Question:
Let $a, b, c \in R$ be all non-zero and satisfy $\mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3}=2$. 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 :
Correct Option: , 4
Solution:
$\mathrm{A}^{\mathrm{T}} \mathrm{A}=\mathrm{I}$
$\Rightarrow a^{2}+b^{2}+c^{2}=1$
and $a b+b c+c a=0$
Now, $(a+b+c)^{2}=1$
$\Rightarrow a+b+c=\pm 1$
So, $a^{3}+b^{3}+c^{3}-3 a b c$
$=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
$=\pm 1(1-0)=\pm 1$
$\Rightarrow 3 \mathrm{abc}=2 \pm 1=3,1$
$\Rightarrow a b c=1, \frac{1}{3}$