Question

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

• A. $0$
• B. $3$
• C. $1$
• D. $4$

Answer 1

The correct option is D $4$

Given, A is circulant matrix of elements $a,b,c$ and $abc=1$
So, $A=\left[\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right]$
So, $detA=a(a^2-bc)-b(ac-b^2)+c(c^2-ab)$
$=a^3+b^3+c^3-3abc$
$detA=a^2+b^3+c^3-3$
and $A^{T}A=I$
$|A^{T}A|=|I|=1$
$|A||A^T|=1$
$|A{|}^2=1$
$|A|=\pm 1$
So, $detA=\pm 1.$

After substituing the value in the $detA=a^3+b63+c^3-3abc$ we get:

$a^3+b^3+c^3=4$ or $a^3+b^3+C^3=-2$