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JEE Main 201322 AprilOnline
Exam
JEE Main
Year
2013
Date
22 April
Mode
Online
Subject
Mathematics
Chapter
Determinants

JEE Main 2013 · 22 April · Online · Mathematics · Determinants

If the system of linear equations : aligned & x_1+2 x_2+3 x_3=6 &...

If the system of linear equations : x1+2x2+3x3=6x1+3x2+5x3=92x1+5x2+ax3=b\begin{aligned} & x_1+2 x_2+3 x_3=6 \\ & x_1+3 x_2+5 x_3=9 \\ & 2 x_1+5 x_2+a x_3=b \end{aligned} is consistent and has infinite number of solutions, then :

Options

  1. Aa=8,ba=8, b can be any real number
  2. Bb=15,ab=15, a can be any real number
  3. Ca∈R−{8}a \in R-\{8\} and b∈R−{15}b \in R-\{15\}
  4. Da=8,b=15a=8, b=15

Correct answer

D. a=8,b=15a=8, b=15

Step-by-step solution

Given system of equations can be written in matrix form as AX=B\mathrm{AX}=\mathrm{B} where \mathrm{A}=\left(12313525a\begin{array}{lll}1 & 2 & 3 \\ 1 & 3 & 5 \\ 2 & 5 & a\end{array}\right) and \mathrm{B}=\left(69b\begin{array}{l}6 \\ 9 \\ b\end{array}\right)Since,systemisconsistentandhasinfinitelymanysolutionsSince, system is consistent and has infinitely many solutions\therefore(\operatorname{adj} . \mathrm{A}) \mathrm{B}=0 ⇒(3a−2515−2a110−aa−6−2−1−11)(69b)=(000)⇒−6−9+b=0⇒b=15 and 6(10−a)+9(a−6)−2(b)=0⇒60−6a+9a−54−30=0⇒3a=24⇒a=8\begin{aligned}& \Rightarrow\left(\begin{array}{ccc}3 a-25 & 15-2 a & 1 \\ 10-a & a-6 & -2 \\ -1 & -1 & 1 \end{array}\right)\left(\begin{array}{l}6 \\ 9 \\ b \end{array}\right)=\left(\begin{array}{l}0 \\ 0 \\ 0 \end{array}\right) \\ & \Rightarrow-6-9+b=0 \Rightarrow b=15 \\ & \text{ and }6(10-a)+9(a-6)-2(b)=0 \\ & \Rightarrow 60-6 a+9 a-54-30=0 \\ & \Rightarrow 3 a=24 \Rightarrow a=8 \end{aligned}Hence, a=8,b=15a=8, b=15.
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