InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
if Delta (x)=|{:(a_(1)+x,,b_(1)+x,,c_(1)+x),(a_(2)+x,,b_(2)+x,,c_(2)+x),(a_(3)+x,,b_(3)+x,,c_(3)+x):}| then show thatDelta (x)=0 andthat Delta (x)=Delta(0)+sx. where sdenotesthe sum of all thecofactors of allthe elements in Delta (0) |
|
Answer» Solution :`Delta (x)=|{:(a_(1)+x,,b_(1)+x,,c_(1)+x),(a_(2)+x,,b_(2)+x,,c_(2)+x),(a_(3)+x,,b_(3)+x,,c_(3)+x):}|` ` :. Delta (x) =|{:(a_(1)+x,,b_(1)+x,,c_(1)+x),(a_(2)+x,,b_(2)+x,,c_(2)+x),(a_(3)+x,,b_(3)+x,,c_(3)+x):}|+|{:(a_(1)+x,,1,,c_(1)+x),(a_(2)+x,,1,,c_(2)+x),(a_(3)+x,,1,,c_(3)+x):}|` `+|{:(a_(1)+x,,b_(1)+x,,1),(a_(2)+x,,b_(2)+x,,1),(a_(3)+x,,b_(3)+x,,1):}|` APPLYING`C_(2) toC_(2)-xC_(1),C_(3) to C_(3) xC_(1)` in thefirstdet. `C_(1) to C_(1) -xC_(2),C_(3) to C_(3) -xC_(3)-xC_(2)` in theseconddet. `" and " C_(1) to C_(1)-xC_(3),C_(2)to C_(2)-xC_(3)` in thethirddet. we get `Delta (x)=|{:(1,,b_(1),,C_(1)),(1,,b_(2),,c_(2)),(1,,b_(3),,c_(3)):}|+|{:(a_(1),,1,,c_(1)),(a_(2),,1,,c_(2)),(a_(3),,1,,c_(3)):}|+|{:(a_(1),,b_(1),,1),(a_(2),,b_(2),,1),(a_(3),,b_(3),,1):}|` `Delta (0)=|{:(a_(1),,b_(1),,c_(1)),(a_(2),,b_(2),,c_(2)),(a_(3),,b_(3),,c_(3)):}|` whichare `b_(2) C_(3) =b_(3)C_(2),C_(2)a_(3)-C_(3)a_(2),a_(2)b_(3)-b_(2)a_(3)` etc Clearly `|{:(1,,b_(1),,c_(1)),(1,,b_(2),,c_(2)),(1,,b_(3),,c_(3)):}| = (b_(2)b_(3) -b_(3)c_(2))+(c_(1)b_(3)-c_(3)b_(1))+(b_(1)c_(2)-b_(2)c_(1))` Whichis the sum ofcofactors of the firstrow ELEMENTS of `Delta (0)` `" similarly "|{:(a_(1),,1,,c_(1)),(a_(2),,1,,c_(2)),(a_(3),,1,,c_(3)):}|" and "|{:(a_(1),,b_(1),,1),(a_(2),,b_(2),,1),(a_(3),,b_(3),,1):}|` are thesum of cofactorsof 2ndrow and3rdelementsrespectively of `Delta (0)` . Hence`Delta(x)=s` where S denotesthe sumof allcofactors of elementsof `Delta(0)` `:. Delta ''(x) =0` SINCE `Delta (x) = s Delta (x) sx+k` So `Delta (0) =k` Hence `Delta(x)= x S+Delta (0)` |
|