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Evaluation of a Determinant

If Δ1 = 1 0 ...

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

A

a c

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B

b d

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C

(b - a) (d - c)

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D

a b c

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Solution

The correct option is B $b d$
$Δ_{2} Δ_{1} = ∣ ∣ ∣ \begin{matrix} 1 & 0 c & d \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 a & b \end{matrix} ∣ ∣ ∣ = ∣ ∣ ∣ \begin{matrix} 1 & 0 c + a d & b d \end{matrix} ∣ ∣ ∣ = b d .$

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Similar questions

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} ∣ ∣ ∣

Δ_{2} Δ_{1}

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}

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}}

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

Δ_{2} = ∣ ∣ ∣ \begin{matrix} x & b a & x \end{matrix} ∣ ∣ ∣

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