Question

If $abc=p$ and $A=\left[\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right]$ such that $A{A}^{\prime }=I$ and $A^{\prime }$ is transpose of $A,$ then $a,b,c$ are the roots of the equation

• A. $x^3-p=0$
• B. $x^3+x^2-p=0$
• C. $x^3+3x^2-p=0$
• D. $x^3+2x^2-p=0$

Answer 1

The correct option is B $x^3+x^2-p=0$

We have, $A=\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|$

$A{A}^{\prime }=\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|\left|\begin{array}{ccc}a& c& b\\ b& a& c\\ c& b& a\end{array}\right|$
$=\left|\begin{array}{ccc}{a}^2+b^2+c^2& ac+ab+bc& ac+ab+bc\\ ac+ab+bc& a^2+b^2+c^2& ac+ab+bc\\ ac+ab+bc& ac+ab+bc& a^2+b^2+c^2\end{array}\right|=I$

Thus $a^2+b^2+c^2=1,ac+ab+bc=0\Rightarrow a+b+c=\pm 1$

Also, $abc=p$

Hence $a,b,c$ are the roots of the equation $x^3\pm x^2-p=0$