$∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ = 5, t h e n t h e v a l u e o f ∣ ∣ ∣ ∣ \begin{matrix} b_{2} c_{3} - b_{3} c_{2} & a_{3} c_{2} - a_{2} c_{3} & a_{2} b_{3} - a_{3} b_{2} b_{3} c_{1} - b_{1} c_{3} & a_{1} c_{3} - a_{3} c_{1} & a_{3} b_{1} - a_{1} b_{3} b_{1} c_{2} - b_{2} c_{1} & a_{2} c_{1} - a_{1} c_{2} & a_{1} b_{2} - a_{2} b_{1} \end{matrix} ∣ ∣ ∣ ∣ i s$

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Solution

The correct option is B
25

Requireddet is |Adj A| where A= $∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣$

$[A^{2}]$ =25

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Similar questions

Q. If

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

= x

then

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

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

$∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ = 0$

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} ∣ ∣ ∣ ∣

Q. Let

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

and

Δ_{2} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} α_{1} & β_{1} & γ_{1} α_{2} & β_{2} & γ_{2} α_{3} & β_{3} & γ_{3} \end{matrix} ∣ ∣ ∣ ∣ ∣

, then

Δ_{1} \times Δ_{2}

can be expressed as the sum of how many determinants?

Three lines given by $L_{1} : a_{1} x + b_{1} y + c_{1} = 0, L_{2} : a_{2} x + b_{2} y + c_{2} = 0 and L_{3} : a_{3} x + b_{3} y + c_{3} = 0$ are always concurrent given that

$∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ = 0$

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∣∣ ∣∣a1b1c1a2b2c2a3b3c3∣∣ ∣∣=5,then the value of∣∣ ∣∣b2c3−b3c2a3c2−a2c3a2b3−a3b2b3c1−b1c3a1c3−a3c1a3b1−a1b3b1c2−b2c1a2c1−a1c2a1b2−a2b1∣∣ ∣∣ is

The correct option is B 25 Requireddet is |Adj A| where A= ∣∣ ∣∣a1b1c1a2b2c2a3b3c3∣∣ ∣∣ [A2]=25

The correct option is B
25

Requireddet is |Adj A| where A= $∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣$

$[A^{2}]$ =25