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

Answer: (D)

since, $A^{T}A=I$
$\Rightarrow \left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
$\Rightarrow \left[\begin{array}{ccc}{a}^2+b^2+c^2& ab+bc+ca& ab+bc+ca\\ ab+bc+ca& a^2+b^2+c^2& ab+bc+ca\\ ab+bc+ca& ab+bc+ca& a^2+b^2+c^2\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
$\Rightarrow a^2+b^2+c^2=1$ and $ab+bc+ca=0$
Now, $(a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)$
$=1+2\cdot 0=1$
$\Rightarrow a+b+c=\mathrm{1..............}(i)$
Now, $(a^3+b^3+c^3)$
$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$
$\Rightarrow a^3+b^3+c^3=1+3=4[using(i)]$