Example 32 - Show that Determinant = 2xyz (x + y + z)^3 - Class 12

Example 32 - Chapter 4 Class 12 Determinants - Part 2
Example 32 - Chapter 4 Class 12 Determinants - Part 3 Example 32 - Chapter 4 Class 12 Determinants - Part 4 Example 32 - Chapter 4 Class 12 Determinants - Part 5 Example 32 - Chapter 4 Class 12 Determinants - Part 6 Example 32 - Chapter 4 Class 12 Determinants - Part 7

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Question 14 Show that Δ = |■8((𝑦+𝑧)2&𝑥𝑦&𝑧𝑥@𝑥𝑦&(𝑥+𝑧)2&𝑦𝑧@𝑥𝑧&𝑦𝑧&(𝑥+𝑦)2)| = 2xyz (x + y + z)3 Solving L.H.S Δ = |■8((𝑦+𝑧)^2&𝑥𝑦&𝑧𝑥@𝑥𝑦&(𝑥+𝑧)2&𝑦𝑧@𝑥𝑧&𝑦𝑧&(𝑥+𝑦)2)| Divide & Multiply by xyz = 𝑥𝑦𝑧/𝑥𝑦𝑧 |■8((𝑦+𝑧)2&𝑥𝑦&𝑧𝑥@𝑥𝑦&(𝑥+𝑧)2&𝑦𝑧@𝑥𝑧&𝑦𝑧&(𝑥+𝑦)2)| = 1/𝑥𝑦𝑧 x. y. z |■8((𝑦+𝑧)2&𝑥𝑦&𝑧𝑥@𝑥𝑦&(𝑥+𝑧)2&𝑦𝑧@𝑥𝑧&𝑦𝑧&(𝑥+𝑦)2)| Multiplying R1 by x , R2 by y & R3 by z = 1/𝑥𝑦𝑧 |■8(𝒙(𝑦+𝑧)2&𝒙(𝑥𝑦)&𝒙(𝑧𝑥)@𝒚(𝑥𝑦)&𝒚(𝑥+𝑧)2&𝒚(𝑦𝑧)@𝒛(𝑥𝑧)&𝑦𝒛2&𝒛(𝑥+𝑦)2)| Taking out x common from C1, y common from C2 & z common from C3 = 𝑥𝑦𝑧/𝑥𝑦𝑧 |■8((𝑦+𝑧)2&𝑥2&𝑥2@𝑦2&(𝑥+𝑧)2&𝑦2@𝑧2&𝑧2&(𝑥+𝑦)2)| Applying C2 → C2 – C1 = |■8((𝑦+𝑧)2&𝑥2−(𝑦+𝑧)2&𝑥2@𝑦2&(𝑥+𝑧)2−𝑦2&𝑦2@𝑧2&𝑧2−𝑧2&(𝑥+𝑦)2)| = |■8((𝑦+𝑧)2&(𝑥−(𝑦+𝑧))(𝑥+(𝑦+𝑧))&𝑥2@𝑦2&((𝑥+𝑧)−𝑦)(𝑥+𝑧+𝑦)&𝑦2@𝑧2&0&(𝑥+𝑦)2)| = |■8((𝑦+𝑧)2&(𝑥−𝑦−𝑧)(𝒙+𝒚+𝒛)&𝑥2@𝑦2&(𝑥+𝑧−𝑦)(𝒙+𝒚+𝒛)&𝑦2@𝑧2&0&(𝑥+𝑦)2)| Taking out (𝒙+𝒚+𝒛) common from C2 = (𝑥+𝑦+𝑧)|■8((y+z)2&𝑥−𝑦−𝑧&x2@y2&𝑥+𝑧−𝑦&y2@𝑧2&0&(x+y)2)| Applying C3 → C3 – C1 = (𝑥+𝑦+𝑧)|■8((y+z)2&𝑥−𝑦−𝑧&𝑥2 −(𝑦+𝑧)2@y2&𝑥+𝑧−𝑦&𝑦2−𝑦2@𝑧2&0&(𝑥+𝑦)2−𝑧2)| = (𝑥+𝑦+𝑧)|■8((y+z)2&𝑥−𝑦−𝑧&(𝒙+𝒚+𝒛)(𝑥−(𝑦+𝑧))@y2&𝑥+𝑧−𝑦&0@𝑧2&0&(𝒙+𝒚+𝒛)((𝑥+𝑦)−𝑧))| Taking out (𝒙+𝒚+𝒛) Common from C3 = (𝑥+𝑦+𝑧)(𝑥+𝑦+𝑧)|■8((y+z)2&𝑥−𝑦−𝑧&𝑥−𝑦−𝑧@y2&𝑥+𝑧−𝑦&0@𝑧2&0&𝑥+𝑦−𝑧)| = (𝑥+𝑦+𝑧)2 |■8((y+z)2&𝑥−𝑦−𝑧&𝑥−𝑦−𝑧@y2&𝑥+𝑧−𝑦&0@𝑧2&0&𝑥+𝑦−𝑧)| Applying R1→ R1 – R2 – R3 = (𝑥+𝑦+𝑧)2|■8((y+z)2−𝑦2−𝑧2&(𝑥−𝑦−𝑧)−(𝑥+𝑧−𝑦)−0&(𝑥−𝑦−𝑧)−0−(𝑥+𝑦−𝑧)@y2&𝑥+𝑧−𝑦&0@𝑧2&0&𝑥+𝑦−𝑧)| = (𝑥+𝑦+𝑧)2|■8(𝑦2+𝑧2+2𝑦𝑧−𝑦2−𝑧2&𝑥−𝑥−𝑦+𝑦−𝑧−𝑧&𝑥−𝑥−𝑦−𝑦−𝑧+𝑧@y2&𝑥+𝑧−𝑦&0@𝑧2&0&𝑥+𝑦−𝑧)| = (𝑥+𝑦+𝑧)2|■8(2𝑦𝑧&−2𝑧&−2𝑦@y2&𝑥+𝑧−𝑦&0@𝑧2&0&𝑥+𝑦−𝑧)| Applying C2→ C2 + 𝟏/𝒚 C1 = (𝑥+𝑧+𝑦)2|■8(2𝑦𝑧&−2𝑧+𝟏/𝒚(𝟐𝒚𝒛)&2𝑦@y2&x−𝑦+𝑧+𝟏/𝒚 (𝒚𝟐)&0@𝑧2&0+𝟏/𝒚(𝒛𝟐)&𝑥+𝑦−𝑧)| = (𝑥+𝑧+𝑦)2|■8(2𝑦𝑧&0&2𝑦@y2&x+𝑧&0@𝑧2&𝑧^2/𝑦&𝑥+𝑦−𝑧)| Applying C3→ C3 + 𝟏/𝒛 C1 = (𝑥+𝑦+𝑧)2|■8(2𝑦𝑧&0&−2𝑦+𝟏/𝒛(𝟐𝒚𝒛)@y2&𝑥+𝑧&0+𝟏/𝒛 (𝒚𝟐)@𝑧2&𝑧^2/𝑦&(𝑥+𝑦−𝑧)+𝟏/𝒛 (𝒛𝟐))| = (𝑥+𝑦+𝑧)2|■8(2𝑦𝑧&0&0@y2&𝑥+𝑧&𝑦^2/𝑧 @𝑧2&𝑧^2/𝑦&𝑥+𝑦)| Expanding Determinant along R1 = (𝑥+𝑦+𝑧)2(2𝑦𝑧|■8(𝑥+𝑧&𝑦^2/𝑧@𝑧^2/𝑦&𝑥+𝑦)|−0|■8(𝑦2&𝑦^2/𝑧@𝑧^2&𝑥+𝑦)|+0|■8(𝑦2&𝑥+𝑦@𝑧^2&𝑧^2/𝑦)|) = (𝑥+𝑦+𝑧)2(2𝑦𝑧|■8(𝑥+𝑧&𝑦^2/𝑧@𝑧^2/𝑦&𝑥+𝑦)|−0+0) = (𝑥+𝑦+𝑧)2 ("2yz " ("(x + z) (x + y) – " 𝑧2/𝑦 " " (𝑦2/𝑧))" – 0 + 0" ) = (𝑥+𝑦+𝑧)2 (2yz ((x + z) (x + y) – zy ) = (𝑥+𝑦+𝑧)2 (2yz) ((x + z) (x + y) – zy ) = (𝑥+𝑦+𝑧)2 (2yz) (x2 + xy + zx + zy – zy) = (𝑥+𝑦+𝑧)2 (2yz) (x2 + xy + xz) = (𝑥+𝑦+𝑧)2 (2yz) . x (x + y + z) = (𝑥+𝑦+𝑧)3 (2xyz) = (2xyz) (𝑥+𝑦+𝑧)^3 = R.H.S Hence Proved

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