Question 1 - Verifying properties of a determinant - Chapter 4 Class 12 Determinants

Last updated at April 16, 2024 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

Transcript

Question 1 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

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

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, Social Science, Physics, Chemistry, Computer Science at Teachoo.