Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Inverse of matrix using elementary transformation
Inverse of a matrix
Finding inverse of a matrix using Elementary Operations
Ex 3.4, 1 (MCQ)
Question 1 Deleted for CBSE Board 2024 Exams You are here
Question 1 Deleted for CBSE Board 2024 Exams You are here
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 Deleted for CBSE Board 2024 Exams
Question 4 Deleted for CBSE Board 2024 Exams
Question 5 Deleted for CBSE Board 2024 Exams
Question 6 Deleted for CBSE Board 2024 Exams
Question 7 Deleted for CBSE Board 2024 Exams
Question 8 Important Deleted for CBSE Board 2024 Exams
Question 9 Deleted for CBSE Board 2024 Exams
Question 10 Deleted for CBSE Board 2024 Exams
Question 11 Deleted for CBSE Board 2024 Exams
Question 13 Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Question 12 Deleted for CBSE Board 2024 Exams
Question 14 Deleted for CBSE Board 2024 Exams
Question 2 Important Deleted for CBSE Board 2024 Exams
Question 15 Important Deleted for CBSE Board 2024 Exams
Question 16 Deleted for CBSE Board 2024 Exams
Question 17 Important Deleted for CBSE Board 2024 Exams
Inverse of matrix using elementary transformation
Last updated at May 29, 2023 by Teachoo
Ex 3.4, 1 Find the inverse of each of the matrices, if it exists. [■8(1&−1@2&3)] Let A = [■8(1&−1@2&3)] We know that A = IA [■8(1&−1@2&3)] = [■8(1&0@0&1)] A R2 → R2 – 2R1 [■8(1&−1@𝟐−𝟐(𝟏)&3−2(−1))] = [■8(1&0@0−2(1)&1−2(0))] A [■8(1&−1@𝟎&5)] = [■8(1&0@−2&1)] A R2 →1/5 R2 [■8(1&−1@0/5&𝟓/𝟓)] = [■8(1&0@(−2)/5&1/5)] A [■8(1&−1@0&𝟏)] = [■8(1&0@(−2)/5 " " &1/5 " " )] A R1 →R1 + R2 [■8(1+0&−𝟏+𝟏@0&1)] = [■8(1−2/5&0+1/5@(−2)/5 " " &1/5 " " )] A [■8(1&𝟎@0&1)] = [■8(3/5&1/5@(−2)/5 " " &1/5 " " )] A I = [■8(3/5&1/5@(−2)/5 " " &1/5 " " )] A This is similar to I = A-1A Thus, A-1 = [■8(𝟑/𝟓&𝟏/𝟓@(−𝟐)/𝟓 " " &𝟏/𝟓 " " )]