

InterviewSolution
Saved Bookmarks
1. |
If `Delta_(1)=|[a,b, c],[x, y, z],[p,q ,r]|"and "Delta_(2) |[q,-b, y],[-p, a, -x],[r,-c ,z]|` then without expanding `Delta_(1) " and "Delta_(2), "prove that "Delta_(1) + Delta_(2) =0` |
Answer» We have `Delta_(2)=|[q,-b, y],[-p, a, -x],[r,-c ,z]|` `=(-1)*|[q,-b, y],[p, -a, x],[r,-c ,z]| ["taking (-1) common from"R_(2)]` `=(-1)(-1)*|[q,b, y],[p, a, x],[r,c ,z]| ["taking (-1) common from"C_(2)]` `=|[q,b, y],[p, a, x],[r,c ,z]|=|[q,p, r],[b, a, c],[y,x ,z]| ["interchanging rows and columns"]` `=(-1)|[p,q, r],[a, b, c],[x,y ,z]|["applying"C_(1) harr C_(2)]` `=(-1)(-1)*|[a,b, c],[p, q, r],[x,y ,z]|["applying"R_(1) harr R_(2)]` `=|[a,b, c],[p, q, r],[x,y ,z]|` `=(-1) *|[a,b, c],[x, y, z],[p,q ,r]| =-Delta_(1) ["applying "R_(2) harr R_(3)]` Thus, `Delta_(2) = -Delta_(1) " and hence `Delta_(1) + Delta_(2) =0` |
|