Consider `Delta=|(alpha,beta,gamma),(alpha^2,beta^2,gamma^2),(beta+gamma,gamma+alpha,alpha+beta)|` Show that `Delta=(alpha-beta)(beta-gamma)(gamma-alpha)(alpha+beta+gamma)`.

Consider Delta=|(alpha,beta,gamma),(alpha^2,beta^2,gamma^2),(beta+gamma,gamma+alpha,alpha+beta)| Apply R_3rarrR_1+R_3

Prove that |(2,alpha+beta+gamma+delta,alphabeta+gammadelta),(alpha+beta+gamma+delta,2(alpha+beta)(gamma+delta),alpha beta(gamma+delta)+gamma delta(gamma+beta)),(alpha beta+gamma delta,alpha beta(gamma+delta)+gammadelta(alpha+beta),2alphabeta gamma delta)| = 0

|(alpha,-beta,0),(0,alpha,beta),(beta,0,alpha)|=0 , then

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)

Let A=[(0,2beta,gamma),(alpha,beta,-gamma),(alpha,-beta,gamma)] be 3xx3 matrix Find A^T

Let A=[(0,2beta,gamma),(alpha,beta,-gamma),(alpha,-beta,gamma)] be 3xx3 matrix If A^T.A=6I, then alpha,beta and gamma is -

If alpha, beta, gamma are the roots of ax^(3) + bx^(2) + cx + d = 0 and |(alpha,beta,gamma),(beta,gamma,alpha),(gamma,alpha,beta)| = 0, alpha != beta != gamma , then find the equation whose roots are alpha + beta - gamma, beta + gamma - alpha and gamma + alpha - beta

If f(x) = a + bx + cx^(2) and alpha, beta, gamma are roots of the equation x^(3) = 1 , then |(a,b,c),(b,c,a),(c,a,b)| is equal to a) f(alpha) + f(beta) + f(gamma) b) f(alpha)f(beta) + f(beta)f(gamma) + f(gamma)f(alpha) c) f(alpha)f(beta)f(gamma) d) -f(alpha)f(beta)f(gamma)

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