Examples

Chapter 4 Class 12 Determinants
Serial order wise

Get live Maths 1-on-1 Classs - Class 6 to 12

### Transcript

Example 32 Show that Ξ = |β 8((π¦+π§)2&π₯π¦&π§π₯@π₯π¦&(π₯+π§)2&π¦π§@π₯π§&π¦π§&(π₯+π¦)2)| = 2xyz (x + y + z)3 Solving L.H.S Ξ = |β 8((π¦+π§)^2&π₯π¦&π§π₯@π₯π¦&(π₯+π§)2&π¦π§@π₯π§&π¦π§&(π₯+π¦)2)| Divide & Multiply by xyz = π₯π¦π§/π₯π¦π§ |β 8((π¦+π§)2&π₯π¦&π§π₯@π₯π¦&(π₯+π§)2&π¦π§@π₯π§&π¦π§&(π₯+π¦)2)| = 1/π₯π¦π§ x. y. z |β 8((π¦+π§)2&π₯π¦&π§π₯@π₯π¦&(π₯+π§)2&π¦π§@π₯π§&π¦π§&(π₯+π¦)2)| Multiplying R1 by x , R2 by y & R3 by z = 1/π₯π¦π§ |β 8(π(π¦+π§)2&π(π₯π¦)&π(π§π₯)@π(π₯π¦)&π(π₯+π§)2&π(π¦π§)@π(π₯π§)&π¦π2&π(π₯+π¦)2)| Taking out x common from C1, y common from C2 & z common from C3 = π₯π¦π§/π₯π¦π§ |β 8((π¦+π§)2&π₯2&π₯[email protected]π¦2&(π₯+π§)2&π¦[email protected]π§2&π§2&(π₯+π¦)2)| Applying C2 β C2 β C1 = |β 8((π¦+π§)2&π₯2β(π¦+π§)2&π₯[email protected]π¦2&(π₯+π§)2βπ¦2&π¦[email protected]π§2&π§2βπ§2&(π₯+π¦)2)| = |β 8((π¦+π§)2&(π₯β(π¦+π§))(π₯+(π¦+π§))&π₯[email protected]π¦2&((π₯+π§)βπ¦)(π₯+π§+π¦)&π¦[email protected]π§2&0&(π₯+π¦)2)| = |β 8((π¦+π§)2&(π₯βπ¦βπ§)(π+π+π)&π₯[email protected]π¦2&(π₯+π§βπ¦)(π+π+π)&π¦[email protected]π§2&0&(π₯+π¦)2)| Taking out (π+π+π) common from C2 = (π₯+π¦+π§)|β 8((y+z)2&π₯βπ¦βπ§&[email protected]&π₯+π§βπ¦&[email protected]π§2&0&(x+y)2)| Applying C3 β C3 β C1 = (π₯+π¦+π§)|β 8((y+z)2&π₯βπ¦βπ§&π₯2 β(π¦+π§)[email protected]&π₯+π§βπ¦&π¦2βπ¦[email protected]π§2&0&(π₯+π¦)2βπ§2)| = (π₯+π¦+π§)|β 8((y+z)2&π₯βπ¦βπ§&(π+π+π)(π₯β(π¦+π§))@y2&π₯+π§βπ¦&[email protected]π§2&0&(π+π+π)((π₯+π¦)βπ§))| Taking out (π+π+π) Common from C3 = (π₯+π¦+π§)(π₯+π¦+π§)|β 8((y+z)2&π₯βπ¦βπ§&π₯βπ¦βπ§@y2&π₯+π§βπ¦&[email protected]π§2&0&π₯+π¦βπ§)| = (π₯+π¦+π§)2 |β 8((y+z)2&π₯βπ¦βπ§&π₯βπ¦βπ§@y2&π₯+π§βπ¦&[email protected]π§2&0&π₯+π¦βπ§)| Applying R1β R1 β R2 β R3 = (π₯+π¦+π§)2|β 8((y+z)2βπ¦2βπ§2&(π₯βπ¦βπ§)β(π₯+π§βπ¦)β0&(π₯βπ¦βπ§)β0β(π₯+π¦βπ§)@y2&π₯+π§βπ¦&[email protected]π§2&0&π₯+π¦βπ§)| = (π₯+π¦+π§)2|β 8(π¦2+π§2+2π¦π§βπ¦2βπ§2&π₯βπ₯βπ¦+π¦βπ§βπ§&π₯βπ₯βπ¦βπ¦βπ§+π§@y2&π₯+π§βπ¦&[email protected]π§2&0&π₯+π¦βπ§)| = (π₯+π¦+π§)2|β 8(2π¦π§&β2π§&β2π¦@y2&π₯+π§βπ¦&[email protected]π§2&0&π₯+π¦βπ§)| Applying C2β C2 + π/π C1 = (π₯+π§+π¦)2|β 8(2π¦π§&β2π§+π/π(πππ)&2π¦@y2&xβπ¦+π§+π/π (ππ)&[email protected]π§2&0+π/π(ππ)&π₯+π¦βπ§)| = (π₯+π§+π¦)2|β 8(2π¦π§&0&2π¦@y2&x+π§&[email protected]π§2&π§^2/π¦&π₯+π¦βπ§)| Applying C3β C3 + π/π C1 = (π₯+π¦+π§)2|β 8(2π¦π§&0&β2π¦+π/π(πππ)@y2&π₯+π§&0+π/π (ππ)@π§2&π§^2/π¦&(π₯+π¦βπ§)+π/π (ππ))| = (π₯+π¦+π§)2|β 8(2π¦π§&0&[email protected]&π₯+π§&π¦^2/π§ @π§2&π§^2/π¦&π₯+π¦)| Expanding Determinant along R1 = (π₯+π¦+π§)2(2π¦π§|β 8(π₯+π§&π¦^2/π§@π§^2/π¦&π₯+π¦)|β0|β 8(π¦2&π¦^2/π§@π§^2&π₯+π¦)|+0|β 8(π¦2&π₯+π¦@π§^2&π§^2/π¦)|) = (π₯+π¦+π§)2(2π¦π§|β 8(π₯+π§&π¦^2/π§@π§^2/π¦&π₯+π¦)|β0+0) = (π₯+π¦+π§)2 ("2yz " ("(x + z) (x + y) β " π§2/π¦ " " (π¦2/π§))" β 0 + 0" ) = (π₯+π¦+π§)2 (2yz ((x + z) (x + y) β zy ) = (π₯+π¦+π§)2 (2yz) ((x + z) (x + y) β zy ) = (π₯+π¦+π§)2 (2yz) (x2 + xy + zx + zy β zy) = (π₯+π¦+π§)2 (2yz) (x2 + xy + xz) = (π₯+π¦+π§)2 (2yz) . x (x + y + z) = (π₯+π¦+π§)3 (2xyz) = (2xyz) (π₯+π¦+π§)^3 = R.H.S Hence Proved