Question

Let $a,b,c\in R$ be all non-zero and satisfy $a^3+b^3+c^3=2$. If the matrix $A=\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]$, satisfies $A^{T}A=I$, then value of $abc$can be:

• A. $\frac{2}{3}$
• B. $3$
• C. $-\frac{1}{3}$
• D. $\frac{1}{3}$

Answer 1

The correct option is D

$\frac{1}{3}$

Explanation for the correct option:

Step 1: Equating $A^{T}A=I$

Given that,

$a^3+b^3+c^3=2...\left(i\right)$,

$A=\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]$

And $A^{T}A=I$

$$\begin{gathered} \Rightarrow {\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]}^T\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]=I \\ \Rightarrow {\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]}^{}\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]\begin{array}{l}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\end{array} \\ \Rightarrow \left[\begin{array}{ccc}{a}^2+b^2+c^2& ab+ac+bc& ab+ac+bc\\ ab+ac+bc& a^2+b^2+c^2& ab+ac+bc\\ ab+ac+bc& ab+ac+bc& a^2+b^2+c^2\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] \end{gathered}$$

Therefore, we have

$$\begin{gathered} a^2+b^2+c^2=1...\left(ii\right) \\ ab+bc+ca=0...\left(iii\right) \end{gathered}$$

Step 2: Finding the value of $abc$ by equating all the equations

Since, $(a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)$

From equations $\left(i\right)\&\left(ii\right)$ we get

$$\begin{gathered} \Rightarrow (a+b+c{)}^2=1+2(0) \\ \Rightarrow a+b+c=1....(iv) \end{gathered}$$

$\because a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

Therefore, from$\left(i\right),\left(ii\right),\left(iii\right)\&\left(iv\right)$ we get

$$\begin{gathered} \Rightarrow 2-3abc=1\left(1-0\right) \\ \Rightarrow abc=\frac{1}{3} \end{gathered}$$

Therefore, option (D) is the correct answer.