# Ex 4.5, 11

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 4.5, 11 Find the inverse of each of the matrices 1000 cos𝛼 sin𝛼0 sin𝛼− cos𝛼 Let A = 1000 cos𝛼 sin𝛼0 sin𝛼− cos𝛼 We know that A–1 = 1|A| (adj A) exists if |A|≠ 0 Step 1: Calculate |A| |A| = 1000 cos𝛼 sin𝛼0 sin𝛼− cos𝛼 = 1 cos𝛼 sin𝛼 sin𝛼− cos𝛼– 0 0 sin𝛼0 − cos𝛼+ 0 0 cos𝛼0𝑠𝑖𝑛 𝛼 = 1(– cos2α – sin2α ) – 0 + 0 = – ( cos2α + sin2α ) = – 1 Since 𝐴≠ 0 Thus A-1 exits Step 2: Calculating adj A adj (A) = A11A21A31A12A22A32A33A23A33 A = 1000 cos𝛼 sin𝛼0 sin𝛼− cos𝛼 M11 = cosα sinα sinα− cosα = – cos2α – sin2α = –( cos2α + 𝑠𝑖𝑛2α )= –1 M12 = 0sin 𝛼0− cos 𝛼 = 0 – 0 = 0 M13 = 0 cos𝛼0sin 𝛼 = 0 – 0 = 0 M21 = 00sin 𝛼− cos𝛼 = 0 – 0 = 0 M22 = 100−cos 𝛼 = –cos α – 0 = – cos α M23 = 100 sin𝛼 = sin α = 0 = sin α M31 = 00cos 𝛼sin 𝛼 = 0 – 0 = – 0 M32 = 100sin 𝛼 = sin α – 0 = sin α M33 = 100cos 𝛼 = cos α + 0 = cos α A11 = (−1)1+1 M11 = (−2)2 ( – 1) = – 1 A12 = (−1)1+2 M12 = (−1)3 0 = 0 A13 = (−1)1+3 M13 = (−4)4 0 = 0 A21 = ( – 1)2+1 M21 = ( – 1)3 0 = 0 A22 = (−1)2+2 M22 = (−1)4(– cos α) = – cos α A23 = ( – 1)2+3 M23 = ( – 1)5s sin α = – sin α A31 = (−1)3+1. M31 = (−1)4( – 1)4 0 = 0 A32 = ( – 1)3+2 sin α = ( – 1)5 sin α = – sin α A33 = ( – 1)3+3 M33 = ( – 1)6 cos α = cos α So, adj (A) = A11A21A31A12A22A32A33A23A33 = −1000− cos𝛼− sin𝛼0− sin𝛼 cos𝛼 Step 3: Calculating inverse Now, A– 1 = 1|A| ( adj (A)) = 1−1 −1000− cos𝛼− sin𝛼0− sin𝛼 cos𝛼 = – −1000− cos𝛼− sin𝛼0− sin𝛼 cos𝛼 = 𝟏𝟎𝟎𝟎 𝒄𝒐𝒔𝜶 𝒔𝒊𝒏𝜶𝟎 𝒔𝒊𝒏𝜶 −𝒄𝒐𝒔𝜶

Chapter 4 Class 12 Determinants

Serial order wise

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .