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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


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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.