Example 24 - Chapter 3 Class 12 Matrices (Term 1)

Transcript

Example 24 Obtain the inverse of the following matrix using elementary operations A = [■8(0&1&2@1&2&3@3&1&1)] Given A = [■8(0&1&2@1&2&3@3&1&1)] We know that A = IA [■8(0&1&2@1&2&3@3&1&1)] = [■8(1&0&0@0&1&0@0&0&1)] A R1↔R2 [■8(𝟏&2&3@0&1&2@3&1&1)] = [■8(0&1&0@1&0&0@0&0&1)] A R3 → R3 – 3R1 [■8(1&2&3@0&1&2@𝟑−𝟑(𝟏)&1−3(2)&1−3(3))] = [■8(0&1&0@1&0&0@0−3(0)&0−3(1)&1−3(0))]A [■8(1&2&3@0&1&2@𝟎&−5&−8)] = [■8(0&1&0@1&0&0@0&−3&1)] R1 → R1 – 2R2 [■8(1−2(0)&𝟐−𝟐(𝟏)&3−2(2)@0&1&2@0&−5&−8)] = [■8(0−2(1)&1−2(0)&0−2(0)@1&0&0@0&−3&1)]A [■8(1&𝟎&−1@0&1&2@0&−5&−8)] = [■8(−2&1&0@1&0&0@0&−3&1)] A R3 → R3 + 5R2 [■8(1&0&−1@0&1&2@0+5(0)&−𝟓+𝟓(𝟏)&−8+5(2))] = [■8(−2&1&0@1&0&0@0+5(1)&−3+5(0)&1+5(0))] A [■8(1&0&−1@0&1&2@0&𝟎&2)] = [■8(−2&1&0@1&0&0@5&−3&1)] A R3 → 1/2 R3 [■8(1&0&−1@0&1&2@0/2&0/2&𝟐/𝟐)] = [■8(−2&1&0@1&0&0@5/2&(−3)/2&1/2)] A R1 → R1 + R3 [■8(1+0&0+0&−𝟏+𝟏@0&1&2@0&0&1)]=[■8(−2+5/2&1+((−3)/2)&0+1/2@1&0&0@5/2&(−3)/2&1/2)] A [■8(1&0&𝟎@0&1&2@0&0&1)] = [■8(1/2&(−1)/2&1/2@1&0&0@5/2&(−3)/2&1/2)] A R2 → R2 – 2R3 [■8(1&0&0@0−2(0)&1−2(0)&𝟐−𝟐(𝟏)@0&0&1)] = [■8(1/2&(−1)/2&1/2@1−2(5/2)&0−2((−3)/2)&0−2(1/2)@5/2&(−3)/2&1/2)]A [■8(1&0&0@0&1&𝟎@0&0&1)] = [■8(1/2&(−1)/2&1/2@−4&3&−1@5/2&(−3)/2&1/2)] A I= [■8(1/2&(−1)/2&1/2@−4&3&−1@5/2&(−3)/2&1/2)] A This is similar to I = A-1 A Hence, A-1 = [■8(𝟏/𝟐&(−𝟏)/𝟐&𝟏/𝟐@−𝟒&𝟑&−𝟏@𝟓/𝟐&(−𝟑)/𝟐&𝟏/𝟐)]