Let $a, b, c \in \mathbb{R}$ be all non-zero and satisfies $a^{3}+b^{3}+c^{3}=2$. If the matrix…
Let $a, b, c \in \mathbb{R}$ be all non-zero and satisfies $a^{3}+b^{3}+c^{3}=2$. If the matrix $A=\begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix}$ satisfies $A^{T}A=I$, then a value of $abc$ can be.
Solution
$A^TA=I$
$\begin{aligned} \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{aligned}$
$\begin{aligned} \begin{bmatrix} a^2+b^2+c^2 & ab+bc+ac & ab+bc+ac \\ ab+bc+ac & a^2+b^2+c^2 & ab+bc+ac \\ ab+bc+ac & ab+bc+ac & a^2+b^2+c^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{aligned}$
On comparing each element both sides, we get
$a^2+b^2+c^2=1$ and $ab+bc+ac=0$
We know that $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$.
$2-3abc=(a+b+c)(1-0)$ (from equation i)
$2-3abc=a+b+c$
Now, we also know that $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)$.
$(a+b+c)^2=1+2(0)$ (from equation i)
$a+b+c=\pm1$
Putting it in equation ii, we get
$2-3abc=\pm1$
$3abc=2\mp1$
$abc=\frac{1}{3}$ or $abc=1$