The given equations are $\begin{aligned}
& x+y+2 z=6 \
& 2 x+3 y+a z=a+1 \
& -x-3 y+b z=2 b, \text { where } a, b, \in \mathbf{R}
\end{aligned}Forinfinitemanysolutions:D=D_1=D_2=D_3=0\begin{aligned}
& \therefore D=\left|\begin{array}{ccc}
1 & 1 & 2 \
2 & 3 & a \
-1 & -3 & b
\end{array}\right|=2 a+b-6 \
& D_1=\left|\begin{array}{ccc}
6 & 1 & 2 \
a+1 & 3 & a \
2 b & -3 & b
\end{array}\right|=12 a+5 b+a b-6 \
& D_2=\left|\begin{array}{ccc}
1 & 6 & 2 \
2 & a+1 & a \
-1 & 2 b & b
\end{array}\right|=-4 a-3 b-a b+2 \
& \text { and } D_3=\left|\begin{array}{ccc}
1 & 1 & 6 \
2 & 3 & a+1 \
-1 & -3 & 2 b
\end{array}\right|=2 a+2 b-16
\end{aligned}fromaboverelations\begin{aligned}
& a=-2, b=10 \
& \therefore 7 a+3 b=16
\end{aligned}$
If the system of linear equations : … wherea, b \in \mathbf{R}, has infinitely many solutions, then7 a+3 b is equal to :
JEE Main 2025 — Mathematics Algebra
If the system of linear equations : $\begin{aligned}
& x+y+2 z=6 \
& 2 x+3 y+\mathrm{a} z=\mathrm{a}+1 \
& -x-3 y+\mathrm{b} z=2 \mathrm{~b}
\end{aligned}wherea, b \in \mathbf{R},hasinfinitelymanysolutions,then7 a+3 b$ is equal to :
Held on 22 Jan 2025 · Verified 6 Jul 2026.
Options
- A
16
- B
12
- C
22
- D
9
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