Question

If matrix $A$ is an circulant matrix whose elements of first row are $a,b,c>0$ such that $abc=1$ and $A^{\tau }A=1$ then $a^3+b^3+c^3$ equals to,

• A. $0$
• B. $4$
• C. $1$
• D. $3$

Answer 1

The correct option is B $4$

Given $A^{T}A=1$
$\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
$\left[\begin{array}{ccc}\sum a^2& \sum ab& \sum ab\\ \sum cb& \sum b^2& \sum bc\\ \sum ac& \sum ac& \sum c^2\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
$\Rightarrow \sum a^2=\sum b^2=\sum c^2=1$ and $\sum ab=\sum bc=\sum ca=0$ ...(1)
Now, $a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$
$a^3+b^3+c^3=(a+b+c)[1-0]+3\left(1\right)$
$a^3+b^3+c^3=(a+b+c)+3$ ....(2)
Now, $\because {(a+b+c)}^2=\sum a^2+2\sum ab=1$
$\Rightarrow a+b+c=1.$
Substitute this value in (2)
$a^3+b^3+c^3=1+3$
$a^3+b^3+c^3=4$