Ex 4.2, 10 - Chapter 4 Class 12 Determinants
Last updated at May 29, 2018 by Teachoo
Last updated at May 29, 2018 by Teachoo
Transcript
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 Ex 4.2 , 10 By using properties of determinants, show that: (ii) y+ y y y y+ y y y+ = k3 (3y + k) Taking L.H.S y+ y y y y+ y y y+ Applying R1 R1 + R2 + R3 = y+ + + y+y+k+y y+y+y+k y y+ y y y+ = + + + y y+ y y y+ Taking out (3y + k) common from R1 = (3y + k) 1 1 1 y y+ y y y+ Applying C1 C1 C2 = (3y + k) 1 1 y+ y y+ = (3y + k) 1 1 y+k y 0 y+ Applying C2 C2 C3 = (3y + k) 0 1 0 y+k y 0 y+ = (3y + k) 0 1 y 0 y+ Expanding Determinant along R1 = (3y + k) 0 + 0 0 + +1 0 = (3y + k) (0 0 + 1 (k2 0)) = (3y + k) (k2) = k2 (3y + k) = R.H.S Hence Proved
Making whole row/column one and simplifying
Ex 4.2, 11 Important
Ex 4.2, 13 Important
Misc 4
Misc 5
Ex 4.2, 12 Important
Misc 9
Misc 11 Important Deleted for CBSE Board 2021 Exams only
Misc. 13 Deleted for CBSE Board 2021 Exams only
Example 15 Important Deleted for CBSE Board 2021 Exams only
Misc 12 Important Deleted for CBSE Board 2021 Exams only
Example 30
Example 16 Important Deleted for CBSE Board 2021 Exams only
Making whole row/column one and simplifying
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