Question

The following system of linear equations
$7x+6y-2z=0$,
$3x+4y+2z=0$
$x-2y-6z=0$, has

• A. infinitely many solutions, $(x,y,z)$ satisfying $y=2z$
• B. infinitely many solutions $(x,y,z)$ satisfying $x=2z$
• C. no solution
• D. only the trivial solution

Answer 1

The correct option is B infinitely many solutions $(x,y,z)$ satisfying $x=2z$

$7x+6y-2z=0$
$3x+4y+2z=0$
$x-2y-6z=0$
As the system of equations are Homogeneous $\Rightarrow$ the system is consistent.
$\Rightarrow \left|\begin{array}{ccc}7& 6& -2\\ 3& 4& 2\\ 1& -2& -6\end{array}\right|=0$
$\Rightarrow$ Infinite solutions exist (both trivial and non-trivial solutions)
When $y=2z$
Let's take $y=2,z=1$
When $(x,2,1)$ is substituted in the system of equations
$\Rightarrow 7x+10=0,3x+10=0,x-10=0$ (which is not possible)
$\therefore y=2x\Rightarrow$ Infinitely many solutions does not exist.
For $x=2z$, lets take $x=2,z=1,y=y$
Substitute $(2,y,1)$ in system of equations
$\Rightarrow y=-2$
$\because$ For each pair of $(x,z)$, we get a value of $y$.
Therefore, for $x=2z$ , infinitely many solutions exists.