Let a, b, c ∈ R be all non-zero and satisfy a^{3}+b^{3}+c^{3}=2 . If matrix A=left(begin{array}{ll

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3 and 4 .Determinants and Matrices

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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

$\frac{1}{3}$

(JEE MAIN-2020)

Solution

$A^{T} A=I$

$\Rightarrow a^{2}+b^{2}+c^{2}=1$

and $a b+b c+c a=0$

Now, $(a+b+c)^{2}=1$

$\Rightarrow a+b+c=\pm 1$

So, $a^{3}+b^{3}+c^{3}-3 a b c$

$=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$

$=\pm 1(1-0)=\pm 1$

$\Rightarrow 3 a b c=2 \pm 1=3,1$

$\Rightarrow \quad a b c=1, \frac{1}{3}$

Standard 12

Mathematics

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