Question 2 - Examples - Chapter 4 Class 12 Determinants

Last updated at April 16, 2024 by

Example 7 - Verify Property 2 for |2 -3 5 6 0 4 1 5 -7| - Examples

Example 7 - Chapter 4 Class 12 Determinants - Part 2
Example 7 - Chapter 4 Class 12 Determinants - Part 3

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Question 2, Verify Property 2 for = 2 3 5 6 0 4 1 5 7 = 2 3 5 6 0 4 1 5 7 = 2 0 4 5 7 - (-3) 6 4 1 7 + 5 6 0 1 5 = 2(0( 7) 5(4) + 3 (6( 7) (1) (4)) +5 (6(5) 1 (0)) = 2(0 20) + 3 (-42 4) + 5 (30 0) = 2 ( 20) +3 ( 46) + 5 (30) = 40 138 + 150 = 28 Interchanging R2 and R3 i.e. R2 R3, We get 1 = 2 3 5 1 5 7 6 0 4 = 2 5 7 0 4 ( 3) 1 7 6 4 + 5 1 5 6 0 = 2(5(4) 0( 7) + 3 (1(4) 6( 7)) + 5(1 (0) 6(5)) = 2 (20 0) + 3 ( 4 + 42) + 5 ( 0 30) = 2 (20) + 3 (46) + 5 ( 30) = 40 + 138 150 = 28 1 = 28 = (-28) = Therefore 1= Hence property 2nd is verified i.e. If any two rows ( or columns) of a determinant are interchanged , then sign of determinant changes.

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo