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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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Transcript

Example 7, 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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.