Step 1

we have to find all the solutions of the system of equations:

x+2y-z=0

2x+y+z=0

x-4y+5z=0

Step 2

the given system of equations is:

x+2y-z=0...(1)

2x+y+z=0...(2)

x-4y+5z=0...(3)

from the equation (1), we get

x+2y-z=0

x=-2y+z

now substitute the value of x in the equation (2) and (3),

2x+y+z=0

2(−2y+z)+y+z=0

−4y+2z+y+z=0

-3y+3z=0

-y+z=0...(4)

Step 3

x−4y+5z=0

(−2y+z)−4y+5z=0

−2y+z−4y+5z=0

−6y+6z=0

-y+z=0...(5)

therefore from equation (4) and (5), we can notice that the given system of equations has infinitely many solutions.

let z=k

therefore, from (4), we get

-y+z=0

-y+k=0

y=k

now substitute the value of y and z in x.

x=-2y+z

x=-2k+k

x=-k

Step 4

therefore, the values of x, y and z are -k, k and k respectively.

therefore all solutions of the given system of equations is x=-k, y=k and z=k

we have to find all the solutions of the system of equations:

x+2y-z=0

2x+y+z=0

x-4y+5z=0

Step 2

the given system of equations is:

x+2y-z=0...(1)

2x+y+z=0...(2)

x-4y+5z=0...(3)

from the equation (1), we get

x+2y-z=0

x=-2y+z

now substitute the value of x in the equation (2) and (3),

2x+y+z=0

2(−2y+z)+y+z=0

−4y+2z+y+z=0

-3y+3z=0

-y+z=0...(4)

Step 3

x−4y+5z=0

(−2y+z)−4y+5z=0

−2y+z−4y+5z=0

−6y+6z=0

-y+z=0...(5)

therefore from equation (4) and (5), we can notice that the given system of equations has infinitely many solutions.

let z=k

therefore, from (4), we get

-y+z=0

-y+k=0

y=k

now substitute the value of y and z in x.

x=-2y+z

x=-2k+k

x=-k

Step 4

therefore, the values of x, y and z are -k, k and k respectively.

therefore all solutions of the given system of equations is x=-k, y=k and z=k