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Prove that: `|(alpha,beta,gamma),(alpha^2,beta^2,gamma^2),(beta+gamma,gamma+alpha,alpha+beta)|``=(alpha-beta)(beta-gamma)(gamma-alpha)(alpha+beta+gamma)`. |
Answer» Let the value of the given determinant be `Delta`. Then. `Delta = |{:(alpha, beta, gamma), (alpha^(2), beta^(2), gamma^(2)), (beta + gamma, gamma + alpha, alpha + beta):}|` `=|{:(alpha, beta, gamma), (alpha^(2), beta^(2), gamma^(2)), (alpha +beta + gamma, alpha + beta +gamma, alpha + beta +gamma):}|["applying "R_(3) to (R_(3) + R_(1))]` `=(alpha + beta +gamma) * |{:(alpha, beta, gamma), (alpha^(2), beta^(2), gamma^(2)), (1, 1, 1):}|["taking" (alpha + beta + gamma)"common from "R_(3)]` `=(alpha + beta +gamma) * |{:(alpha-gamma, beta-gamma, gamma), (alpha^(2)-gamma^(2), beta^(2)-gamma^(2), gamma^(2)), (0, 0, 1):}|["applying"C_(1)to (C_(1)-C_(3))"and "C_(2) to (C_(2)-C_(3))]` `=(alpha + beta +gamma)(alpha-gamma)(beta-gamma) * |{:(1, 1, gamma), (alpha+gamma, beta+gamma, gamma^(2)), (0, 0, 1):}|` `["taking" (alpha-gamma) "common from "C_(1) " and "(beta-gamma) "common from"C_(2)]` `= (alpha + beta + gamma)(alpha - gamma)(beta -gamma) *1* |{:(1, 1), (alpha + gamma, beta + gamma):}| " "["expanded by"R_(3)]` `=(alpha + beta + gamma) (alpha- gamma) (beta -gamma) [(beta + gamma) - (alpha + gamma)]` `=(alpha + beta + gamma)(alpha -gamma)(beta-gamma)(beta -alpha)` `=(alpha -beta)(beta-gamma)(gamma-alpha) (alpha + beta +gamma).` Hence, `Delta = (alpha-beta)(beta-gamma)(gamma-alpha)(alpha + beta + gamma).` |
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