436 CHAPTER 20. THEORY OF DETERMINANTS∗

If A has two equal columns or two equal rows, then switching them results in the samematrix. Therefore, det(A) =−det(A) and so det(A) = 0.

It remains to verify the last assertion.

det(A)≡ ∑(k1,··· ,kn)

sgn(k1, · · · ,kn)a1k1 · · ·(xaki + ybki

)· · ·ankn

= x ∑(k1,··· ,kn)

sgn(k1, · · · ,kn)a1k1 · · ·aki · · ·ankn

+y ∑(k1,··· ,kn)

sgn(k1, · · · ,kn)a1k1 · · ·bki · · ·ankn

≡ xdet(A1)+ ydet(A2) .

The same is true of columns because det(AT)= det(A) and the rows of AT are the columns

of A.

20.2.5 Linear Combinations and DeterminantsLinear combinations have been discussed already. However, here is a review and some newterminology.

Definition 20.2.7 A vector w, is a linear combination of the vectors

{v1, · · · ,vr}

if there exists scalars, c1, · · ·cr such that w= ∑rk=1 ckvk. This is the same as saying

w ∈ span(v1, · · · ,vr) .

The following corollary is also of great use.

Corollary 20.2.8 Suppose A is an n × n matrix and some column (row) is a linearcombination of r other columns (rows). Then det(A) = 0.

Proof: Let A =(a1 · · · an

)be the columns of A and suppose the condition that

one column is a linear combination of r of the others is satisfied. Then by using Corollary20.2.6 the determinant of A is zero if and only if the determinant of the matrix B, which hasthis special column placed in the last position, equals zero. Thus an = ∑

rk=1 ckak and so

det(B) = det(a1 · · · ar · · · an−1 ∑

rk=1 ckak

).

By Corollary 20.2.6

det(B) =r

∑k=1

ck det(a1 · · · ar · · · an−1 ak

)= 0.

because there are two equal columns. The case for rows follows from the fact that det(A) =det(AT).