Question 14 - Inverse of matrix using elementary transformation - Chapter 3 Class 12 Matrices
Last updated at April 16, 2024 by Teachoo
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 2025 Exams
Question 1 Deleted for CBSE Board 2025 Exams
Question 2 Deleted for CBSE Board 2025 Exams
Question 3 Deleted for CBSE Board 2025 Exams
Question 4 Deleted for CBSE Board 2025 Exams
Question 5 Deleted for CBSE Board 2025 Exams
Question 6 Deleted for CBSE Board 2025 Exams
Question 7 Deleted for CBSE Board 2025 Exams
Question 8 Important Deleted for CBSE Board 2025 Exams
Question 9 Deleted for CBSE Board 2025 Exams
Question 10 Deleted for CBSE Board 2025 Exams
Question 11 Deleted for CBSE Board 2025 Exams
Question 13 Deleted for CBSE Board 2025 Exams
Question 3 Important Deleted for CBSE Board 2025 Exams
Question 12 Deleted for CBSE Board 2025 Exams
Question 14 Deleted for CBSE Board 2025 Exams You are here
Question 2 Important Deleted for CBSE Board 2025 Exams
Question 15 Important Deleted for CBSE Board 2025 Exams
Question 16 Deleted for CBSE Board 2025 Exams
Question 17 Important Deleted for CBSE Board 2025 Exams
Inverse of matrix using elementary transformation
Last updated at April 16, 2024 by Teachoo
Ex 3.4, 14 Find the inverse of each of the matrices, if it exists.[■8(2&1@4&2)] Let A = [■8(2&1@4&2)] We know that A = IA [■8(2&1@4&2)] = [■8(1&0@0&1)] A R1 →"R1"−1/2 R2 [■8(𝟐−𝟏/𝟐(𝟒)&1−1/2(2)@4&2)] = [■8(1&0@0&1)] A [■8(0&0@4&2)] = [■8(1&−1/2@0&1)] A Since we have all zeros in the first row of the left hand side matrix of the above equation. Therefore, A−1 does not exist.