every linear dependency among the rows of $A$ . Here $rank (A) = 2$ with 4 equations, so there are two such dependencies: row 2 equals $2 \times$ row 1, and $6 R_{1} - 3 R_{3} + R_{4} = 0$ . The right-hand side has to obey the same two relations.

WATCH OUT FOR

Spotting only the obvious condition $b_{2} = 2 b_{1}$ and then guessing the second one, or getting a sign wrong when eliminating. Solving equations 1 and 3 gives $x_{2} = b_{3} - 3 b_{1}$ and $x_{1} = 7 b_{1} - 2 b_{3}$ . Substituting into equation 4 gives $3 b_{3} - 6 b_{1} = b_{4}$ , which is $6 b_{1} - 3 b_{3} + b_{4} = 0$ , not $3 b_{1} - 6 b_{3} + b_{4} = 0$ . Also watch for option (B), which writes $b_{3} = 2 b_{1}$ in place of $b_{2} = 2 b_{1}$ .

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GATE 2020 EC Q.26: solution and answer

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