Q 33 - Teachoo.jpg
Using the properties of determinants, prove that

| (y + z) 2   x 2   x 2   y 2 (z + x) 2 y 2   z 2   z 2 (x + y) 2 | = 2xyz (x + y + z) 3

 

Question 33 (OR 1st Question) - CBSE Class 12 Sample Paper for 2020 Boards - Part 2

Question 33 (OR 1st Question) - CBSE Class 12 Sample Paper for 2020 Boards - Part 3
Question 33 (OR 1st Question) - CBSE Class 12 Sample Paper for 2020 Boards - Part 4
Question 33 (OR 1st Question) - CBSE Class 12 Sample Paper for 2020 Boards - Part 5 Question 33 (OR 1st Question) - CBSE Class 12 Sample Paper for 2020 Boards - Part 6 Question 33 (OR 1st Question) - CBSE Class 12 Sample Paper for 2020 Boards - Part 7 Question 33 (OR 1st Question) - CBSE Class 12 Sample Paper for 2020 Boards - Part 8 Question 33 (OR 1st Question) - CBSE Class 12 Sample Paper for 2020 Boards - Part 9

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Question 33 (OR 1st Question) Using the properties of determinants, prove that [ (y + z)2 x2 x2 y2 (z + x)2 y2 z2 z2 (x + y)2 ] 2xyz (x + y + z)3 [ (y + z)2 x2 x2 y2 (z + x)2 y2 z2 z2 (x + y)2 ] Applying C2 -> C2 - C1 = | (y + z)2 x2 - (y + z)2 x2 y2 (z + x)2 - y2 y2 z2 z2 - z2 (x + y)2 = | y + z |2 (x - (y + z)) (x + (y + z)) x2 y2 ((z + x) - y ((z + x ) + y y2 z2 0 (x + y)2 = | (y + z)2 (x - y - z) (x + y + z) x2 y2 (x + z - y) (x + y + z) y2 z2 0 (x + y)2 Applying C3-> C3 - C1 | (y + z)2 (x - y - z) (x + y + z) x2 - (y + z)2 y2 (x + z - y) (x + y + z) y2 - y2 z2 0 (x + y)2 - z2 = (y + z)2 (x - y - z) (x + y + z) (x - (y + z) (x + (y + z) y2 (x + z - y) (x + y + z) 0 z2 0 ((x + y) - z) ((x + y) + z) = | (y + z)2 (x - y - z) (x + y + z) (x - y - z) (x + y + z) y2 (x + z - y) (x + y + z) 0 z2 0 (x + y - z) (x + y + z) Taking (x + y + z) common from both c2 and c3 = (x + y + z)2 | (y + z)2 (x - y - z) (x - y - z) y2 (x + z - y) 0 z2 0 (x + y - z) Applying R1 -> R1 - R2 = (x + y + z)2 | (y + z)2 - y2 (x - y - z) - (x + z - y) (x - y - z) - 0 y2 (x + z - y) 0 z2 0 (x + y - z) = (x + y + z)2 | (y + z) - y (y + z) + y) -2z (x - y - z) y2 (x + z - y) 0 z2 0 (x + y - z) = (x + y + z)2 | z (2y + z) - 2z (x - y - z) y2 (x + z - y) 0 z2 0 (x + y - z) = (x + y + z)2 | 2yz2 - z2 (x - y - z) y2 (x + z - y) 0 z2 0 (x + y - z) Applying R1 -> R1 - R3 = (x + y + z)2 | 2yz +z2 - Z2 -2z - 0 (x - y - z) - (x + y - z) y2 (x + z - y) 0 z2 0 (x + y - z) = (x + y + z)2 |2yz -2z -2y y2 (x + z - y) 0 z2 0 (x + y - z) Applying C2 -> C2 + 1/y + C1 = (x + y + z) | 2yz -2z + 1/y + 2yz -2y y2 (x + z - y) + 1/y x y2 0 z2 0 + 1/y + z2 (x + y - z) = (x + y + z) | 2yz -2z + 2z -2y y2 (x + z - y) + y 0 z2 z2/y (x + y - z) | Applying C3 -> C3 + 1/z + C1 = (x + y + z )2 | 2yz 0 -2y + 1/z + 2yz y2 (x + z) 0 + 1/z x y2 z2 z2/y (x + y - z) + 1/z x z2 = (x + y + z)2 | 2yz 0 -2y + 2y y2 (x + z) y2/z z2 z2/y (x + y - z) + z = (x + y + z)2 |2yz 0 0 y2 (x + z) y2/z z2 z2/y (x + y) | Expanding Determinant along R1 = (𝑥 + 𝑦 + 𝑧)^2 × [𝟐𝒚𝒛 ((𝒙 + 𝒛)(𝒙 + 𝒚) − 𝒛^𝟐/𝒚 × 𝒚^𝟐/𝒛) − 𝟎 − 𝟎] = (𝑥 + 𝑦 + 𝑧)^2 × 2𝑦𝑧 ((𝑥 + 𝑧)(𝑥 + 𝑦) − 𝑧𝑦) = (𝑥 + 𝑦 + 𝑧)^2 × 2𝑦𝑧 (𝑥(𝑥 + 𝑦) + 𝑧(𝑥 + 𝑦) − 𝑧𝑦) = (𝑥 + 𝑦 + 𝑧)^2 × 2𝑦𝑧 (𝑥^2 + 𝑥𝑦 + 𝑧𝑥 + 𝑧𝑦 − 𝑧𝑦) = (𝑥 + 𝑦 + 𝑧)^2 × 2𝑦𝑧 (𝑥^2 + 𝑥𝑦 + 𝑧𝑥) = (𝑥 + 𝑦 + 𝑧)^2 × 2𝑥𝑦𝑧 × (𝑥 + 𝑦 + 𝑧) = 2𝑥𝑦𝑧(𝑥 + 𝑦 + 𝑧)^3 Hence proved

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo