40 CHAPTER 1. REVIEW OF SOME LINEAR ALGEBRA

=C(n,m)

∑k=1

∑{r1,··· ,rm}=Ik

1m! ∑

(i1···im)sgn(i1 · · · im)Bi1r1 Bi2r2 · · ·Bimrm ·

∑( j1··· jm)

sgn( j1 · · · jm)Ar1 j1Ar2 j2 · · ·Arm jm

=C(n,m)

∑k=1

∑{r1,··· ,rm}=Ik

1m!

sgn(r1 · · ·rm)2 det(Bk)det(Ak) =

C(n,m)

∑k=1

det(Bk)det(Ak)

since there are m! ways of arranging the indices {r1, · · · ,rm}. ■

1.9.5 Expansion Using Cofactors

Lemma 1.9.15 Suppose a matrix is of the form

M =

(A ∗0 a

)or(

A 0∗ a

)(1.24)

where a is a number and A is an (n−1)× (n−1) matrix and ∗ denotes either a columnor a row having length n− 1 and the 0 denotes either a column or a row of length n− 1consisting entirely of zeros. Then det(M) = adet(A) .

Proof: Denote M by (mi j) . Thus in the first case, mnn = a and mni = 0 if i ̸= n while inthe second case, mnn = a and min = 0 if i ̸= n. From the definition of the determinant,

det(M)≡ ∑(k1,··· ,kn)

sgnn (k1, · · · ,kn)m1k1 · · ·mnkn

Letting θ denote the position of n in the ordered list, (k1, · · · ,kn) then using the earlierconventions used to prove Lemma 1.9.1, det(M) equals

∑(k1,··· ,kn)

(−1)n−θ sgnn−1

(k1, · · · ,kθ−1,

θ

kθ+1, · · · ,n−1kn

)m1k1 · · ·mnkn

Now suppose the second case. Then if kn ̸= n, the term involving mnkn in the above expres-sion equals zero. Therefore, the only terms which survive are those for which θ = n or inother words, those for which kn = n. Therefore, the above expression reduces to

a ∑(k1,··· ,kn−1)

sgnn−1 (k1, · · ·kn−1)m1k1 · · ·m(n−1)kn−1 = adet(A) .

To get the assertion in the first case, use Corollary 1.9.8 to write

det(M) = det(MT )= det

((AT 0∗ a

))= adet

(AT )= adet(A) .■

In terms of the theory of determinants, arguably the most important idea is that ofLaplace expansion along a row or a column. This will follow from the above definition ofa determinant.