7. _Decomposition of a Determinant into complementary

Determinants._--Consider, for simplicity, a determinant of the fifth order, 5 = 2 + 3, and let the top two lines be a , b , c , d , e a', b', c', d', e' then, if we consider how these elements enter into the determinant, it is at once seen that they enter only through the determinants of the second order |a , b |, &c., which can be formed by selecting any two |a', b'| columns at pleasure. Moreover, representing the remaining three lines by a" , b" , c" , d" , e" a"', b"', c"', d"', e"' a"", b"", c"", d"", e"" it is further seen that the factor which multiplies the determinant formed with any two columns of the first set is the determinant of the third order formed with the complementary three columns of the second set; and it thus appears that the determinant of the fifth order is a sum of all the products of the form = |a , b | |c" , d" , e" |, |a', b"| |c"', d"', e"'| |c"", d"", e""| the sign ± being in each case such that the sign of the term ± ab'c"d'"e"" obtained from the diagonal elements of the component determinants may be the actual sign of this term in the determinant of the fifth order; for the product written down the sign is obviously +. Observe that for a determinant of the n-th order, taking the decomposition to be 1 + (n - 1), we fall back upon the equations given at the commencement, in order to show the genesis of a determinant.