Generally, a determinant A is defined as e_{ijk..n}a_{1i}a_{2j}a_{3k}.. Then a co-factor A_{rs} is defined and it is shown (arduously) that A=a_{Li}A_{Li}. Or, a determinant is defined inductiveley in terms of defined A_{rs}, in which case lABl = lAllBl is stated without proof, proved quite arduously, or, in one case, assigned as an exercise!!
(convention- sum over lower case letters)
Determinant and Derived Cofactor:
A=e_{LMN}e_{ijk}a_{Li}a_{Mj}a_{Nk}
A=a_{Li}{(-1)^{(L+i)}e_{MN}e_{jk}a_{Mj}a_{Nk}}^{*}
A_{Li}=(-1)^{(L+i)}e_{MN}e_{jk}a_{Mj}a_{Nk}
A=a_{Li}A_{Li}
a_{Qi}A_{Li=}e_{LMN}e_{ijk}a_{Qi}a_{Mj}a_{Nk}=0 if Q unequal L because then Q=M or N
End
Definition:
e_{abc…}=1 with a,b,c,.. any positive integers in sequential order.
e_{abc…}= +1 or -1 according as abc..are an even or odd permutation of the sequential order.
^{*}Proof of e_{LMN}e_{ijk}=(-1)^{(L+i)}e_{MN}e_{jk:}
Suppose abc..L.. are in sequential order and L is in rth place. Then
e_{Labc....} = (-1)^{(r-1)}e_{abc…}
If now abc.. are permuted, both sides still hold.
If LMN are put in sequential order, L will be in Lth place and then in general
e_{LMN}=(-1)^{(L-1)}e_{MN, }and similarly e_{ijk}=(-1)^{(i-1)}e_{jk} so that e_{LMN}e_{ijk}=(-1)^{(L+i)}e_{MN}e_{jk}
Note: For general case of cofactor, replace LMN with LMN,…, and ijk wiith ijk,…, in derivation.