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Question

Using the property of determinants , prove that,
∣ ∣b+cp+ry+zc+ar+pz+xa+bp+qx+y∣ ∣=2∣ ∣apxbqycrz∣ ∣

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Solution

=∣ ∣b+cp+ry+zc+ar+pz+xa+bp+qx+y∣ ∣=∣ ∣bp+ry+zcr+pz+xap+qx+y∣ ∣+∣ ∣cp+ry+zar+pz+xbp+qx+y∣ ∣
Again Splitting
=∣ ∣bpy+zcrz+xapx+y∣ ∣+∣ ∣cry+zapz+xbqx+y∣ ∣+∣ ∣cry+zapz+xbqx+y∣ ∣
On splitting all above Determinants together, we get
=∣ ∣bcaqrpyzx∣ ∣+∣ ∣bcbqrqyzy∣ ∣+∣ ∣baaqppyxx∣ ∣+∣ ∣babqpqyxy∣ ∣+∣ ∣caarppzxx∣ ∣+∣ ∣cabrpqzxy∣ ∣=∣ ∣bcaqrpyzx∣ ∣+∣ ∣cabrpqzxy∣ ∣(1)C1C3&C12(2)=∣ ∣acbprqxzy∣ ∣∣ ∣acbprqxzy∣ ∣C2C3=2∣ ∣abcpqrxyz∣ ∣
By taking transpose, we get
=2∣ ∣apxbqycrz∣ ∣

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