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

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

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

x+2y-2z=5,3x-y+z=8,x+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)

The values of x, y, z for the equations x+y+z=6, 3x-y+2z=7, 5x+5y-4z=3 are

Solve for x by Cramers rule x+y+z=6 , x-y+z=2 3x+2y-4z=-5

Solve for x by Cramers rule x+y+z=6 , x-y+z=2 3x+2y-4z=-5

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