If P = [ cases a & b & c b & c & a c & a & b cases ] , abc = 1 , P^ T P = I then the value of a^ 3 + b^ 3 + c^ 3 is

Correct answer: A. 2 Given,& P = [ cases a & b & c b & c & a c & a & b cases ] ∴ P^ T = [ cases a & b & c b & c & a c & a & b cases ] Also given~ P^ T P = I ∴ |P^ T P| = |I| = 1 or~| cases a & b & c b & c & a c & a & b cases |2 = 1 ⇒ | cases a & b & c b & c & a c & a & b cases | = ± 1 or~ 3abc - a^ 3 - b^ 3 - c^ 3 = ± 1 or~ 3 ∓ 1 = a^ 3 + b^ 3 + c^ 3 ∴ a^ 3 + b^ 3 + c^ 3 = 2 ~or& 4

BITSAT 2018

Exam: BITSAT
Year: 2018
Subject: Mathematics
Chapter: Matrices

BITSAT 2018 · Mathematics · Matrices

If P = [ cases a & b & c b & c & a c & a & b cases ] , abc = 1 , P^...

P = \left[

\begin{cases}a & b & c \\ b & c & a \\ c & a & b\end{cases}

\right]

abc = 1

P^{T}P = I

then the value of

a^{3} + b^{3} + c^{3}

Options

A $2$
B $1$
C $0$
D $5$

Correct answer

A. $2$

Step-by-step solution

Given,& $P = \left[$ \begin{cases}a & b & c \\ b & c & a \\ c & a & b\end{cases} $\right]$

$∴ P^{T} = \left[$ \begin{cases}a & b & c \\ b & c & a \\ c & a & b\end{cases} $\right]$

Also given~ $P^{T}P = I$

$∴ \left|P^{T}P\right| = \left|I\right| = 1$

or~\left| $\begin{cases}a & b & c \\ b & c & a \\ c & a & b\end{cases}$ \right|2 = 1 ⇒ \left| $\begin{cases}a & b & c \\ b & c & a \\ c & a & b\end{cases}$ \right| = \pm 1 $or~$ 3abc - a^{3} - b^{3} - c^{3} = \pm 1 $or~$ 3 ∓ 1 = a^{3} + b^{3} + c^{3} $$ ∴ a^{3} + b^{3} + c^{3} = 2 $~or&$ 4$

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