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

Answer: (B)

$7x+6y-2z=0....\left(1\right)$
$3x+4y+2z=0....\left(2\right)$
$x-2y-6z=0....\left(3\right)$
$\Delta =\left|\begin{array}{ccc}7& 6& -2\\ 3& 4& 2\\ 1& -2& -6\end{array}\right|=0\Rightarrow$ infinite solutions
Now $\left(1\right)+\left(2\right)\Rightarrow y=-x$ put in $\left(1\right),\left(2\right)\&\left(3\right)$ all will lead to $x=2z$