Maths Q3 – JEE-MAIN 2026

If the system of equations

x + 5 y + 6 z = 4

2 x + 3 y + 4 z = 7

x + 6 y + a z = b

has infinitely many solutions, then the point

(a, b)

lies on the line

Clue 1 of 3: System of Linear Equations

For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero, and the system must be consistent.

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● Video Walkthrough

Step 1: Formulate Matrices and Conditions for Infinite Solutions✦ Active

The given system of equations is: $x + 5 y + 6 z = 4$ $2 x + 3 y + 4 z = 7$ $x + 6 y + a z = b$ For a system of linear equations to have infinitely many solutions, two conditions must be met: the determinant of the coefficient matrix $A$ must be zero ( $d e t (A) = 0$ ), and the rank of the coefficient matrix must be equal to the rank of the augmented matrix $[A | B]$ , and this rank must be less than the number of variables (3 in this case).

A = (\begin{matrix} 1 & 5 & 6 \\ 2 & 3 & 4 \\ 1 & 6 & a \end{matrix}), [A | B] = (\begin{matrix} 1 & 5 & 6 & 4 \\ 2 & 3 & 4 & 7 \\ 1 & 6 & a & b \end{matrix})

Step 2: Calculate Determinant and Find 'a'○ Expand

First, set the determinant of the coefficient matrix $A$ to zero:

d e t (A) = 1 (3 a - 24) - 5 (2 a - 4) + 6 (12 - 3)

d e t (A) = 3 a - 24 - 10 a + 20 + 54

d e t (A) = - 7 a + 50

Setting

d e t (A) = 0

gives:

- 7 a + 50 = 0 ⟹ 7 a = 50 ⟹ a = \frac{50}{7}

💡 Teacher's Secret Hint

Remember to expand the determinant carefully to avoid algebraic errors.

Step 3: Use Row Operations to Find 'b' and Identify the Line○ Expand

Next, perform row operations on the augmented matrix $[A | B]$ to ensure consistency and find the value of $b$ . We aim for the last row to be all zeros after reduction.

(\begin{matrix} 1 & 5 & 6 & 4 \\ 2 & 3 & 4 & 7 \\ 1 & 6 & a & b \end{matrix}) \overset{R_{2} \to R_{2} - 2 R_{1}, R_{3} \to R_{3} - R_{1}}{\to} (\begin{matrix} 1 & 5 & 6 & 4 \\ 0 & - 7 & - 8 & - 1 \\ 0 & 1 & a - 6 & b - 4 \end{matrix})

\overset{R_{2} \leftrightarrow R_{3}}{\to} (\begin{matrix} 1 & 5 & 6 & 4 \\ 0 & 1 & a - 6 & b - 4 \\ 0 & - 7 & - 8 & - 1 \end{matrix})

\overset{R_{3} \to R_{3} + 7 R_{2}}{\to} (\begin{matrix} 1 & 5 & 6 & 4 \\ 0 & 1 & a - 6 & b - 4 \\ 0 & 0 & - 8 + 7 (a - 6) & - 1 + 7 (b - 4) \end{matrix})

For infinitely many solutions, the last row must be entirely zeros. We already confirmed $- 8 + 7 (a - 6) = - 8 + 7 (\frac{50}{7} - 6) = - 8 + 50 - 42 = 0$ . Now, set the last element to zero:

- 1 + 7 (b - 4) = 0

- 1 + 7 b - 28 = 0

7 b - 29 = 0 ⟹ 7 b = 29 ⟹ b = \frac{29}{7}

So, the point $(a, b) = (\frac{50}{7}, \frac{29}{7})$ . Now, check which option line this point satisfies: 1. $y - x = 3 ⟹ \frac{29}{7} - \frac{50}{7} = - \frac{21}{7} = - 3 \neq 3$ 2. $x - y = 3 ⟹ \frac{50}{7} - \frac{29}{7} = \frac{21}{7} = 3$ . This option is correct. 3. $x + y = 11 ⟹ \frac{50}{7} + \frac{29}{7} = \frac{79}{7} \neq 11$ 4. $x + y = 12 ⟹ \frac{50}{7} + \frac{29}{7} = \frac{79}{7} \neq 12$ Therefore, the point $(a, b)$ lies on the line $x - y = 3$ .

💡 Teacher's Secret Hint

Ensure both conditions for infinite solutions (determinant zero and consistency) are used to find both $a$ and $b$ .

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Maths Question 3 – JEE-MAIN 2026

Clue 1 of 3: System of Linear Equations

🧠 Full Solution Path