Question

The system of equations
$x+3y+2z=6x+ay+2z=7x+3y+2z=b$
has

• A. a unique solution, if $a=2$ and $b\ne 6$
• B. infinitely many solutions, if $a=4$ and $b=6$
• C. no solution, if $a=5$ and $b=7$
• D. no solution, if $a=3$ and $b=5$

Answer 1

Answer: (B), (C), (D)

The correct options are
B infinitely many solutions, if $a=4$ and $b=6$
C no solution, if $a=5$ and $b=7$
D no solution, if $a=3$ and $b=5$

$P_1:x+3y+2z=6P_2:x+ay+2z=7P_3:x+3y+2z=b$
This system of equations represents three planes in $R^3.$
Clearly, $P_1$ and $P_3$ are parallel.

Case $1:$ $a=3$
In this case, $P_1$ and $P_2$ are parallel (not coincident).
So, the system has no solution.

Case $2:$ $a\ne 3$
I. When $b=6,$ $P_1$ and $P_3$ coincide.
Hence, the system has infinitely many solutions.

II. When $b\ne 6,$ $P_1$ and $P_3$ are parallel (not coincident).
Hence, the system has no solution.