Let -1 - a1 a2 a3b1 b2 b3c1 c2 c3- -2 -6a1 2a2 2a33b1 b2 b312c1 4c2 4c3 -3 - 3a1 -2b1 -4c1 3a2 -2b2 -4c2 3a3 - 2b3 -4c3 3b1 3b2 3b3 3c1 3c2 3c3 then -3 - -2 is

Δ_3 = 3a_1 +2b_1 4c_1 amp; 3a_2 +2b_2 4c_2 amp; 3a_3 + 2b_3 4c_3 nbsp; 3b_1 amp; 3b_2 amp; 3b_3 nbsp; 3c_1 amp; 3c_2 amp; 3c_3 Δ_3 = ...

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Properties of Determinants

Let Δ1 = a1 ...

Question

Let $Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣, Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} 6 a_{1} & 2 a_{2} & 2 a_{3} 3 b_{1} & b_{2} & b_{3} 12 c_{1} & 4 c_{2} & 4 c_{3} \end{matrix} ∣ ∣ ∣ ∣$
& $Δ_{3} = ∣ ∣ ∣ ∣ \begin{matrix} 3 a_{1} + 2 b_{1} - 4 c_{1} & 3 a_{2} + 2 b_{2} - 4 c_{2} & 3 a_{3} + 2 b_{3} - 4 c_{3} 3 b_{1} & 3 b_{2} & 3 b_{3} 3 c_{1} & 3 c_{2} & 3 c_{3} \end{matrix} ∣ ∣ ∣ ∣$ then $Δ_{3} - Δ_{2}$ is

27 Δ_{1}

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9 Δ_{1}

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3 Δ_{1}

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Solution

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Similar questions

Δ_{1} + Δ_{2} + Δ_{3} =

△

A M N, B N L

C L M

Δ_{1}, Δ_{2}

Δ_{3}

Δ_{1} + Δ_{2} + Δ_{3}

Δ_{1} + Δ_{2} + Δ_{3}

I f x^{a} y^{b} = e^{m}, x^{c} y^{d} = e^{n}, Δ_{1} = ∣ ∣ ∣ \begin{matrix} m & b n & d \end{matrix} ∣ ∣ ∣ Δ_{2} = ∣ ∣ ∣ \begin{matrix} a & m c & n \end{matrix} ∣ ∣ ∣

Δ_{3} = ∣ ∣ ∣ \begin{matrix} a & b c & d \end{matrix} ∣ ∣ ∣,

Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} x & y & x + y y & x + y & x x + y & x & y \end{matrix} ∣ ∣ ∣ ∣

Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & y 1 & x + y & y 1 & x & x + y \end{matrix} ∣ ∣ ∣ ∣

Δ_{3} = ∣ ∣ ∣ ∣ \begin{matrix} x & x + y & x + 2 y - x & x & 0 0 & - x & x \end{matrix} ∣ ∣ ∣ ∣

Δ_{4} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & x^{2} 1 & y & y^{2} 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

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