1.4 The Product of Two Determinants — 1 5
Comparing this result with (1.2.5),
|aij|n=
n
∑
k=
aikAik (1.3.14)
which is the expansion of|aij|nby elements from rowiand their cofactors.
From (1.3.1) and noting (1.3.5),
x 1 x 2 ···xn=(y 1 +a 1 jej)(y 2 +a 2 jej)···(yn+anjej)
=a 1 jejy 2 y 3 ···yn+a 2 jy 1 ejy 3 ···yn
+···+anjy 1 y 2 ···yn− 1 ej
=(a 1 jA 1 j+a 2 jA 2 j+···+anjAnj)e 1 e 2 ···en
=
[
n
∑
k=
akjAkj
]
e 1 e 2 ···en. (1.3.15)
Comparing this relation with (1.2.5),
|aij|n=
n
∑
k=
akjAkj (1.3.16)
which is the expansion of |aij|nby elements from columnj and their
cofactors.
1.4 The Product of Two Determinants — 1...........
Put
xi=
n
∑
k=
aikyk,
yk=
n
∑
j=
bkjej.
Then,
x 1 x 2 ···xn=|aij|ny 1 y 2 ···yn,
y 1 y 2 ···yn=|bij|ne 1 e 2 ···en.
Hence,
x 1 x 2 ···xn=|aij|n|bij|ne 1 e 2 ···en. (1.4.1)
But,
xi=
n
∑
k=
aik
n
∑
j=
bkjej