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Summation of Determinant

If Δ = a1 ...

Question

If $Δ = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣$ and $A_{2}, B_{2}, C_{2}$ are respectively cofactors of $a_{2}, b_{2}, c_{2}$ then $a_{1} A_{2} + b_{1} B_{2} + c_{1} C_{2}$ is equal to

A

- Δ

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B

0

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C

Δ

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D

none of these

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Solution

The correct option is C 0
$Δ = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ A_{2} = - ∣ ∣ ∣ \begin{matrix} b_{1} & c_{1} b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ = b_{3} c_{1} - b_{1} c_{3} B_{2} = ∣ ∣ ∣ \begin{matrix} a_{1} & c_{1} a_{3} & c_{3} \end{matrix} ∣ ∣ ∣ = c_{3} a_{1} - c_{1} a_{3} C_{3} = - ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} a_{3} & b_{3} \end{matrix} ∣ ∣ ∣ = a_{3} b_{1} - a_{1} b_{3} ∴ a_{1} A_{2} + b_{1} B_{2} + c_{1} C_{3} = a_{1} b_{3} c_{1} - a_{1} b_{1} c_{3} + a_{1} b_{1} c_{3} - a_{3} b_{1} c_{1} + a_{3} b_{1} c_{1} - a_{1} b_{3} c_{1} = 0$

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Similar questions

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

A_{2}

B_{2}

C_{2}

are respectively cofactors of

a_{2}, b_{2}, c_{2}

a_{1} A_{2} + b_{1} B_{2} + c_{1} C_{2}

Two lines whose direction ratios are a₁, b₁, c₁ and a₂, b₂, c₂ are perpendicular, if

△ = ⎡ ⎢ ⎣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ⎤ ⎥ ⎦

A_{2}, B_{2}, C_{2}

are respectively cofactors of

a_{2}, b_{2}, c_{2}

a_{1} A_{2} + b_{1} B_{2} + c_{1} C_{2}

(a_{1}, a_{2}, a_{3})

(b_{1}, b_{2}, b_{3})

(c_{1}, c_{2}, c_{3})

are linearly independent and

x (a_{1}, a_{2}, a_{3}) + y (b_{1}, b_{2}, b_{3}) + z (c_{1}, c_{2}, c_{3}) = 0

Q. If each pair of equations

a_{1} x^{2} + b_{1} x + c_{1} = 0, a_{2} x^{2} + b_{2} x + c_{2} = 0

a_{3} x^{2} + b_{3} x + c_{3} = 0

has a common root, then show that
(i)

\frac{c_{1} a_{2} + c_{2} a_{1}}{c_{1} a_{2} - c_{2} a_{1}} + \frac{a_{1} a_{2} b_{3}}{a_{3} (a_{1} b_{2} - a_{2} b_{1})} = 0

{(\frac{a_{1} b_{2} - a_{2} b_{1}}{a_{1} c_{2} - a_{2} c_{1}})}^{2} = \frac{a_{1} a_{2} c_{3}}{c_{1} c_{2} a_{3}}

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