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  1. Chapter 3 Class 12 Matrices
  2. Serial order wise

Transcript

Ex 3.4, 16 (Method 1) Find the inverse of each of the matrices, if it exists. [1 3 −2 −3 0 −5 2 5 0)] Let A = [1 3 −2 −3 0 −5 2 5 0)] A = IA [1 3 −2 −3 0 −5 2 5 0)] = [1 0 0 0 1 0 0 0 1)] A R2 → R2 + 3R1 [1 3 −2 −𝟑+𝟑(𝟏) 0+3(3) −5+3(−2) 2 5 0)] = [1 0 0 0+3(1) 1+3(0) 0+3(0) 0 0 1)] A [1 3 −2 𝟎 9 −11 2 5 0)] = [1 0 0 3 1 0 0 0 1)] A R3 → R3 − 2R1 [1 3 −2 0 9 −11 𝟐−𝟐(𝟏) 5−2(3) 0−2(−2))] = [1 0 0 3 1 0 0−2(1) 0−2(0) 1−2(0))] A [1 3 −2 0 9 −11 𝟎 −1 4)] = [1 0 0 3 1 0 −2 0 1)] A R2 → 1/9 R2 [1 3 −2 0/3 𝟗/𝟗 (−11)/9 0 −1 4)]= [1 0 0 3/9 1/9 0/9 −2 0 1)]A [1 3 −2 0 𝟏 (−11)/9 0 −1 4)] = [1 0 0 1/3 1/9 0 −2 0 1)] A R3 → R3 − R2 [1 3 −2 0 1 (−11)/9 0+0 −𝟏+𝟏 4+((−11)/9) )] = [1 0 0 1/3 1/9 0 −2+1/3 0+1/9 1+0)] A [1 3 −2 0 1 (−11)/9 0 𝟎 25/9)] = [1 0 0 1/3 1/9 0 (−5)/3 1/9 1)] A R1 → R1 − 3R2 [1−3(0) 𝟑−𝟑(𝟏) −2−3(11/3) 0 1 (−11)/9 0 0 25/3)] =[1−3(1/3) 0−3(1/9) 0−3(0) 1/3 1/9 0 (−5)/3 1/9 1)] A [1 𝟎 5/3 0 1 (−11)/9 0 0 25/9)] = [0 (−1)/3 0 1/3 1/9 0 (−5)/3 1/9 1)] A R3 → 9/25 R3 [1 0 5/3 0 1 (−11)/9 0 0 𝟏)] = [0 (−1)/3 0 1/3 1/9 0 (−3)/5 1/25 9/25)] A R2 → R2 + 11/9 R3 [1 0 5/3 0+11/9 (0) 1+11/9(0) (−𝟏𝟏)/𝟗+𝟏𝟏/𝟗(𝟏) 0 0 1)] =[0 (−1)/3 0 1/3+11/9 ((−3)/5) 1/9+11/9 (1/25) 0+11/9 (9/25) (−5)/3 1/25 9/25)] A [1 0 5/3 0 1 𝟎 0 0 1)] = [0 (−1)/3 0 −(−2)/5 1/9+1/(9×25) 1/25 (−3)/5 1/25 9/25)] A R1 → R1 – 5/3 R3 [1+ 5/3 (0) 0+ 5/3 (0) (−𝟓)/𝟑+ 𝟓/𝟑 (𝟏) 0 1 0 0 0 1)] =[0−5/3 ((−3)/5) (−1)/3− 5/3 ((−3)/5) 0−5/3 (9/25) 2/5 4/25 1/25 (−3)/5 1/25 9/25)] A [1 0 𝟎 0 1 0 0 0 1)] = [5/5 (−1)/3− 1/( 5×3) 3/5 2/5 (−4)/25 11/25 (−3)/5 1/25 9/25)] A I = [1 (−2)/5 3/5 2/5 (−4)/25 1/25 (−3)/5 1/25 9/25)] A This is similar to I = A-1 A Hence A-1 = [1 (−2)/5 3/5 2/5 (−4)/25 1/25 (−3)/5 1/25 9/25)] Ex 3.4, 16 (Method 2) Find the inverse of each of the matrices, if it exists. [1 3 −2 −3 0 −5 2 5 0)] Let A = [1 3 −2 −3 0 −5 2 5 0)] Now, A = IA [1 3 −2 −3 0 −5 2 5 0)] = [1 0 0 0 1 0 0 0 1)] A C2 → C2 − 3C1 [1 𝟎 −2 −3 9 −5 2 −1 0)] = [1 −3 0 0 1 0 0 0 1)] A C3 → C3 + 2C1 [1 0 𝟎 −3 9 −11 2 −1 4)] = [1 −3 2 0 1 0 0 0 1)] A C2 → 𝐶_2/9 [1 0 0 −3 𝟏 −11 2 (−1)/9 4)] = [1 −1/3 2 0 1/9 0 0 0 1)] A C1 → C1 + 3C2 [1 0 0 𝟎 1 −11 5/3 (−1)/9 4)] = [0 (−1)/3 2 1/3 1/9 0 0 0 1)] A C3 → C3 + 11C2 [1 0 0 0 1 𝟎 5/3 (−1)/9 25/9)] = [0 (−1)/3 (−5)/3 1/3 1/9 11/9 0 0 1)] A C3 → 9/25 C3 [1 0 0 0 1 0 5/3 (−1)/9 𝟏)] = [0 (−1)/3 (−3)/5 1/3 1/9 11/25 0 0 9/25)] A C1 → C1 – 5/3 C3 [1 0 0 0 1 0 𝟎 (−1)/9 1)] = [1 (−1)/3 (−3)/5 (−2)/5 1/9 11/25 (−3)/5 0 9/25)] A C2 → C2 + 1/9 C3 [1 0 0 0 1 0 0 𝟎 1)] = [1 (−2)/5 (−3)/5 (−2)/5 4/25 11/25 (−3)/5 1/25 9/25)] A I = [1 (−2)/5 3/5 2/5 (−4)/25 1/25 (−3)/5 1/25 9/25)] A This is similar to I = A-1 A Hence A-1 = [1 (−2)/5 3/5 2/5 (−4)/25 1/25 (−3)/5 1/25 9/25)]

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.