Question 3 Using the property of determinants and without expanding, prove that: |■8(2&7&65@3&8&75@5&9&86)| = 0 |■8(2&7&65@3&8&75@5&9&86)| Applying C3 → C3 − C1 = |■8(2&7&𝟔𝟓−𝟐@3&8&𝟕𝟓−𝟑@5&9&𝟖𝟔−𝟓)| = |■8(2&7&63@3&8&72@5&9&81)| Rough 65 – 2 = 63, 63/7 = 9 75 – 3 = 72, 72/8 = 9 86 – 5 = 81, 81/9 = 9 = |■8(2&7&𝟗 × 7@3&8&𝟗 ×8@5&9&𝟗 × 9)| Taking out 9 common from C3 = 9 |■8(2&𝟕&𝟕@3&𝟖&𝟖@5&𝟗&𝟗)| Here, C2 and C3 are identical = 9 × 0 = 0 Thus, |■8(2&7&65@3&8&75@5&9&86)| = 0 Hence proved Using Property: If any two row or column are identical, then value of determinant is zero

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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