If p+q+r=a+b+c=0 , then the determinant | array lll p a & q b & r c q c & ra & p b r b & p c & q a array | equals

Correct answer: A. 0 | array ccc pa & qb & rc qc & ra & pb rb & pc & qa array |=pa(rqa^2-p^2 bc)-qb(q^2 ac-prb^2)+ aligned & rc(q qcc^2-r^2 a b) & =pqra^3-p^3 a b c-q^3 a b c+pqrb^3+pqrc^3-r^3 a b c & =pqr(a^3+b^3+c^3)-abc(p^3+q^3+r^3) & Given, p+q+r=a+b+c=0 aligned aligned & p^3+q^3+r^3=3 pqr and a^3+b^3+c^3=3 abc . & pqr(a^3+b^3+c^3)-abc(p^3+q^3+r^3) & =pqr(3 abc)-abc(3 pqr) & =0 aligned

NDA 2017

Exam: NDA
Year: 2017
Subject: Mathematics
Chapter: Determinants

NDA 2017 · Mathematics · Determinants

If p+q+r=a+b+c=0 , then the determinant | array lll p a & q b & r c...

p+q+r=a+b+c=0

, then the determinant \left|

\begin{array}{lll}p a & q b & r c \\ q c & \text{ ra } & p b \\ r b & p c & q a\end{array}

\right|$ equals

Options

A $0$
B $1$
C $\mathrm{pa}+\mathrm{qb}+\mathrm{rc}$
D $\mathrm{pa}+\mathrm{qb}+\mathrm{rc}+\mathrm{a}+\mathrm{b}+\mathrm{c}$

Correct answer

A. $0$

Step-by-step solution

\left|

\begin{array}{ccc}\mathrm{pa} & \mathrm{qb} & \mathrm{rc} \\ \mathrm{qc} & \mathrm{ra} & \mathrm{pb} \\ \mathrm{rb} & \mathrm{pc} & \mathrm{qa}\end{array}

\right|=\mathrm{pa}\left(\mathrm{rqa}^2-\mathrm{p}^2 \mathrm{bc}\right)-\mathrm{qb}\left(\mathrm{q}^2 \mathrm{ac}-\mathrm{prb}^2\right)+

\begin{aligned} & \operatorname{rc}\left(q \mathrm{qcc}^2-\mathrm{r}^2 a b\right) \\ & =\mathrm{pqra}^3-\mathrm{p}^3 a b c-\mathrm{q}^3 a b c+\mathrm{pqrb}^3+\mathrm{pqrc}^3-\mathrm{r}^3 a b c \\ & =\mathrm{pqr}\left(\mathrm{a}^3+\mathrm{b}^3+\mathrm{c}^3\right)-\mathrm{abc}\left(\mathrm{p}^3+\mathrm{q}^3+\mathrm{r}^3\right) \\ & \text{Given, } \mathrm{p}+\mathrm{q}+\mathrm{r}=\mathrm{a}+\mathrm{b}+\mathrm{c}=0\end{aligned}

$\begin{aligned} & \Rightarrow \mathrm{p}^3+\mathrm{q}^3+\mathrm{r}^3=3 \mathrm{pqr} \text{\text{ and } } \mathrm{a}^3+\mathrm{b}^3+\mathrm{c}^3=3 \mathrm{abc} . \\ & \therefore \operatorname{pqr}\left(\mathrm{a}^3+\mathrm{b}^3+\mathrm{c}^3\right)-\mathrm{abc}\left(\mathrm{p}^3+\mathrm{q}^3+\mathrm{r}^3\right) \\ & =\operatorname{pqr}(3 \mathrm{abc})-\mathrm{abc}(3 \mathrm{pqr}) \\ & =0\end{aligned}

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