# Ex 4.6, 8 - Chapter 4 Class 12 Determinants (Term 1)

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Step 2 Calculate |A| |A| = 2 1 3 4 = 2(4) ( 1)(3) = 8 + 3 = 11 |A| 0 Thus, System of equation is consistent & has a unique solution Now, AX = B X = A-1 B Step 3 Calculate X = A-1 B A-1 = 1 |A| adj (A) A = 2 1 3 4 adj A = 2 1 3 4 = 4 1 3 2 Now, A-1 = 1 |A| adj A A-1 = 1 11 4 1 3 2 Thus, X = A-1 B = 1 11 4 1 3 2 2 3 = 1 11 4 2 +1(3) 3 2 +2(3) = 1 11 8+3 6+6 = 1 11 5 12 = 5 11 12 11 Hence, x = & y =

Ex 4.6

Ex 4.6, 1
Deleted for CBSE Board 2022 Exams

Ex 4.6, 2 Deleted for CBSE Board 2022 Exams

Ex 4.6, 3 Important Deleted for CBSE Board 2022 Exams

Ex 4.6, 4 Deleted for CBSE Board 2022 Exams

Ex 4.6, 5 Important Deleted for CBSE Board 2022 Exams

Ex 4.6, 6 Deleted for CBSE Board 2022 Exams

Ex 4.6, 7

Ex 4.6, 8 You are here

Ex 4.6, 9

Ex 4.6, 10 Important

Ex 4.6, 11

Ex 4.6, 12

Ex 4.6, 13 Important

Ex 4.6, 14 Important

Ex 4.6, 15 Important

Ex 4.6, 16 Important

Chapter 4 Class 12 Determinants (Term 1)

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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.