If D- - a1 b1 c1- a2 b2 c2- a3 b3 c3- - - and D- - a1-pb1 b1-qc1 c1-ra1- a2-pb2 b2-qc2 c2-ra2- a3-pb3 b3-qc3 c3-ra3- - - thenprove that D-D1-pqr-

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Definition of a Determinant

If D= |[ a1...

Question

If D= $∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣$
and D'= $∣ ∣ ∣ ∣ \begin{matrix} a_{1} + p b_{1} & b_{1} + q c_{1} & c_{1} + r a_{1} a_{2} + p b_{2} & b_{2} + q c_{2} & c_{2} + r a_{2} a_{3} + p b_{3} & b_{3} + q c_{3} & c_{3} + r a_{3} \end{matrix} ∣ ∣ ∣ ∣$ , then
prove that D'=D(1+pqr).

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Δ =

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

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} ∣ ∣ ∣ ∣

Δ_{1} =

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} ∣ ∣ ∣ ∣

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

D_{0} = ∣ ∣ ∣ ∣ \begin{matrix} k a_{1} & k b_{1} & k c_{1} k a_{2} & k b_{2} & k c_{2} k a_{3} & k b_{3} & k c_{3} \end{matrix} ∣ ∣ ∣ ∣

D_{0} = k^{3} D

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

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

⎡ ⎢ ⎣ \begin{matrix} a & b & c a^{2} & b^{2} & c^{2} a^{3} & b^{3} & c^{3} \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} d d^{2} d^{3} \end{matrix} ⎤ ⎥ ⎦, a \neq b \neq c \neq 0,

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