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)

$7x + 6y - 2z = 0\quad.... \left(1\right)$
$3x +4y + 2z = 0\quad .... \left(2\right)$
$x - 2y - 6z = 0\quad .... \left(3\right)$
$\Delta = \begin{vmatrix}7&6&-2\\ 3&4&2\\ 1&-2&-6\end{vmatrix} = 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 $x = 2z$