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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

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Transcript

Example 32 Show that Ξ” = |β– 8((𝑦+𝑧)2&π‘₯𝑦&𝑧π‘₯@π‘₯𝑦&(π‘₯+𝑧)2&𝑦𝑧@π‘₯𝑧&𝑦𝑧&(π‘₯+𝑦)2)| = 2xyz (x + y + z)3 Solving L.H.S Ξ” = |β– 8((𝑦+𝑧)^2&π‘₯𝑦&𝑧π‘₯@π‘₯𝑦&(π‘₯+𝑧)2&𝑦𝑧@π‘₯𝑧&𝑦𝑧&(π‘₯+𝑦)2)| Divide & Multiply by xyz = π‘₯𝑦𝑧/π‘₯𝑦𝑧 |β– 8((𝑦+𝑧)2&π‘₯𝑦&𝑧π‘₯@π‘₯𝑦&(π‘₯+𝑧)2&𝑦𝑧@π‘₯𝑧&𝑦𝑧&(π‘₯+𝑦)2)| = 1/π‘₯𝑦𝑧 x. y. z |β– 8((𝑦+𝑧)2&π‘₯𝑦&𝑧π‘₯@π‘₯𝑦&(π‘₯+𝑧)2&𝑦𝑧@π‘₯𝑧&𝑦𝑧&(π‘₯+𝑦)2)| Multiplying R1 by x , R2 by y & R3 by z = 1/π‘₯𝑦𝑧 |β– 8(𝒙(𝑦+𝑧)2&𝒙(π‘₯𝑦)&𝒙(𝑧π‘₯)@π’š(π‘₯𝑦)&π’š(π‘₯+𝑧)2&π’š(𝑦𝑧)@𝒛(π‘₯𝑧)&𝑦𝒛2&𝒛(π‘₯+𝑦)2)| Taking out x common from C1, y common from C2 & z common from C3 = π‘₯𝑦𝑧/π‘₯𝑦𝑧 |β– 8((𝑦+𝑧)2&π‘₯2&π‘₯2@𝑦2&(π‘₯+𝑧)2&𝑦2@𝑧2&𝑧2&(π‘₯+𝑦)2)| Applying C2 β†’ C2 – C1 = |β– 8((𝑦+𝑧)2&π‘₯2βˆ’(𝑦+𝑧)2&π‘₯2@𝑦2&(π‘₯+𝑧)2βˆ’π‘¦2&𝑦2@𝑧2&𝑧2βˆ’π‘§2&(π‘₯+𝑦)2)| = |β– 8((𝑦+𝑧)2&(π‘₯βˆ’(𝑦+𝑧))(π‘₯+(𝑦+𝑧))&π‘₯2@𝑦2&((π‘₯+𝑧)βˆ’π‘¦)(π‘₯+𝑧+𝑦)&𝑦2@𝑧2&0&(π‘₯+𝑦)2)| = |β– 8((𝑦+𝑧)2&(π‘₯βˆ’π‘¦βˆ’π‘§)(𝒙+π’š+𝒛)&π‘₯2@𝑦2&(π‘₯+π‘§βˆ’π‘¦)(𝒙+π’š+𝒛)&𝑦2@𝑧2&0&(π‘₯+𝑦)2)| Taking out (𝒙+π’š+𝒛) common from C2 = (π‘₯+𝑦+𝑧)|β– 8((y+z)2&π‘₯βˆ’π‘¦βˆ’π‘§&x2@y2&π‘₯+π‘§βˆ’π‘¦&y2@𝑧2&0&(x+y)2)| Applying C3 β†’ C3 – C1 = (π‘₯+𝑦+𝑧)|β– 8((y+z)2&π‘₯βˆ’π‘¦βˆ’π‘§&π‘₯2 βˆ’(𝑦+𝑧)2@y2&π‘₯+π‘§βˆ’π‘¦&𝑦2βˆ’π‘¦2@𝑧2&0&(π‘₯+𝑦)2βˆ’π‘§2)| = (π‘₯+𝑦+𝑧)|β– 8((y+z)2&π‘₯βˆ’π‘¦βˆ’π‘§&(𝒙+π’š+𝒛)(π‘₯βˆ’(𝑦+𝑧))@y2&π‘₯+π‘§βˆ’π‘¦&0@𝑧2&0&(𝒙+π’š+𝒛)((π‘₯+𝑦)βˆ’π‘§))| Taking out (𝒙+π’š+𝒛) Common from C3 = (π‘₯+𝑦+𝑧)(π‘₯+𝑦+𝑧)|β– 8((y+z)2&π‘₯βˆ’π‘¦βˆ’π‘§&π‘₯βˆ’π‘¦βˆ’π‘§@y2&π‘₯+π‘§βˆ’π‘¦&0@𝑧2&0&π‘₯+π‘¦βˆ’π‘§)| = (π‘₯+𝑦+𝑧)2 |β– 8((y+z)2&π‘₯βˆ’π‘¦βˆ’π‘§&π‘₯βˆ’π‘¦βˆ’π‘§@y2&π‘₯+π‘§βˆ’π‘¦&0@𝑧2&0&π‘₯+π‘¦βˆ’π‘§)| Applying R1β†’ R1 – R2 – R3 = (π‘₯+𝑦+𝑧)2|β– 8((y+z)2βˆ’π‘¦2βˆ’π‘§2&(π‘₯βˆ’π‘¦βˆ’π‘§)βˆ’(π‘₯+π‘§βˆ’π‘¦)βˆ’0&(π‘₯βˆ’π‘¦βˆ’π‘§)βˆ’0βˆ’(π‘₯+π‘¦βˆ’π‘§)@y2&π‘₯+π‘§βˆ’π‘¦&0@𝑧2&0&π‘₯+π‘¦βˆ’π‘§)| = (π‘₯+𝑦+𝑧)2|β– 8(𝑦2+𝑧2+2π‘¦π‘§βˆ’π‘¦2βˆ’π‘§2&π‘₯βˆ’π‘₯βˆ’π‘¦+π‘¦βˆ’π‘§βˆ’π‘§&π‘₯βˆ’π‘₯βˆ’π‘¦βˆ’π‘¦βˆ’π‘§+𝑧@y2&π‘₯+π‘§βˆ’π‘¦&0@𝑧2&0&π‘₯+π‘¦βˆ’π‘§)| = (π‘₯+𝑦+𝑧)2|β– 8(2𝑦𝑧&βˆ’2𝑧&βˆ’2𝑦@y2&π‘₯+π‘§βˆ’π‘¦&0@𝑧2&0&π‘₯+π‘¦βˆ’π‘§)| Applying C2β†’ C2 + 𝟏/π’š C1 = (π‘₯+𝑧+𝑦)2|β– 8(2𝑦𝑧&βˆ’2𝑧+𝟏/π’š(πŸπ’šπ’›)&2𝑦@y2&xβˆ’π‘¦+𝑧+𝟏/π’š (π’šπŸ)&0@𝑧2&0+𝟏/π’š(π’›πŸ)&π‘₯+π‘¦βˆ’π‘§)| = (π‘₯+𝑧+𝑦)2|β– 8(2𝑦𝑧&0&2𝑦@y2&x+𝑧&0@𝑧2&𝑧^2/𝑦&π‘₯+π‘¦βˆ’π‘§)| Applying C3β†’ C3 + 𝟏/𝒛 C1 = (π‘₯+𝑦+𝑧)2|β– 8(2𝑦𝑧&0&βˆ’2𝑦+𝟏/𝒛(πŸπ’šπ’›)@y2&π‘₯+𝑧&0+𝟏/𝒛 (π’šπŸ)@𝑧2&𝑧^2/𝑦&(π‘₯+π‘¦βˆ’π‘§)+𝟏/𝒛 (π’›πŸ))| = (π‘₯+𝑦+𝑧)2|β– 8(2𝑦𝑧&0&0@y2&π‘₯+𝑧&𝑦^2/𝑧 @𝑧2&𝑧^2/𝑦&π‘₯+𝑦)| Expanding Determinant along R1 = (π‘₯+𝑦+𝑧)2(2𝑦𝑧|β– 8(π‘₯+𝑧&𝑦^2/𝑧@𝑧^2/𝑦&π‘₯+𝑦)|βˆ’0|β– 8(𝑦2&𝑦^2/𝑧@𝑧^2&π‘₯+𝑦)|+0|β– 8(𝑦2&π‘₯+𝑦@𝑧^2&𝑧^2/𝑦)|) = (π‘₯+𝑦+𝑧)2(2𝑦𝑧|β– 8(π‘₯+𝑧&𝑦^2/𝑧@𝑧^2/𝑦&π‘₯+𝑦)|βˆ’0+0) = (π‘₯+𝑦+𝑧)2 ("2yz " ("(x + z) (x + y) – " 𝑧2/𝑦 " " (𝑦2/𝑧))" – 0 + 0" ) = (π‘₯+𝑦+𝑧)2 (2yz ((x + z) (x + y) – zy ) = (π‘₯+𝑦+𝑧)2 (2yz) ((x + z) (x + y) – zy ) = (π‘₯+𝑦+𝑧)2 (2yz) (x2 + xy + zx + zy – zy) = (π‘₯+𝑦+𝑧)2 (2yz) (x2 + xy + xz) = (π‘₯+𝑦+𝑧)2 (2yz) . x (x + y + z) = (π‘₯+𝑦+𝑧)3 (2xyz) = (2xyz) (π‘₯+𝑦+𝑧)^3 = R.H.S 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.