[x+y+z=6],[x-y+z=2],[2x+y-z=1]

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

The augmented matrix of x+y+z=6, 2x-y+z=3, 2y-z+x=2 is

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)

show that |[y+z ,x, y],[ z+x, z, x],[x+y, y ,z]|=(x+y+z)(x-z)^2

If cos^(-1)x+cos^(-1)y+cos^(-1)z=pi,t h e n x^2+y^2+z^2+x y z=0 x^2+y^2+z^2+2x y z=0 x^2+y^2+z^2+x y z=1 x^2+y^2+z^2+2x y z=1

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

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

The value of x, y, z for the following system of equations x + y + z = 6, x - y + 2z = 5, 2x + y - z = 1 are

2x + 3y-5z = 7, x + y + z = 6,3x-4y + 2z = 1, then x =