Ex 3.4, 4

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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" " )]