Page 87 - A Course in Linear Algebra with Applications

3.2: Basic Properties of Determinants 71

Proof
The proof is by mathematical induction. The statement is
certainly true if n = 1 since then A T = A. Let n > 1 and
assume that the theorem is true for all matrices with n — 1
rows and columns. Expansion by row 1 gives

n

3 = 1

T
Let B denote the matrix A . Then a^ = bji. By induction on
n, the determinant A±j equals its transpose. But this is just
the (j, 1) cofactor Bji of B. Hence A\j = Bji and the above
equation becomes

n
det(A) = ^2bjiBji.

However the right hand side of this equation is simply the
expansion of det(B) by column 1; thus det(A) = det(B).

A useful feature of this result is that it sometimes enables
us to deduce that a property known to hold for the rows of a
determinant also holds for the columns.

Theorem 3.2.2
A determinant with two equal rows (or two equal columns) is
zero.

Proof
Suppose that the n x n matrix A has its th and kth rows
j
equal. We have to show that det(A) = 0. Let ii,i^, • •., i n be
a permutation of 1, ,..., n; the corresponding term in the ex-
2
.
pansion of det(i4) is sign(ii, i-2, .. , «n)aii 102i 2 • • -ctni n- Now
if we switch ij and i^ in this product, the sign of the permuta-
tion is changed, but the product of the a's remains the same