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Let a, b, c ∈R be all non-zero and satisfy a3 + b3 + c3 = 2. If the matrixsatisfies ATA = I, then a value of abc can be :(1) \(\frac{2}{3}\)(2) \(\frac{1}{3}\)(3) 3(4) \(\frac{1}{3}\) |
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Answer» Answer is (4) \(\frac{1}{3}\) ATA = I ⇒ a2 + b2 + c2 = 1 and ab + bc + ca = 0 Now, (a + b + c) 2 = 1 ⇒a + b + c = ± 1 So, a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) = ± 1 (1 – 0) = ± 1 ⇒ 3 abc = 2 ± 1 = 3, 1 ⇒ abc = 1,\(\frac{1}{3}\) |
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