If - a b c x y z p q r then ka kb kc kx ky kz kp kq kr is equal to

The correct option is B k^3Δ D= ka amp; kb amp; kc kx amp; ky amp; kz kp amp; kq amp; kr = k ^ 3 a amp; b amp; c ...

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Properties of Determinants

If Δ= a b...

Question

If $Δ = ∣ ∣ ∣ ∣ \begin{matrix} a & b & c x & y & z p & q & r \end{matrix} ∣ ∣ ∣ ∣$ then $∣ ∣ ∣ ∣ \begin{matrix} k a & k b & k c k x & k y & k z k p & k q & k r \end{matrix} ∣ ∣ ∣ ∣$ is equal to

k Δ

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3 k Δ

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k^{3} Δ

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k Δ^{3}

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Solution

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Similar questions

Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} a & b & c x & y & z p & q & r \end{matrix} ∣ ∣ ∣ ∣

Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} y & b & q x & a & p z & c & r \end{matrix} ∣ ∣ ∣ ∣

Δ_{1}

A B C

Δ,

b^{2} sin 2 C + c^{2} sin 2 B = k Δ,

k

Δ = ∣ ∣ ∣ ∣ \begin{matrix} a & 0 & 0 b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ ∣ \begin{matrix} p^{2} a & 0 & 0 p b & c & a p c & a & b \end{matrix} ∣ ∣ ∣ ∣ ∣

A = ∣ ∣ ∣ ∣ \begin{matrix} a & b & c x & y & z p & q & r \end{matrix} ∣ ∣ ∣ ∣

B = ∣ ∣ ∣ ∣ \begin{matrix} q & - b & y - p & a & - x r & - c & z \end{matrix} ∣ ∣ ∣ ∣

Δ

Δ A B C

b^{2} s i n 2 C + c^{2} s i n 2 B

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