If - - a1 b1 c1 a2 b2 c2 a3 b3 c3 and - 1- a1-pb1 b1-qc1 c1-ra1 a2-pb2 b2-qc2 c2-ra2 a3-pb3 b3-qc3 c3-ra3 then - 1-

The correct option is A Δ ( 1+pqr ) Given nbsp; Δ = a_ 1 amp; b_ 1 amp; c_ 1 a_ 2 amp; b_ 2 amp; c_ 2 a_ 3 amp; b_ 3 ...

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Properties of Determinants

If Δ = a1...

Question

If $Δ =$ $∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣$ and $Δ_{1}$ = $∣ ∣ ∣ ∣ \begin{matrix} a_{1} + p b_{1} & b_{1} + q c_{1} & c_{1} + r a_{1} a_{2} + p b_{2} & b_{2} + q c_{2} & c_{2} + r a_{2} a_{3} + p b_{3} & b_{3} + q c_{3} & c_{3} + r a_{3} \end{matrix} ∣ ∣ ∣ ∣$ then $Δ_{1} =$

Δ (1 + p q r)

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Δ (1 + p + q + r)

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Δ (1 - p q r)

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D = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣

D^{'} = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} + p b_{1} & b_{1} + q c_{1} & c_{1} + r a_{1} a_{2} + p b_{2} & b_{2} + q c_{2} & c_{2} + r a_{2} a_{3} + p b_{3} & b_{3} + q c_{3} & c_{2} + r a_{3} \end{matrix} ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ \begin{matrix} a_{1} + p b_{1} & b_{1} + q c_{1} & c_{1} + r a_{1} a_{2} + p b_{2} & b_{2} + q c_{2} & c_{2} + r a_{2} a_{3} + p b_{3} & b_{3} + q c_{3} & c_{3} + r a_{3} \end{matrix} ∣ ∣ ∣ ∣

If ∆ = |\begin{matrix} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{matrix}|, ∆_{1} = |\begin{matrix} 1 & 1 & 1 \\ y z & z x & x y \\ x & y & z \end{matrix}|, then prove that ∆ + ∆_{1} = 0 .

Δ_{1} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} a_{1}^{2} + b_{1} + c_{1} & a_{1} a_{2} + b_{2} + c_{2} & a_{1} a_{3} + b_{3} + c_{3} b_{1} b_{2} + c_{1} & b_{2}^{2} + c_{2} & b_{2} b_{1} + c_{3} c_{3} c_{1} & c_{3} c_{2} & c_{3}^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣

\frac{Δ_{1}}{Δ_{2}}

∆_{1} = |\begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{matrix}|, ∆_{2} = |\begin{matrix} 1 & b c & a \\ 1 & c a & b \\ 1 & a b & c \end{matrix}|, then

∆_{1} + ∆_{2} = 0

∆_{1} + 2 ∆_{2} = 0

∆_{1} = ∆_{2}

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