If p - q - r -0 and begin vmatrix pa - qb - rcllq c - r a - p b lrb - pc - qa lend vmatrix - k vmatrix a - b - cc - a - b lb - c - alendvmatrix - then K -1 A- 0B- abcC- pqrD- a-b-c

The correct option is C pqrp + q + r = 0 ⇒ p^3 + q^3 + r^3 = 3pqr pa amp; qb amp; rc qc amp; rb amp; pb rb amp; pc amp; qa = pqr (a^3 + ...

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Byju's Answer

Standard XII

Mathematics

Sarrus Rule

If p + q + r ...

Question

If p + q + r = 0 and $∣ ∣ ∣ ∣ \begin{matrix} p a & q b & r c q c & r a & p b r b & p c & q a \end{matrix} ∣ ∣ ∣ ∣ = k ∣ ∣ ∣ ∣ \begin{matrix} a & b & c c & a & b b & c & a \end{matrix} ∣ ∣ ∣ ∣, t h e n K =$

abc

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pqr

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a + b +c

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Solution

The correct option is B pqr
$p + q + r = 0 \Rightarrow p^{3} + q^{3} + r^{3} = 3 p q r ∣ ∣ ∣ ∣ \begin{matrix} p a & q b & r c q c & r b & p b r b & p c & q a \end{matrix} ∣ ∣ ∣ ∣ = p q r (a^{3} + b^{3} + c^{3} - 3 a b c) p q r ∣ ∣ ∣ ∣ \begin{matrix} a & b & c c & a & b b & c & a \end{matrix} ∣ ∣ ∣ ∣ \Rightarrow k = p q r$

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Similar questions

If p + q + r = a + b + c = 0, then the determinant
$Δ = ∣ ∣ ∣ ∣ \begin{matrix} p a & q b & r c q c & r a & p b r b & p c & q a \end{matrix} ∣ ∣ ∣ ∣$ equals

Q. If p + q + r = 0 = a + b + c, then the value of the determinant

∣ ∣ ∣ ∣ \begin{matrix} p a & q b & r c q c & r a & p b r b & p c & q a \end{matrix} ∣ ∣ ∣ ∣

Q. If

p + q + r = 0

, then prove that

∣ ∣ ∣ ∣ \begin{matrix} p a & q b & r c q c & r a & p b r b & p c & q a \end{matrix} ∣ ∣ ∣ ∣ = p q r ∣ ∣ ∣ ∣ \begin{matrix} a & b & c c & a & b b & c & a \end{matrix} ∣ ∣ ∣ ∣

Q. If

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

, then the determinant

∣ ∣ ∣ ∣ \begin{matrix} p a & q b & r c q c & r a & p b r b & p c & q a \end{matrix} ∣ ∣ ∣ ∣

equals.

If p + q + r = a + b + c = 0, then the determinant
$Δ = ∣ ∣ ∣ ∣ \begin{matrix} p a & q b & r c q c & r a & p b r b & p c & q a \end{matrix} ∣ ∣ ∣ ∣$ equals

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If p + q + r = 0 and ∣∣ ∣∣paqbrcqcrapbrbpcqa∣∣ ∣∣=k∣∣ ∣∣abccabbca∣∣ ∣∣,thenK=

The correct option is B pqrp+q+r=0⇒p3+q3+r3=3pqr∣∣ ∣∣paqbrcqcrbpbrbpcqa∣∣ ∣∣=pqr(a3+b3+c3−3abc)pqr∣∣ ∣∣abccabbca∣∣ ∣∣⇒k=pqr

If p + q + r = 0 and $∣ ∣ ∣ ∣ \begin{matrix} p a & q b & r c q c & r a & p b r b & p c & q a \end{matrix} ∣ ∣ ∣ ∣ = k ∣ ∣ ∣ ∣ \begin{matrix} a & b & c c & a & b b & c & a \end{matrix} ∣ ∣ ∣ ∣, t h e n K =$