Question

If matrix $A=\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]$ , where $a,b$ and $c$ are real positive number, $abc=1$ and $A^{T}A=I$ , then which of the following is true

• A. $a^3+b^3+c^3=4$
• B. $a^3+b^3+c^3=3$
• C. $a^3+b^3+c^3=6$
• D. $a^3+b^3+c^3=0$

Answer 1

Answer: (A)

The given matrix is
$A=\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]$
$A^T=\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]$
$\Rightarrow A^T=A$
$\Rightarrow A^T\cdot A=A^2$
$A=\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]$
$\therefore \left|A^T\cdot A\right|=\left|A^2\right|$
It is given that $A^{\top }A=I$
$|I|=|A^2|$
$I=|A^2|$
Now, $\left|A\right|=\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|$
Applying the transformation $R_1\to R_1+R_2+R_3$ , we get
$\left|A\right|=\left|\begin{array}{ccc}a+b+c& a+b+c& a+b+c\\ b& c& a\\ c& a& b\end{array}\right|$
$=\left(a+b+c\right)\left|\begin{array}{ccc}1& 1& 1\\ b& c& a\\ c& a& b\end{array}\right|$
Now, apply $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$ , we get
$\left|A\right|=\left(a+b+c\right)\left|\begin{array}{ccc}1& 0& 0\\ b& c-b& a-b\\ c& a-c& b-c\end{array}\right|$
$=\left(a+b+c\right)\left|\begin{array}{cc}c-b& a-b\\ a-c& b-c\end{array}\right|$
$=\left(a+b+c\right)\left[bc-c^2-b^2+bc-\left(a^2-ac-ab+bc\right)\right]$
$=-\left(a+b+c\right)\left[a^2+b^2+c^2-ab-ac-bc\right]$
$=-\left(a^3+b^3+c^3-3abc\right)$
$\Rightarrow \left|A^2\right|={\left(a^3+b^3+c^3-3\right)}^2=1\left(\because abc=1\right)\dots \left(i\right)$
$\because a,b$ and $c$ are positive.
$\frac{a^3+b^3+c^3}{3}>\sqrt[3]{a^3{b}^3{c}^3}=abc=1$
$a^3+b^3+c^3>3$
So, from equation $\left(i\right)$ we have
$a^3+b^3+c^3>3$
$a^3+b^3+c^3-3=1$
$a^3+b^3+c^3=4$