Question

If $A=\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right],abc=I,A^{T}A=1$, then the value of $a^3+b^3+c^3$ can be

• A. $3$
• B. $0$
• C. $1$
• D. $4$

Answer 1

Answer: (D)

$4$

Given:- $A=\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right],abc=I,A^{T}A=I$

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

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

So, $A^T=A$, from the matrices.

$A^{T}A=I$

$A^2=I[A^T=A]$

$\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}{a}^2+b^2+c^2& ab+bc+ca& ab+bc+ca\\ ab+bc+ca& a^2+b^2+c^2& ab+bc+ca\\ ab+bc+ca& ab+bc+ca& 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]$

from the matrix $a^2+b^2+c^2=1,ab+bc+ca=0$

$(a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca=0)$ [formula]

$(a+b+c{)}^2=1[\because ab+bc+ca=0,a^2+b^2+c^2=1]$

$(a+b+c)=\pm 1$

We know

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

$a^3+b^3+c^3-3\left(1\right)=(\pm 1)(1-0)[\because abc=I]$

$a^3+b^3+c^3=3\pm 1$

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

$a^3+b^3+c^3=4$

Hence, option D is correct.