Prove the following
Question:

Let $a, b, c \in \mathbf{R}$ 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 $A^{T} A=I$, then a value of $a b c$ can be $:$

  1. (1) $-\frac{1}{3}$

  2. (2) $\frac{1}{3}$

  3. (3) 3

  4. (4) $\frac{2}{3}$


Correct Option: , 2

Solution:

Given: $A^{T} A=I$

$\Rightarrow\left[\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right]\left[\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$\Rightarrow\left[\begin{array}{ccc}\Sigma a^{2} & \Sigma a b & \Sigma a b \\ \Sigma a b & \Sigma a^{2} & \Sigma a b \\ \Sigma a b & \Sigma a b & \Sigma a^{2}\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

So, $\Sigma a^{2}=1$ and $\Sigma a b=0$

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

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

$=(a+b+c)(1-0)$

$=\sqrt{(a+b+c)^{2}}=\sqrt{\Sigma a^{2}+2 \Sigma a b}=\pm 1$

$\Rightarrow 2-3 a b c=1 \Rightarrow a b c=\frac{1}{3}$

or $2-3 a b c=-1 \Rightarrow a b c=1$

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