Question

If $A=\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]$, where $a,b,c$ are real positive numbers $abc=1$ and $A^{T}A=1$ then:

• A. $ab+bc+ca=0$
• B. $a^2+b^2+c^2=1$
• C. $a^3+b^3+c^3=4$
• D. $a+b+c=2$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $ab+bc+ca=0$
B $a^2+b^2+c^2=1$
C $a^3+b^3+c^3=4$

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

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

$A^T=A$

$A^T\cdot A=1$

$\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]\times \left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$\left[\begin{array}{ccc}{a}^2+b^2+c^2& ab+bc+ca& ac+ab+bc\\ ab+bc+ca& a^2+b^2+c^2& ac+ab+bc\\ ac+ab+bc& ab+bc+ca& a^2+b^2+c^2\end{array}\right]$

$a^2+b^2+c^2=1$

$ab+bc+ca=0$

${(a+b+c)}^2=a^2+b^2+c^2+2(ab+bc+ca$

or ${(a+b+c)}^2=1+0=1$

or $a+b+c=1$

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

$a^3+b^3+c^3=1\times (1-0)+3\times 1=4$

Answer $\left(A\right)ab+bc+ca=0\left(B\right)a^2+b^2+c^2=1\left(C\right)a^3+b^3+c^3=4$