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.