If - - - arg z1 arg z2 arg z3 arg z2 arg z3 arg z1 arg z3 arg z1 arg z2 - then - is divisible by

The correct options are A arg z_1z_2z_3 C arg z_1+arg z_2+arg z_3 Apply nbsp; C_1+C_2+C_3 and take nbsp; ∑ arg z_1 common. Δ =∑ ( ...

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Byju's Answer

Standard XII

Mathematics

Summation of Determinant

If Δ = | ar...

Question

If $Δ = ∣ ∣ ∣ ∣ \begin{matrix} a r g z_{1} & a r g z_{2} & a r g z_{3} a r g z_{2} & a r g z_{3} & a r g z_{1} a r g z_{3} & a r g z_{1} & a r g z_{2} \end{matrix} ∣ ∣ ∣ ∣$ , then $Δ$ is divisible by

arg

(z_{1} + z_{2} + z_{3})

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arg

z_{1} z_{2} z_{3}

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a r g z_{1} + a r g z_{2} + a r g z_{3}

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Solution

The correct options are
A arg $z_{1} z_{2} z_{3}$
C $a r g z_{1} + a r g z_{2} + a r g z_{3}$
Apply $C_{1} + C_{2} + C_{3}$ and take $\sum a r g z_{1}$ common.
$Δ = \sum (a r g z_{1}) ∣ ∣ ∣ ∣ \begin{matrix} 1 & a r g z_{2} & a r g z_{3} 1 & a r g z_{3} & a r g z_{1} 1 & a r g z_{1} & a r g z_{2} \end{matrix} ∣ ∣ ∣ ∣$
Hence $Δ$ is divisible by $\sum a r g z_{1}$ $= a r g z_{1} + a r g z_{2} + a r g z_{3} = a r g (z_{1} z_{2} z_{3}) .$

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Similar questions

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\frac{z_{1} - z_{2}}{z_{1} - z_{3}} = \frac{π}{2}

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If Δ=∣∣ ∣∣argz1argz2argz3argz2argz3argz1argz3argz1argz2∣∣ ∣∣, then Δ is divisible by

The correct options are A arg z1z2z3 C argz1+argz2+argz3Apply C1+C2+C3 and take ∑argz1 common. Δ=∑(argz1)∣∣ ∣∣1argz2argz31argz3argz11argz1argz2∣∣ ∣∣ Hence Δ is divisible by ∑argz1 =argz1+argz2+argz3=arg(z1z2z3).

If $Δ = ∣ ∣ ∣ ∣ \begin{matrix} a r g z_{1} & a r g z_{2} & a r g z_{3} a r g z_{2} & a r g z_{3} & a r g z_{1} a r g z_{3} & a r g z_{1} & a r g z_{2} \end{matrix} ∣ ∣ ∣ ∣$ , then $Δ$ is divisible by