# Ex 3.4, 11 - Chapter 3 Class 12 Matrices (Term 1)

Last updated at Jan. 17, 2020 by Teachoo

Ex 3.4

Ex 3.4, 1
Deleted for CBSE Board 2023 Exams

Ex 3.4, 2 Deleted for CBSE Board 2023 Exams

Ex 3.4, 3 Deleted for CBSE Board 2023 Exams

Ex 3.4, 4 Deleted for CBSE Board 2023 Exams

Ex 3.4, 5 Deleted for CBSE Board 2023 Exams

Ex 3.4, 6 Deleted for CBSE Board 2023 Exams

Ex 3.4, 7 Deleted for CBSE Board 2023 Exams

Ex 3.4, 8 Important Deleted for CBSE Board 2023 Exams

Ex 3.4, 9 Deleted for CBSE Board 2023 Exams

Ex 3.4, 10 Deleted for CBSE Board 2023 Exams

Ex 3.4, 11 Deleted for CBSE Board 2023 Exams You are here

Ex 3.4, 12 Deleted for CBSE Board 2023 Exams

Ex 3.4, 13 Deleted for CBSE Board 2023 Exams

Ex 3.4, 14 Deleted for CBSE Board 2023 Exams

Ex 3.4, 15 Important Deleted for CBSE Board 2023 Exams

Ex 3.4, 16 Deleted for CBSE Board 2023 Exams

Ex 3.4, 17 Important Deleted for CBSE Board 2023 Exams

Ex 3.4, 18 (MCQ) Deleted for CBSE Board 2023 Exams

Chapter 3 Class 12 Matrices

Serial order wise

Last updated at Jan. 17, 2020 by Teachoo

Ex 3.4, 11 Find the inverse of each of the matrices, if it exists.[■8(2&−[email protected]&−2)] Let A =[■8(2&−[email protected]&−2)] We know that A = IA [■8(2&−[email protected]&−2)] = A [■8(1&[email protected]&1)] R1 → R1 – R2 [■8(𝟐−𝟏&−6−(−2)@1&−2)] = [■8(1−0&0−[email protected]&1)] A [■8(𝟏&−[email protected]&−2)] = [■8(1&−[email protected]&1)] A R2 → R2 – R1 [■8(1&−[email protected]𝟏−𝟏&−2−(−4))] = [■8(1&−[email protected]−1&1−(−1))] A [■8(1&−[email protected]𝟎&2)] = [■8(1&−[email protected]−1&2)] A R2 → 1/2 R2 [■8(1&−[email protected]𝟎/𝟐&2/2)] = [■8(1&−[email protected](−1)/2&2/2)] A [■8(1&−[email protected]𝟎&1)] = [■8(1&−[email protected](−1)/2&1)] A R1 → R1 + 4R2 [■8(1+4(0)&−𝟒+𝟒(𝟏)@0&1)] = [■8(1+4((−1)/2)&−1+4(1)@(−1)/2&1)] A [■8(1&𝟎@0&1)] = [■8(−1&[email protected](−1)/2&1)] A I = [■8(−1&[email protected](−1)/2&1)] A This is similar to I = A-1A Thus, A-1 = [■8(−𝟏&𝟑@(−𝟏)/𝟐&𝟏)]