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Question

Show that ∣ ∣111abca2b2c2∣ ∣=(ab)(bc)(ca).

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Solution

∣ ∣ ∣111abca2b2c2∣ ∣ ∣=(ab)(bc)(ca)
Solving determinant, we have
∣ ∣ ∣111abca2b2c2∣ ∣ ∣
Applying C2C2C1 and C3C3C1
=∣ ∣ ∣100abacaa2b2a2c2a2∣ ∣ ∣
Taking (ca) and (ba) as common from C3 and C2 respectively.
=(ba)(ca)∣ ∣100a11a2b+ac+a∣ ∣
Expanding the determinant along R1, we have
=(ba)(ca)(1(c+aba)0+0)
=(ba)(ca)(cb)
=(ab)(bc)(ca)
= R.H.S.
Hence proved.A

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