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.
Also $\left|\begin{array}{ccc}7& 6& -2\\ 3& 4& 2\\ 1& -2& -6\end{array}\right|=0$
$\Rightarrow$ Infinite solutions exist (both trivial and non-trivial solutions)
By replacing $y$ value from equation(iii) i.e $2y=x-6z$ in equation $3x+4y+2z=0,$ we have $x=2z$
Therefore, for $x=2z$ , infinitely many solutions exists.
If we replace $x$ value,then we get$y=-2z$
Therefore, for $y=-2z,$ infinitely many solutions exist