[x-y+z=1],[2x+y-z=2],[x-2y-z=4]

2x+y+z=1x-y+2z=-1,3x+2y-z=4

Show that: |[x-y-z,2x,2x],[2y,y-z-x,2y],[2z,2z,z-x-y]|=(x+y+z)^3

2x-y+z=6,x+2y+3z=3,3x+y-z=4

Prove the identities: |[z, x, y],[ z^2,x^2,y^2],[z^4,x^4,y^4]|=|[x, y, z],[ x^2,y^2,z^2],[x^4,y^4,z^4]|=|[x^2,y^2,z^2],[x^4,y^4,z^4],[x, y, z]| =x y z (x-y)(y-z)(z-x)(x+y+z)

Prove the identities: |[z, x, y],[ z^2,x^2,y^2],[z^4,x^4,y^4]|=|[x, y, z],[ x^2,y^2,z^2],[x^4,y^4,z^4]|=|[x^2,y^2,z^2],[x^4,y^4,z^4],[x, y, z]| =x y z (x-y)(y-z)(z-x)(x+y+z)

The values of x, y, z for the equations x-y+z=1, 2x-y=1, 3x+3y-4z=2 are

The solution of 2x+y+z=1,x-2y-3z=1,3x+2y+4z=5 is

x+y+z=4 2x-y+z=-1 2x+y-3z=-9

2x+y-z=1 x-y+z=2 3x+y-2z=-1