If A 1- B 1- C 1- are respectively the co-factors of the elements a 1- b 1- c 1- of the determinant - a 1 b 1 c 1- a 2 b 2 c 2- a 3 b 3 c 3 - then - B 2 C 2- B 3 C 3 -A- None of theseB- a 1-C- a 1 a 3-D- a 1- b 1 -

The correct option is B a1ΔB_2 = a_1 amp;c_1 a_3 amp;c_3 = a_1c_3 c_1a_3 C_2 = a_1 amp;b_1 a_3 amp;b_3 = (a_1b_3 a_3b_1) B_3 = a_1 am ...

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Byju's Answer

Standard XII

Mathematics

Evaluation of a Determinant

If A 1, B 1, ...

Question

If $A_{1}, B_{1}, C_{1} . . . .$ are respectively the co-factors of the elements $a_{1}, b_{1}, c_{1} . . . .$ of the determinant $Δ = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣,$ then $∣ ∣ ∣ \begin{matrix} B_{2} & C_{2} B_{3} & C_{3} \end{matrix} ∣ ∣ ∣ =$

a₁Δ

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a₁a₃Δ

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(a₁ + b₁)Δ

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None of these

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Solution

The correct option is A a₁Δ
$B_{2} = ∣ ∣ ∣ \begin{matrix} a_{1} & c_{1} a_{3} & c_{3} \end{matrix} ∣ ∣ ∣ = a_{1} c_{3} - c_{1} a_{3} C_{2} = - ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} a_{3} & b_{3} \end{matrix} ∣ ∣ ∣ = - (a_{1} b_{3} - a_{3} b_{1}) B_{3} = - ∣ ∣ ∣ \begin{matrix} a_{1} & c_{1} a_{2} & c_{2} \end{matrix} ∣ ∣ ∣ = - (a_{1} c_{2} - a_{2} c_{1}) C_{3} = ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} a_{2} & b_{2} \end{matrix} ∣ ∣ ∣ = (a_{1} b_{2} - a_{2} b_{1}) ∣ ∣ ∣ \begin{matrix} B_{2} & C_{2} B_{3} & C_{3} \end{matrix} ∣ ∣ ∣ = ∣ ∣ ∣ \begin{matrix} a_{1} c_{3} - a_{3} c_{1} & - (a_{1} b_{3} - a_{3} b_{1}) - (a_{1} c_{2} - a_{2} c_{1}) & a_{1} b_{2} - a_{2} b_{1} \end{matrix} ∣ ∣ ∣ = ∣ ∣ ∣ \begin{matrix} a_{1} c_{3} & - a_{1} b_{3} - a_{1} c_{2} & a_{1} b_{2} \end{matrix} ∣ ∣ ∣ + ∣ ∣ ∣ \begin{matrix} a_{1} c_{3} & a_{3} b_{1} - a_{1} c_{2} & - a_{2} b_{1} \end{matrix} ∣ ∣ ∣ + ∣ ∣ ∣ \begin{matrix} - a_{3} c_{1} & - a_{1} b_{3} a_{2} c_{1} & a_{1} b_{2} \end{matrix} ∣ ∣ ∣ + ∣ ∣ ∣ \begin{matrix} - a_{3} c_{1} & a_{3} b_{1} a_{2} c_{1} & - a_{2} b_{1} \end{matrix} ∣ ∣ ∣ = {a_{1}}^{2} (b_{2} c_{3} - b_{3} c_{2}) + a_{1} b_{1} (- c_{3} a_{2} + a_{3} c_{2}) + a_{1} c_{1} (- a_{3} b_{2} + a_{2} b_{3}) + c_{1} b_{1} (a_{3} a_{2} - a_{2} a_{3}) = a_{1} Δ .$

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Similar questions

Q. 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_{1}, B_{1}, C_{1}

denote the co-factors of

a_{1}, b_{1}, c_{1}

respectively, then the value of the determinant

∣ ∣ ∣ ∣ \begin{matrix} A_{1} & B_{1} & C_{1} A_{2} & B_{2} & C_{2} A_{3} & B_{3} & C_{3} \end{matrix} ∣ ∣ ∣ ∣

If A1, B1, C1 … are the cofactors of the elements a1, b1, c1... of the matrix

$⎡ ⎢ ⎣ \begin{matrix} a 1 & b 1 & c 1 a 2 & b 2 & c 2 a 3 & b 3 & c 3 \end{matrix} ⎤ ⎥ ⎦$ then $∣ ∣ ∣ \begin{matrix} B 2 & C 2 B 3 & C 3 \end{matrix} ∣ ∣ ∣ =$

Q. If

A_{1}, B_{1}, C_{1} . . . .

are respectively the co-factors of the elements

a_{1}, b_{1}, c_{1} . . . .

of the determinant

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

then

∣ ∣ ∣ \begin{matrix} B_{2} & C_{2} B_{3} & C_{3} \end{matrix} ∣ ∣ ∣ =

Q. If in the determinant

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

etc. be the co-factors of

a_{1}, b_{1}, c_{1}

etc., then which of the following relations is incorrect

Q. If

A_{1}, B_{1}, C_{1} . .

are respectively the co-factor of the elements

a_{1}, b_{1}, c_{1}

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

, then

∣ ∣ ∣ \begin{matrix} B_{2} & C_{2} B_{3} & C_{3} \end{matrix} ∣ ∣ ∣

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If A1, B1, C1.... are respectively the co-factors of the elements a1, b1, c1.... of the determinant Δ = ∣∣ ∣∣a1b1c1a2b2c2a3b3c3∣∣ ∣∣, then ∣∣∣B2C2B3C3∣∣∣ =