-1- 1 0- a b - and -2- 1 0- c d - then -2-1 is equal toA- b dB- b-ad-cC- abcD- ac

The correct option is A bdΔ_2Δ_1 = 1 amp;0 c amp;d 1 amp;0 a amp;b = 1 amp;0 c+ad amp;bd = bd. ...

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Evaluation of a Determinant

Δ1=|[ 1 0; a ...

Question

If $Δ_{1} = ∣ ∣ ∣ \begin{matrix} 1 & 0 a & b \end{matrix} ∣ ∣ ∣$ and $Δ_{2} = ∣ ∣ ∣ \begin{matrix} 1 & 0 c & d \end{matrix} ∣ ∣ ∣$ , then $Δ_{2} Δ_{1}$ is equal to

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(b - a)(d - c)

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abc

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Δ_{1} = ∣ ∣ ∣ \begin{matrix} 1 & 0 a & b \end{matrix} ∣ ∣ ∣

Δ_{2} = ∣ ∣ ∣ \begin{matrix} 1 & 0 c & d \end{matrix} ∣ ∣ ∣

Δ_{2} Δ_{1}

Δ_{1} = ∣ ∣ ∣ \begin{matrix} 1 & 0 a & b \end{matrix} ∣ ∣ ∣

Δ_{2} = ∣ ∣ ∣ \begin{matrix} 1 & 0 c & d \end{matrix} ∣ ∣ ∣

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

Δ_{1}

If AB is perpendicular bisector of CD at P then

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