JEE Main 2025 — Determinants Question with Solution
JEE Main 2025 (22 Jan Shift 2)
Question
If the system of linear equations :
$\begin{aligned}
& x+y+2 z=6 \\
& 2 x+3 y+\mathrm{a} z=\mathrm{a}+1 \\
& -x-3 y+\mathrm{b} z=2 \mathrm{~b}
\end{aligned}$
where , has infinitely many solutions, then is equal to :
Choose an option
Show full solutionCorrect option: A
Correct answer
A16
Step-by-step explanation
The given equations are
$\begin{aligned}
& x+y+2 z=6 \\
& 2 x+3 y+a z=a+1 \\
& -x-3 y+b z=2 b, \text { where } a, b, \in \mathbf{R}
\end{aligned}$
For infinite many solutions:
$\begin{aligned}
& \therefore D=\left|\begin{array}{ccc}
1 & 1 & 2 \\
2 & 3 & a \\
-1 & -3 & b
\end{array}\right|=2 a+b-6 \\
& D_1=\left|\begin{array}{ccc}
6 & 1 & 2 \\
a+1 & 3 & a \\
2 b & -3 & b
\end{array}\right|=12 a+5 b+a b-6 \\
& D_2=\left|\begin{array}{ccc}
1 & 6 & 2 \\
2 & a+1 & a \\
-1 & 2 b & b
\end{array}\right|=-4 a-3 b-a b+2 \\
& \text { and } D_3=\left|\begin{array}{ccc}
1 & 1 & 6 \\
2 & 3 & a+1 \\
-1 & -3 & 2 b
\end{array}\right|=2 a+2 b-16
\end{aligned}$
from above relations
$\begin{aligned}
& a=-2, b=10 \\
& \therefore 7 a+3 b=16
\end{aligned}$
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