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Maths Question 3 – JEE-MAIN 2026

If the system of linear equations: x+y+z=6 x+2y+5z=10 2x+3y+λz=μ has infinitely many solutions, then the value of λ+μ equals :

For a system of linear equations AX=B to have infinitely many solutions, the determinant of the coefficient matrix |A| must be zero, and the system must be consistent (i.e., rank of A equals rank of augmented matrix [A|B] and is less than the number of variables).

Step 1: Form Coefficient Matrix and Apply Determinant Condition✦ Active

The given system of linear equations is: x+y+z=6 x+2y+5z=10 2x+3y+λz=μ For infinitely many solutions, the determinant of the coefficient matrix A must be zero.

A=(11112523λ) |A|=1(2λ15)1(λ10)+1(34) |A|=2λ15λ+101 |A|=λ6 Setting |A|=0 gives: λ6=0λ=6
Step 2: Apply Consistency Condition using Row Operations○ Expand

Substitute λ=6 into the augmented matrix [A|B] and perform row operations to find μ for consistency (infinitely many solutions).

(111|6125|10236|μ) R2R2R1 and R3R32R1: (111|6014|4014|μ12) R3R3R2: (111|6014|4000|(μ12)4) For infinitely many solutions, the last row must be all zeros, so the last element must be zero: (μ12)4=0 μ16=0μ=16
💡 Teacher's Secret Hint

Alternatively, one could use Cramer's rule and set Dx=0 (or Dy=0, Dz=0) to find μ after finding λ.

Step 3: Calculate λ+μ○ Expand

Now, calculate the value of λ+μ.

λ+μ=6+16=22
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