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 $a, b \in \mathbf{R}$, has infinitely many solutions, then $7 a+3 b$ is equal to :

Question

Aicharya · Accepted Answer

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:
$D=D_1=D_2=D_3=0$
$\begin{aligned}
& \therefore D=\left|\begin{array}{ccc}
1 & 1 & 2 \\ 2 & 3 & a \\ -1 & -3 & b
\end{array}\right|=2 a+b-6 \\ & D_1=\left|\begin{array}{ccc}
6 & 1 & 2 \\ a+1 & 3 & a \\ 2 b & -3 & b
\end{array}\right|=12 a+5 b+a b-6 \\ & D_2=\left|\begin{array}{ccc}
1 & 6 & 2 \\ 2 & a+1 & a \\ -1 & 2 b & b
\end{array}\right|=-4 a-3 b-a b+2 \\ & \text { and } D_3=\left|\begin{array}{ccc}
1 & 1 & 6 \\ 2 & 3 & a+1 \\ -1 & -3 & 2 b
\end{array}\right|=2 a+2 b-16
\end{aligned}$
from above relations
$\begin{aligned}
& a=-2, b=10 \\ & \therefore 7 a+3 b=16
\end{aligned}$

If the system of linear equations : $\begin{aligned} & x+y+2 z=6 \\ & 2 x+3 y+\mathrm{a} z=\mathrm{a}+1 \\ &…

Solution