Question

Prove these assertions concerning the effects of row operations on the determinant:

1) multiplying one row by k multiplies the determinant by k.
2) interchanging two rows changes the sign of the determinant.
3) adding to one row a multiple of another row has no effect on the determinant.

Answer 1

All statement implies from formula $\det A = \sum (-1)^{\left|i_1,i_2,\dots ,i_n\right|}a_{i_1!}a_{i_2,2}\dots a_{i_n,n}$ .

1) if we substitute $a_{i_k k}$ by $t a_{i_k k}$ , (row is multiplied by $t$ ) then we can bring $t$ out of sum, and we are done.

2) as any transposition changes sign of permutation, then $\left|i_1,i_2,\dots ,i_n\right|$ will be changed from odd to even and vice versa in every summand, so sign of det A will be opposite.

3) if one of rows is equal to another one, then their permutation change sign of det A, and in general det A will be the same. It can be iff det $\mathrm{A} = 0$ . So, if we have

$\sum (-1)^{\left|i_1,i_2,\dots ,i_n\right|}a_{i_1!}a_{i_2,2}\dots \left(a_{i_kk} + ta_{i_l!}\right)\dots a_{i_nn} = \sum (-1)^{\left|i_1,i_2,\dots ,i_n\right|}a_{i_1!}a_{i_2,2}\dots \left(a_{i_kk}\right)\dots a_{i_nn} + t\sum (-1)^{\left|i_1,i_2,\dots ,i_n\right|}a_{i_1!}a_{i_2,2}\dots \left(a_{i_ll}\right)\dots a_{i_nn}$ then in first summand all i's are distinct, and in second 1 is being used twice, so second summand is zero.