Question

If matrix $A=\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]$ where a, b, c are real positive numbers, abc = 1 and $A^{T}A=I,$ then the value of $a^3+b^3+c^3$ is ___.

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is D 4

Given, $A=\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right],abc=1and{A}^{T}A=INow,A^{T}A=1\Rightarrow \left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}{a}^2+b^2+c^2& ab+bc+ca& ab+bc+ca\\ ab+bc+ca& a^2+b^2+c^2& ab+bc+ca\\ ab+bc+ca& ab+bc+ca& a^2+b^2+c^2\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\Rightarrow a^2+b^2+c^2=1andab+bc+ca=\mathrm{0.....}\left(ii\right)weknow{a}^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\Rightarrow a^3+b^3+c^3=(a+b+c)(1-0)+3[FromEqs.(i\left)and\right(iii\left)\right]\therefore a^3+b^3+c^3=(a+b+c)+\mathrm{3.......}\left(iii\right)Now,(a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)=1......(iv)From,Eq.(iii),a^3+b^3+c^3=1+3\Rightarrow a^3+b^3+c^3=4.$