Question

If $A=\begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix},abc=1$ and $A^{T}A=I,$ then the value of $a^{3}+b^{3}+c^{3}$ (where $a,b,c>0$ and $A^{T}$ is transpose of matrix $A$ ) is

• A. $2$
• B. $0$
• C. $1$
• D. $4$

Answer 1

Answer: (D)

$A^{T}A=I$
As $A^{T}=A$
$\Rightarrow A^{2}=I$
$\Rightarrow a^{2}+b^{2}+c^{2}=1$ and $ab+bc+ca=0$
As $\left(a + b + c\right)^{2}=a^{2}+b^{2}+c^{2}+2\left(a b + b c + c a\right)$
$\Rightarrow a+b+c=1$
Now, $a^{3}+b^{3}+c^{3}-3abc=\left(a + b + c\right)\left(a^{2} + b^{2} + c^{2} - a b - b c - c a\right)$
$\Rightarrow a^{3}+b^{3}+c^{3}-3=\left(a + b + c\right)$
$\Rightarrow a^{3}+b^{3}+c^{3}=3+1$
$\Rightarrow a^{3}+b^{3}+c^{3}=4$