Ex 3.4, 5

Transcript

Ex3.4, 5 Find the inverse of each of the matrices, if it exists. [■8(2&1@7&4)] Let A = [■8(2&1@7&4)] We know that A = IA [■8(2&1@7&4)] = [■8(1&0@0" " &1)] A R1→ R1 – 1/7 R2 [■8(𝟐−𝟏/𝟕(𝟕)&1−1/7(4)@7&4)] = [■8(1−1/7(0)&0−1/7(1)@0" " &1)] A [■8(𝟐−𝟏&1−4/7@7&4)] = [■8(1−0&(−1)/7@0" " &1)] A [■8(𝟏&3/7@7&4)] = [■8(1&(−1)/7@0" " &1)] A R2→ R2 – 7R1 [■8(1&3/7@𝟕−𝟕(𝟏)&4−7(3/7) )] = [■8(1&(−1)/7@0−7(1)&1−7((−1)/7) )] A [■8(1&3/7@𝟕−𝟕&4−3)] = [■8(1&(−1)/7@0−7&1+1)] A [■8(1&3/7@𝟎&1)] = [■8(1&(−1)/7@−7&2)] A R1→ R1 – 3/7R2 [■8(1−3/7(0)&𝟑/𝟕−𝟑/𝟕 (𝟏)@0&1)] = [■8(1−3/7(−7)&(−1)/7−3/7(2)@−7&2)] A [■8(1−0&𝟑/𝟕−𝟑/𝟕@0&1)] = [■8(1+3&(−1)/7−6/7@−7&2)] A [■8(1&𝟎@0&1)] = [■8(4&−1@−7" " &2)] A I = [■8(4&−1@−7" " &2)] A This is similar to I = A-1A Thus, A-1 = [■8(4&−1@−7" " &2)] A