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MCQMediumJEE 2026Solving Linear Equations (Matrix Method)

JEE Mathematics 2026 Question with Solution

The system of linear equations

x+y+z=6x + y + z = 6

2x+5y+az=362x + 5y + az = 36

x+2y+3z=bx + 2y + 3z = b

has

  • A

    infinitely many solutions for a=8a=8 and b=14b=14

  • B

    infinitely many solutions for a=8a=8 and b=16b=16

  • C

    unique solution for a=8a=8 and b=16b=16

  • D

    unique solution for a=8a=8 and b=14b=14

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The system is

x+y+z=62x+5y+az=36x+2y+3z=b\begin{aligned} x + y + z &= 6 \\ 2x + 5y + az &= 36 \\ x + 2y + 3z &= b \end{aligned}

Find: For which values of aa and bb the system has infinitely many solutions.

Form the augmented matrix:

(111625a36123b)\left( \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 2 & 5 & a & 36 \\ 1 & 2 & 3 & b \end{array} \right)

Apply row operations R2→R2−2R1R_2 \to R_2 - 2R_1 and R3→R3−R1R_3 \to R_3 - R_1:

(111603a−224012b−6)\left( \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 3 & a-2 & 24 \\ 0 & 1 & 2 & b-6 \end{array} \right)

Swap R2R_2 and R3R_3:

(1116012b−603a−224)\left( \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & 2 & b-6 \\ 0 & 3 & a-2 & 24 \end{array} \right)

Now apply R3→R3−3R2R_3 \to R_3 - 3R_2:

(1116012b−600a−2−624−3(b−6))\left( \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & 2 & b-6 \\ 0 & 0 & a-2-6 & 24 - 3(b-6) \end{array} \right)

So the last row becomes

(0  0  a−8  ∣  42−3b)\left( 0 \; 0 \; a-8 \; | \; 42-3b \right)

For infinitely many solutions, the rank of the coefficient matrix must equal the rank of the augmented matrix, and both must be less than the number of variables. Therefore the entire last row must be zero:

a−8=0a - 8 = 0

and

42−3b=042 - 3b = 0

Hence,

a=8,3b=42  ⟹  b=14a = 8, \qquad 3b = 42 \implies b = 14

Therefore, the system has infinitely many solutions for a=8a=8 and b=14b=14. The correct option is A.

Rank Condition Explanation

A system of three linear equations in three variables has infinitely many solutions when it becomes dependent after elimination. That means after row reduction, one row must reduce to all zeros, so that the rank is less than 33 but the system remains consistent.

Here, consistency with dependence requires

(a−8)=0and(42−3b)=0(a-8) = 0 \quad \text{and} \quad (42-3b) = 0

If only a−8=0a-8=0 but 42−3b≠042-3b \ne 0, the last row would be of the form

0x+0y+0z=non-zero0x + 0y + 0z = \text{non-zero}

which is inconsistent. If a−8≠0a-8 \ne 0, then the rank becomes 33 and the system has a unique solution.

Common mistakes

  • Setting only a−8=0a-8=0 and stopping there is incorrect. For infinitely many solutions, the entire last row of the augmented matrix must become zero. You must also impose 42−3b=042-3b=0.

  • Confusing the condition for unique solution with infinite solutions is a common error. A unique solution occurs when the coefficient matrix has full rank 33. Infinite solutions require rank less than 33 but equal to the augmented rank.

  • After applying R3→R3−3R2R_3 \to R_3 - 3R_2, students may simplify the constant term incorrectly. Since 24−3(b−6)=24−3b+18=42−3b24 - 3(b-6) = 24 - 3b + 18 = 42 - 3b, careful expansion is necessary.

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