[(1)1+alpha^(2)+beta gamma=0],[(2)1-alpha^(2)+beta gamma=0],[(3)1+alpha^(2)-beta gamma=0],[(4)1-alpha^(2)-beta gamma=0]

If A=[[alpha,betagamma,alpha]] is such that A^(2)=I, then (A)1+alpha^(2)+beta gamma=0(B)1-alpha^(2)+beta gamma,=0(C)1-alpha^(2)-beta gamma,=0(D)1+alpha^(2)-beta gamma=0

If A=[alpha beta gamma-alpha] is such that A^(2)=I, then 1+alpha^(2)+beta gamma=0( b) 1-alpha^(2)+beta gamma=0 (c) 1-alpha^(2)-beta gamma=0 (d) 1+alpha^(2)-beta gamma=0

If A=[[alpha ,beta],[ gamma , -alpha]] is such that A^2=I , then A) 1+alpha^2+beta gamma=0 B) 1-alpha^2+beta gamma=0 C) 1-alpha^2-beta gamma=0^ D) 1+alpha^2-beta gamma=0

If alpha,beta,gamma are such that alpha+beta+gamma=2alpha^(2)+beta^(2)+gamma^(2)=6,alpha^(3)+beta^(3)+gamma^(3)=8, then alpha^(4)+beta^(4)+gamma^(4)

Prove that |[[alpha,beta,gamma],[alpha^2,beta^2,gamma^2],[alpha^3,beta^3,gamma^3]]|= alpha beta gamma (alpha-beta)(beta-gamma)(gamma-alpha)

Show that | (1,1,1), (alpha ^ 2, beta ^ 2, gamma ^ 2), (alpha ^ 3, beta ^ 3, gamma ^ 3) | = (alpha-beta) (beta-gamma) (gamma-alpha) (alphabeta + betagamma + gammaalpha) |

Using properties of determinants prove that : {:|( alpha , alpha ^(2) , beta +gamma ),( beta , beta ^(2) , gamma +alpha ),( gamma , gamma ^(2) ,alpha +beta ) |:} =(beta -gamma ) (gamma -alpha ) (alpha -beta ) (alpha +beta +gamma )