Ex 4.2

Ex 4.2, 1
Deleted for CBSE Board 2022 Exams

Ex 4.2, 2 Important Deleted for CBSE Board 2022 Exams

Ex 4.2, 3 Deleted for CBSE Board 2022 Exams

Ex 4.2, 4 Deleted for CBSE Board 2022 Exams

Ex 4.2, 5 Important Deleted for CBSE Board 2022 Exams

Ex 4.2, 6 Important Deleted for CBSE Board 2022 Exams

Ex 4.2, 7 Important Deleted for CBSE Board 2022 Exams

Ex 4.2, 8 (i) Important Deleted for CBSE Board 2022 Exams

Ex 4.2, 8 (ii) Deleted for CBSE Board 2022 Exams

Ex 4.2, 9 Important Deleted for CBSE Board 2022 Exams

Ex 4.2, 10 (i) Deleted for CBSE Board 2022 Exams You are here

Ex 4.2, 10 (ii) Important Deleted for CBSE Board 2022 Exams

Ex 4.2, 11 (i) Deleted for CBSE Board 2022 Exams

Ex 4.2, 11 (ii) Important Deleted for CBSE Board 2022 Exams

Ex 4.2, 12 Important Deleted for CBSE Board 2022 Exams

Ex 4.2, 13 Important Deleted for CBSE Board 2022 Exams

Ex 4.2, 14 Important Deleted for CBSE Board 2022 Exams

Ex 4.2, 15 (MCQ) Important Deleted for CBSE Board 2022 Exams

Ex 4.2, 16 (MCQ) Deleted for CBSE Board 2022 Exams

Chapter 4 Class 12 Determinants (Term 1)

Serial order wise

Last updated at Aug. 18, 2021 by Teachoo

Hello! Teachoo has made this answer with days (even weeks!) worth of effort and love. Since your board exams are coming, why not help Teachoo create more videos and content by supporting us? Please click on this link to make a donation

Hello! Teachoo has made this answer with days (even weeks!) worth of effort and love. Since your board exams are coming, why not help Teachoo create more videos and content by supporting us? Please click on this link to make a donation

Ex 4.2, 10 By using properties of determinants, show that: (i) x+4 2x 2x 2x x+4 2x 2x 2 x+4 = (5x + 4) (4 x)2 Taking L.H.S x+4 2x 2x 2x x+4 2x 2x 2 x+4 Applying R1 R1 + R2 + R2 = x+4+2 +2 2x+x+4+2x 2x+2x+x+4 2x x+4 2x 2x 2 x+4 = + + + 2x x+4 2x 2x 2 x+4 Taking out (5x + 4) common from R1 = (5x + 4) 1 1 1 2x x+4 2x 2x 2 x+4 Applying C1 C1 C2 = (5x + 4) 1 1 1 1 2x x 4 x+4 2x 2x 2x 2 x+4 = (5x + 4) 1 1 x 4 x+4 2x 0 2 x+4 Applying C2 C2 C3 = (5x + 4) 0 1 x 4 x+4 2x 2x 0 2 4 x+4 = (5x + 4) 0 1 x 4 ( 4) 2x 0( 4) ( 4) x+4 Taking common (x 4) from C1 & C2 = (5x + 4) (x 4) (x 4) 0 0 1 1 1 2x 0 1 x+4 Expanding Determinant along R1 = (5x + 4) (x 4) (x 4) 0 1 2 1 +4 0 1 2 0 +4 +1 1 1 0 1 = (5x 4) (x 4)2 (0 0 + (1 0)) = (5x 4) (x 4)2 (1) = (5x 4) (x 4)2 = R.H.S Hence Proved