Question

If matrix $A=\begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix}$ , where $a,b$ and $c$ are real positive number, $abc=1$ and $A^{T}A=I$ , then which of the following is true

• A. $a^{3}+b^{3}+c^{3}=4$
• B. $a^{3}+b^{3}+c^{3}=3$
• C. $a^{3}+b^{3}+c^{3}=6$
• D. $a^{3}+b^{3}+c^{3}=0$

Answer 1

Answer: (A)

The given matrix is
$A=\begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix}$
$A^{T}=\begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix}$
$\Rightarrow A^{T}=A$
$\Rightarrow A^{T}\cdot A=A^{2}$
$A=\begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix}$
$\therefore \left|A^{T} \cdot A\right|=\left|A^{2}\right|$
It is given that $A^{\top}A=I$
$\left|\right.I\left|\right.=\left|\right.A^{2}\left|\right.$
$I=\left|\right.A^{2}\left|\right.$
Now, $\left|A\right|=\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}$
Applying the transformation $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$ , we get
$\left|A\right|=\begin{vmatrix} a+b+c & a+b+c & a+b+c \\ b & c & a \\ c & a & b \end{vmatrix}$
$=\left(a + b + c\right)\begin{vmatrix} 1 & 1 & 1 \\ b & c & a \\ c & a & b \end{vmatrix}$
Now, apply $C_{2} \rightarrow C_{2}-C_{1}$ and $C_{3} \rightarrow C_{3}-C_{1}$ , we get
$\left|A\right|=\left(a + b + c\right)\begin{vmatrix} 1 & 0 & 0 \\ b & c-b & a-b \\ c & a-c & b-c \end{vmatrix}$
$=\left(a + b + c\right)\begin{vmatrix} c-b & a-b \\ a-c & b-c \end{vmatrix}$
$=\left(a + b + c\right)\left[b c - c^{2} - b^{2} + b c - \left(a^{2} - a c - a b + b c\right)\right]$
$=-\left(a + b + c\right)\left[a^{2} + b^{2} + c^{2} - a b - a c - b c\right]$
$=-\left(a^{3} + b^{3} + c^{3} - 3 a b c\right)$
$\Rightarrow \left|A^{2}\right|=\left(a^{3} + b^{3} + c^{3} - 3\right)^{2}=1\left(\because a b c = 1\right)\ldots \left(i\right)$
$\because a,b$ and $c$ are positive.
$\frac{a^{3} + b^{3} + c^{3}}{3}>\sqrt[3]{a^{3} b^{3} c^{3}}=abc=1$
$a^{3}+b^{3}+c^{3}>3$
So, from equation $\left(i\right)$ we have
$a^{3}+b^{3}+c^{3}>3$
$a^{3}+b^{3}+c^{3}-3=1$
$a^{3}+b^{3}+c^{3}=4$