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Example 6 - Chapter 4 Class 12 Determinants (Term 1)

Last updated at March 16, 2023 by

Example 6 - Verify Property 1 for |2 -3 5 6 0 4 1 5 -7| - Verifying properties of a determinant

Example 6 - Chapter 4 Class 12 Determinants - Part 2
Example 6 - Chapter 4 Class 12 Determinants - Part 3

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Transcript

Example 6 Verify Property 1 for = 2 3 5 6 0 4 1 5 7 Expanding the determinant along first row, = 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 By interchanging rows and columns , we get 1 = 2 6 1 3 0 5 5 4 7 1 = 2 0 5 4 7 6 3 5 5 7 + 1 3 0 5 4 = 2(0( 7) 4(5) 6( 3 ( 7) 5 (5)) +1 ( 3 (4) 5 (0)) = 2 (0 20) 6 (+ 21 25 ) +1 ( 12 + 0) = 2 ( 20) 6 ( 4) + 1 ( 12) = 40 + 24 12 = 52 + 24 = 28 Hence 1 = Hence 1 = Property 1 is verified. i.e. If we interchange row & Columns, value of determinant remains unchanged

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.