The number of solutions of equations x+4y-z=0,3x-4y-z=0,x-3y+z=0is

Answer 1

x+4y-z = 0

3x-4y-z = 0

x-3y+z=0

Subtract the second equation from the first to get: -2x+8y = 0

Add the first equation to the third to get: 2x + y = 0

Adding these equations, we have 9y = 0. So, y = 0

Back substitute to obtain x = y = z = 0