If D - - - a 1 b 1 c 1- a 2 b 2 c 2- a 3 b 3 c 3 - and D 0 - - - k a 1 k b 1 k c 1- k a 2 kb 2 kc 2- ka 3 kb 3 k c 3 - then show that D 0 - k3 D

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Skew Symmetric Matrix

If D = | [ ...

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If $D = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣$ and $D_{0} = ∣ ∣ ∣ ∣ \begin{matrix} k a_{1} & k b_{1} & k c_{1} k a_{2} & k b_{2} & k c_{2} k a_{3} & k b_{3} & k c_{3} \end{matrix} ∣ ∣ ∣ ∣$ then show that $D_{0} = k^{3} D$

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

A

A^{- 1} = \frac{A d j A}{d e t 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}, B_{2}, C_{2}

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

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

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

| A | \neq 0

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

a_{3} x + b_{3} y + c_{3} z = 0

∣ ∣ ∣ ∣ \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} & 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} ∣ ∣ ∣ ∣

The fraction of a floating object of volume $V_{0}$ and density $d_{0}$ above the surface of a liquid of density d will be

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