Maths Q3 – JEE-MAIN 2026

If the system of linear equations:

x + y + z = 6

x + 2 y + 5 z = 10

2 x + 3 y + λ z = μ

has infinitely many solutions, then the value of

λ + μ

equals :

Clue 1 of 3: Condition for Infinitely Many Solutions

For a system of linear equations $A X = B$ 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).

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● Video Walkthrough

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

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

A = (\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 5 \\ 2 & 3 & λ \end{matrix})

| A | = 1 (2 λ - 15) - 1 (λ - 10) + 1 (3 - 4)

| A | = 2 λ - 15 - λ + 10 - 1

| A | = λ - 6

Setting

| A | = 0

gives:

λ - 6 = 0 ⟹ λ = 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).

(\begin{matrix} 1 & 1 & 1 & | & 6 \\ 1 & 2 & 5 & | & 10 \\ 2 & 3 & 6 & | & μ \end{matrix})

R_{2} \to R_{2} - R_{1}

and

R_{3} \to R_{3} - 2 R_{1}

(\begin{matrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 4 & | & 4 \\ 0 & 1 & 4 & | & μ - 12 \end{matrix})

R_{3} \to R_{3} - R_{2}

(\begin{matrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 4 & | & 4 \\ 0 & 0 & 0 & | & (μ - 12) - 4 \end{matrix})

For infinitely many solutions, the last row must be all zeros, so the last element must be zero:

(μ - 12) - 4 = 0

μ - 16 = 0 ⟹ μ = 16

💡 Teacher's Secret Hint

Alternatively, one could use Cramer's rule and set $D_{x} = 0$ (or $D_{y} = 0$ , $D_{z} = 0$ ) 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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Maths Question 3 – JEE-MAIN 2026

Clue 1 of 3: Condition for Infinitely Many Solutions

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