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Inverse of matrix using elementary transformation
Inverse of a matrix
Finding inverse of a matrix using Elementary Operations
Ex 3.4, 18 (MCQ)
Example 23
Ex 3.4, 1
Ex 3.4, 2
Ex 3.4, 3
Ex 3.4, 4 You are here
Ex 3.4, 5
Ex 3.4, 6
Ex 3.4, 7
Ex 3.4, 8 Important
Ex 3.4, 9
Ex 3.4, 10
Ex 3.4, 11
Ex 3.4, 13
Example 25 Important
Ex 3.4, 12
Ex 3.4, 14
Example 24 Important
Ex 3.4, 15 Important
Ex 3.4, 16
Ex 3.4, 17 Important
Inverse of matrix using elementary transformation
Last updated at Sept. 24, 2018 by Teachoo
Ex3.4, 4 Find the inverse of each of the matrices, if it exists. [■8(2&3@5&7)] Let A = [■8(2&3@5&7)] We know that A = IA [■8(2&3@5&7)] = [■8(1&0@0&1)] A R1→1/2R1 [■8(𝟐/𝟐&3/2@5&7)] = [■8(1/2&0/2@0&1)] A [■8(𝟏&3/2@5" " &7" " )] = [■8(1/2&0@0&1)] A R2 →R2 – 5R1 [■8(1&3/2@𝟓−𝟓" " &7−5 (3/2)" " )] = [■8(1/2&0@0−5/2&1−5(0))] A [■8(1&3/2@𝟎" " &−1/2)] = [■8(1/2&0@(−5)/2&1)] A R2 → -2R1 [■8(1&3/2@−2(0)" " &−𝟐((−𝟏)/𝟐) )] = [■8(1/2&0@−2((−5)/2)&−2(1))] A [■8(1&3/2@0" " &𝟏)] = [■8(1/2&0@5&−2)] A R1 →R1 – 3/2 R2 [■8(1−0(3/2)&𝟑/𝟐−𝟑/𝟐(𝟏)@0" " &1)] = [■8(1/2−3/2(5)&0−3/2(−2)@5&−2)] A [■8(1&𝟎@0" " &1)] = [■8(−7&3@5&−2)] A This is similar to I = A-1A Thus, A-1 = [■8(−7&3@5&−2" " )]