If p-q-r-a-b-c-0- then the determinantpa -qb -rcllqc -ra -pb 1 -rb -pc -qa lend vmatrix equalsA- 0B- 1C- pa - qb - rcD- none of these

The correct option is A 0 We have Δ =pqr (a^3+b^3+c^3) abc (p^3+q^3+r^3) But a + b + c = 0 ⇒ (a+b)^3 = c^3 ⇒ a^3+b^3+3ab(a+b)+c^3=0 ⇒ a^3+b^3+c^3 = 3ab( c)=3ab ...

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Byju's Answer

Standard XII

Mathematics

Special Determinants

If p+q+r=a+b+...

Question

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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pa + qb + rc

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none of these

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Solution

The correct option is A
0

We have $Δ = p q r (a^{3} + b^{3} + c^{3}) - a b c (p^{3} + q^{3} + r^{3})$
But a + b + c = 0
$\Rightarrow (a + b)^{3} = - c^{3} \Rightarrow a^{3} + b^{3} + 3 a b (a + b) + c^{3} = 0 \Rightarrow a^{3} + b^{3} + c^{3} = - 3 a b (- c) = 3 a b c$
Similarly, $p^{3} + q^{3} + r^{3} = 3 p q r$
Thus, $Δ$ = pqr (3 abc) - abc (3pqr) = 0
Hence the correct option is (a)

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If p + q + r = a + b + c = 0, then the determinant Δ=∣∣ ∣∣paqbrcqcrapbrbpcqa∣∣ ∣∣ equals

The correct option is A 0 We have Δ=pqr (a3+b3+c3)−abc(p3+q3+r3) But a + b + c = 0 ⇒(a+b)3=−c3⇒a3+b3+3ab(a+b)+c3=0⇒a3+b3+c3=−3ab(−c)=3abc Similarly, p3+q3+r3=3pqr Thus, Δ = pqr (3 abc) - abc (3pqr) = 0 Hence the correct option is (a)

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