Question

The number of solution (s) of the system of equation $x+4y-z=0,3x-4y-z=0$ and $x-3y+z=0$ is

• A. Zero
• B. $1$
• C. $2$
• D. Infinitely many

Answer 1

The correct option is D Infinitely many

$x+y+z=0-\left(1\right)$

$2x+3y+z=0-\left(2\right)$

$x+2y=0-\left(3\right)$

From equation (3), we have

$x=-2y$

Putting this value of $x$ in equations (1), we

get $-2y+y+z=0⟹y=z$

Hence $x=-2z$

Thus, the solution of the given system of equations is $(-2z,z,z),$ where z is a parameter $(Z\in R)$.

Hence the system has infinite number of solutions including zero solution.