# Example 27 - Chapter 4 Class 12 Determinants (Term 1)

Last updated at Jan. 16, 2020 by Teachoo

Last updated at Jan. 16, 2020 by Teachoo

Transcript

A = 2532 = 2 × 2 – 5 × 3 = 4 – 15 = – 11 Since 𝑨 ≠ 0, System is consistent and the system of equations has a unique solution AX = B X = A-1 B Step 3 Calculating X = A-1 B Here, A-1 = 1|A| adj (A) adj A = 2532 = 2−5−32 Now, A-1 = 1|A| adj A Putting values = 1−11 2−5−32 & B = 17 Now X = A-1 B 𝑥𝑦 = −111 2−5−32 17 = −111 2 1+ −57−3 1+2(7) = −111 2−35−3+14 𝑥𝑦 = −111 −3311 = −33 × −11111 × −111 𝑥𝑦 = 3−1 Hence x = 3 & y = – 1

Examples

Example 1

Example 2

Example 3

Example 4

Example 5 Important

Example 6 Deleted for CBSE Board 2022 Exams

Example 7 Deleted for CBSE Board 2022 Exams

Example 8 Deleted for CBSE Board 2022 Exams

Example 9 Important Deleted for CBSE Board 2022 Exams

Example 10 Important Deleted for CBSE Board 2022 Exams

Example 11 Deleted for CBSE Board 2022 Exams

Example 12 Deleted for CBSE Board 2022 Exams

Example 13 Deleted for CBSE Board 2022 Exams

Example 14 Important Deleted for CBSE Board 2022 Exams

Example 15 Important Deleted for CBSE Board 2022 Exams

Example 16 Important Deleted for CBSE Board 2022 Exams

Example 17

Example 18 Important

Example 19

Example 20

Example 21

Example 22 Important

Example 23

Example 24 Important

Example 25

Example 26 Important

Example 27 You are here

Example 28 Important

Example 29

Example 30 Deleted for CBSE Board 2022 Exams

Example 31 Important Deleted for CBSE Board 2022 Exams

Example 32 Important Deleted for CBSE Board 2022 Exams

Example 33 Important

Example 34 Important Deleted for CBSE Board 2022 Exams

Chapter 4 Class 12 Determinants (Term 1)

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.