# Misc. 2 - Chapter 4 Class 12 Determinants

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc. 2 Without expanding the determinant, prove that aa2bcbb2cacc2ab = 1a2a31b2b31c2c3 Taking L.H.S aa2bcbb2cacc2ab Multiplying and dividing by abc = abcabc aa2bcbb2cacc2ab Multiplying a to R1, b to R2 & c to R3 = 1abc a(𝑎)𝑎(a2)a(bc)b(𝑏)b(b2)b (ca)c 𝑐𝑐(c2)c (ab) Multiplying a to R1, b to R2 & c to R3 = 1abc a(𝑎)𝑎(a2)a(bc)b(𝑏)b(b2)b (ca)c 𝑐𝑐(c2)c (ab) = 1abc a2a3𝑎𝑏𝑐b2b3𝑎𝑏𝑐c2c3𝑎𝑏𝑐 Taking abc common from C3 = 𝑎𝑏𝑐𝑎𝑏𝑐 a2a31b2b31c2c31 = a2a31b2b31c2c31 Interchange C1 ↔ C3 = ( – 1) 1a3a21b3b21c3c2 Interchange C2 ↔ C3 = ( – 1) ( – 1) 1a2a31b2b31c2c3 = 1a2a31b2b31c2c3 = R.H.S. Hence Proved.

Chapter 4 Class 12 Determinants

Serial order wise

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.