InterviewSolution
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InterviewSolution
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Prove that`|[b+c,a,b],[c+a,c,a],[a+b,b,c]|=(a+b+c)(a-c)^2` |
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Answer» Let the given determinant be `Delta`. Then, `Delta = |{:(b+c, a, b), (c+a, c, a), (a+b, b, c):}|` `= |{:(b, a, b), (c, c, a), (a, b, c):}| + |{:(c, a, b), (a, c, a), (b, b, c):}|` `= |{:(a+b+c, a+b+c, a+b+c), (" "c, " "c, " "a), (" "a, " "b, " "c):}| + |{:(a+b+c, a+b+c, a+b+c), (" "a, " "c, " "a), (" "b, " "b, " "c):}| [R_(1) to (R_(1) + R_(2) + R_(3))" in each determinant"]` `=(a+b+c) *|{:(1, 1, 1), (c, c, a), (a, b, c):}| + (a+b+c) * |{:(1, 1, 1), (a, c, a),(b, b,c):}| ["taking out(a+b+c) common from"R_(1) "in each determinant"]` `=(a+b+c) * |{:(1, " "0, " "0), (c, " "0, a-c), (a, b-a, c-a):}| +(a+b+c)* |{:(1, " "0, 0), (a, c-a, 0), (b, " "0, c-b):}| [C_(1) to (C_(2) - C_(1)) "and "C_(3) to (C_(3)-C_(1))"in each"]` `=(a+b+c) *1* |{:(0, a-c), (b-a, c-a):}| +(a+b+c)*1* |{:(c-a, " "0), (" "0, c-b):}| ["each det. expanded by"R_(1)]` `=(a+b+c)*[0-(b-a)(a-c)] + (a+b+c)(c-a)(c-b)` `=(a+b+c)(a-b)(a-c) - (a+b+c) (a-c)(c-b)` `=(a+b+c)(a-c){(a-b)-(c-b)}` `=(a+b+c)(a-c)^(2)`, Hence, `Delta = (a+b+c)(a-c)^(2)` |
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