72 CHAPTER 5. SOME IMPORTANT LINEAR ALGEBRA

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.

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

Definition 5.4.7 A vector w, is a linear combination of the vectors {v1, · · · ,vr} if thereexists 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 5.4.8 Suppose A is an n×n matrix and some column (row) is a linear combina-tion of r other columns (rows). Then det(A) = 0.

Proof: Let A =(

a1 · · · an)

be the columns of A and suppose the condition thatone column is a linear combination of r of the others is satisfied. Then by using Corollary5.4.6 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 5.4.6

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).