Let a, b, c∈R be all non-zero satisfying a3+b3+c3=2. If the matrix A=a b cb c ac a b satisfies ATA=I, then a value of abc can be

It is given thatATA=I⇒abcbcacababcbcacab=100010001⇒a2+b2+c2ab+bc+caab+bc+caab+bc+caa2+b2+c2ab+bc+caab+bc+caab+bc+caa2+b2+c2=100010001⇒a2+b2+c2=1 and ab+bc+ca=0⇒(a+b+c)2=a2+b2+c2+2(ab+bc+ca)⇒(a+b+c)2=1⇒a+b+c=1∴a3+b3+c3=2⇒a3+b3+c3−3abc=2−3abc ⇒ (a+b+c)a2+b2+c2−ab−bc−ca=2−3abc⇒ 1(1−0)=2−3abc⇒1=2−3abc⇒abc=13

Let a, b, c∈R be all non-zero satisfying a3+b3+c3=2. If the matrix A=a b cb c ac a b satisfies ATA=I, then a value of abc can be

It is given thatATA=I⇒abcbcacababcbcacab=100010001⇒a2+b2+c2ab+bc+caab+bc+caab+bc+caa2+b2+c2ab+bc+caab+bc+caab+bc+caa2+b2+c2=100010001⇒a2+b2+c2=1 and ab+bc+ca=0⇒(a+b+c)2=a2+b2+c2+2(ab+bc+ca)⇒(a+b+c)2=1⇒a+b+c=1∴a3+b3+c3=2⇒a3+b3+c3−3abc=2−3abc ⇒ (a+b+c)a2+b2+c2−ab−bc−ca=2−3abc⇒ 1(1−0)=2−3abc⇒1=2−3abc⇒abc=13

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Let $a, b, c \in R$ be all non-zero satisfying $a^{3} + b^{3} + c^{3} = 2 . If the matrix A = [\begin{array}{lll} a b c \\ b c a \\ c a b \end{array}] satisfies A^{T} A = I, then$ a value of $a b c$ can be

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{1}{3}$

$- \frac{1}{3}$

$\frac{2}{3}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

It is given that

$\begin{array}{l} A^{T} A = I \\ \Rightarrow [\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}] [\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ \Rightarrow [\begin{array}{lll} a^{2} + b^{2} + c^{2} & a b + b c + c a & a b + b c + c a \\ a b + b c + c a & a^{2} + b^{2} + c^{2} & a b + b c + c a \\ a b + b c + c a & a b + b c + c a & a^{2} + b^{2} + c^{2} \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \\ \Rightarrow a^{2} + b^{2} + c^{2} = 1 and a b + b c + c a = 0 \\ \Rightarrow (a + b + c)^{2} = (a^{2} + b^{2} + c^{2}) + 2 (a b + b c + c a) \\ \Rightarrow (a + b + c)^{2} = 1 \Rightarrow a + b + c = 1 \\ ∴ a^{3} + b^{3} + c^{3} = 2 \\ \Rightarrow a^{3} + b^{3} + c^{3} - 3 a b c = 2 - 3 a b c \end{array} \begin{array}{l} \Rightarrow (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) = 2 - 3 a b c \\ \Rightarrow 1 (1 - 0) = 2 - 3 a b c \Rightarrow 1 = 2 - 3 a b c \Rightarrow a b c = \frac{1}{3} \end{array}$