JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
Let \(a, b, c \in R\) be all non-zero and satisfy \(a^{3}+b^{3}+c^{3}=2 .\) If the matrix \(A=\left(\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right)\) satisfies \(\mathrm{A}^{\mathrm{T}} \mathrm{A}=\mathrm{I},\) then a value of \(abc\) can be
- A \(\frac{2}{3}\)
- B \(-\frac{1}{3}\)
- C \(3\)
- D \(\frac{1}{3}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
\(A^{T} A=I\) \(\Rightarrow a^{2}+b^{2}+c^{2}=1\) and \(a b+b c+c a=0\) Now, \((a+b+c)^{2}=1\) \(\Rightarrow a+b+c=\pm 1\) So, \(a^{3}+b^{3}+c^{3}-3 a b c\) \(=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)\) \(=\pm 1(1-0)=\pm 1\) \(\Rightarrow 3 a b c=2 \pm 1=3,1\)…
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