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`


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