If `Delta=|(a,p,x),(b,q,y),(c,r,z)|=16` then `Delta_(1)=|(p+x,a+x,a+p),(q+y,b+y,b+q),(r+z,c+z,c+r)|=32`

Answer

N/a

Explanation

True
We have `Delta=|(a,p,x),(b,q,y),(c,r,z)|=16`
and we have to prove `Delta_(1)=|(p+x,a+x,a+p),(q+y,b+y,b+q),(r+z,c+z,c+r)|=32`
`Delata_(1)=|(2p+2x+2a,a+x,a+p),(2q+2y+2b,b+y,b+q),(2r+2z+2c,c+z,c+r)| [ :' C_(1)toC_(1)+C_(2)+C_(3)]`
`=2|(p,x-p,a+p),(q,y-q,b+q),(r,z-r,c+r)|`
[taking 2 common from `C_(1)` and then `C_(1)toC_(1)-C_(2),C_(2)toC_(2)-C_(3)`]
`=2[(p,x,a+p),(q,y,b+q),(r,z,c+r)]-[(p,p,a+p),(q,q,b+q),(r,r,c+r)]`
`=2|(p,x,a+p),(q,y,b+q),(r,z,c+r)|-0`
[since two column `C_(1)` and `C_(2)` are identicals]
`=2|(p,x,a),(q,y,b),(r,z,c)|+|(p,x,p),(q,y,q),(r,z,r)|`
`=2|(a,p,x),(b,q,y),(c,r,z)|+0`
[since `C_(1)` and `C_(3)` are identical in second determinant and in first determinant `C_(1)hArrC_(2) and then C_(1)hArrC_(3)`]
`=2xx16 [ :' Delta=16]`
`=32` Hence proved.