If `Delta_(1) = |(1,1,1),(a,b,c),(a^(2),b^(2),c^(2))|, Delta_(2) = |(1,bc,a),(1,ca,b),(1,ab,c)|`, thenA. `Delta_(1) + Delta_(2) =0`B. `Delta_(1) + 2Delta_(2) = 0`C. `Delta_(1) = Delta_(2)`D. none of these

If `Delta_(1) = |(1,1,1),(a,b,c),(a^(2),b^(2),c^(2))|, Delta_(2) = |(1

1.	If `Delta_(1) = \|(1,1,1),(a,b,c),(a^(2),b^(2),c^(2))\|, Delta_(2) = \|(1,bc,a),(1,ca,b),(1,ab,c)\|`, thenA. `Delta_(1) + Delta_(2) =0`B. `Delta_(1) + 2Delta_(2) = 0`C. `Delta_(1) = Delta_(2)`D. none of these
Answer» Correct Answer - A We have, `Delta_(2) = \|(1,bc,a),(1,ca,b),(1,ab,c)\|` ltrbgt `rArr Delta_(2) = (1)/(abc)\|(a,abc,a^(2)),(b,abc,b^(2)),(c,abc,c^(2))\| "Applying " R_(1) rarr R_(1) (a) " " R_(2) rarr R_(2) (b) " " R_(3) rarr R_(3) (c)` `rArr Delta_(2) = \|(a,1,a^(2)),(b,1,b^(2)),(c,1,c^(2))\|` [Taking abc common from `C_(2)`] `rArr Delta_(2) = -\|(1,a,a^(2)),(1,b,b^(2)),(1,c,c^(2))\|` [Interchanging `C_(1) and C_(2)`] `rArr Delta_(2) = - \|(1,1,1),(a,b,c),(a^(2),b^(2),c^(2))\| = - Delta_(1) rArr Delta_(1) + Delta_(2) = 0`