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JEE Main 202522 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
22 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Determinants

JEE Main 2025 · 22 January · Evening Shift · Mathematics · Determinants

If the system of linear equations : aligned & x+y+2 z=6 & 2 x+3 y+a...

If the system of linear equations : x+y+2z=62x+3y+az=a+1−x−3y+bz=2 b\begin{aligned} & x+y+2 z=6 \\ & 2 x+3 y+\mathrm{a} z=\mathrm{a}+1 \\ & -x-3 y+\mathrm{b} z=2 \mathrm{~b} \end{aligned} where a,b∈Ra, b \in \mathbf{R}, has infinitely many solutions, then 7a+3b7 a+3 b is equal to :

Options

  1. A16
  2. B12
  3. C22
  4. D9

Correct answer

A. 16

Step-by-step solution

The given equations are x+y+2z=62x+3y+az=a+1−x−3y+bz=2b, where a,b,∈R\begin{aligned} & x+y+2 z=6 \\ & 2 x+3 y+a z=a+1 \\ & -x-3 y+b z=2 b, \text{ where } a, b, \in \mathbf{R} \end{aligned} For infinite many solutions: D=D1=D2=D3=0D=D_1=D_2=D_3=0∴D=∣11223a−1−3b∣=2a+b−6D1=∣612a+13a2b−3b∣=12a+5b+ab−6D2=∣1622a+1a−12bb∣=−4a−3b−ab+2 and D3=∣11623a+1−1−32b∣=2a+2b−16\begin{aligned} & \therefore D=\left|\begin{array}{ccc} 1 & 1 & 2 \\ 2 & 3 & a \\ -1 & -3 & b \end{array}\right|=2 a+b-6 \\ & D_1=\left|\begin{array}{ccc} 6 & 1 & 2 \\ a+1 & 3 & a \\ 2 b & -3 & b \end{array}\right|=12 a+5 b+a b-6 \\ & D_2=\left|\begin{array}{ccc} 1 & 6 & 2 \\ 2 & a+1 & a \\ -1 & 2 b & b \end{array}\right|=-4 a-3 b-a b+2 \\ & \text{ and } D_3=\left|\begin{array}{ccc} 1 & 1 & 6 \\ 2 & 3 & a+1 \\ -1 & -3 & 2 b \end{array}\right|=2 a+2 b-16 \end{aligned}fromaboverelationsfrom above relationsa=−2,b=10∴7a+3b=16\begin{aligned} & a=-2, b=10 \\ & \therefore 7 a+3 b=16 \end{aligned}$
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