JEE Main 2020 — Mathematics Algebra

$$\because {A}^{T}A=I$$∵ATA=I

$$\Rightarrow [\begin{matrix}a & b & c \\ b & c & a \\ c & a & b\end{matrix}][\begin{matrix}a & b & c \\ b & c & a \\ c & a & b\end{matrix}]=[\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}]$$⇒[abc​bca​cab​][abc​bca​cab​]=[100​010​001​]

$$\Rightarrow [\begin{matrix}{a}^{2}+{b}^{2}+{c}^{2} & ab+bc+ac & ab+bc+ac \\ ab+bc+ac & {a}^{2}+{b}^{2}+{c}^{2} & ab+bc+ac \\ ab+bc+ac & ab+bc+ac & {a}^{2}+{b}^{2}+{c}^{2}\end{matrix}]=[\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}]$$⇒[a2+b2+c2ab+bc+acab+bc+ac​ab+bc+aca2+b2+c2ab+bc+ac​ab+bc+acab+bc+aca2+b2+c2​]=[100​010​001​]

On comparing each element both sides, we get

{a}^{2}+{b}^{2}+{c}^{2}=1&ab+bc+ca=0.........(i)

We know that ${a}^{3}+{b}^{3}+{c}^{3}-3abc=(a+b+c)({a}^{2}+{b}^{2}+{c}^{2}-ab-bc-ac)$a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ac).

$\Rightarrow 2-3abc=(a+b+c)(1-0)$⇒2−3abc=(a+b+c)(1−0) (from equation $(i)$(i))

$$\Rightarrow 2-3abc=(a+b+c).........(\mathrm{ii})$$⇒2−3abc=(a+b+c).........(ii)

Now, we also know that ${(a+b+c)}^{2}={a}^{2}+{b}^{2}+{c}^{2}+2(ab+bc+ca)$(a+b+c)2=a2+b2+c2+2(ab+bc+ca).

$\Rightarrow {(a+b+c)}^{2}=1+2(0)$⇒(a+b+c)2=1+2(0) (from equation $(i)$(i))

$$\Rightarrow a+b+c=\pm 1$$⇒a+b+c=±1

Putting it in equation $(\mathrm{ii})$(ii), we get

$$2-3abc=\pm 1$$2−3abc=±1

$$\Rightarrow 3abc=2\mp 1$$⇒3abc=2∓1

$$\Rightarrow abc=\frac{1}{3}\text{or}abc=1$$⇒abc=31​orabc=1