Kenneth Kuttler

3.3. THE MATHEMATICAL THEORY OF DETERMINANTS 95

Theorem 3.3.14 Let A be an n×m matrix with n ≥ m and let B be a m×n matrix. Alsolet Ai

i = 1, · · · , C (n,m)

be the m×m submatrices of A which are obtained by deleting n−m rows and let Bi be them×m submatrices of B which are obtained by deleting corresponding n−m columns. Then

det (BA) =

C(n,m)∑k=1

det (Bk) det (Ak)

Proof: This follows from a computation. By Corollary 3.3.8 on Page 93, det (BA) =

∑(i1···im)

∑(j1···jm)

sgn (i1 · · · im) sgn (j1 · · · jm) (BA)i1j1 (BA)i2j2 · · · (BA)imjm

∑(i1···im)

∑(j1···jm)

sgn (i1 · · · im) sgn (j1 · · · jm) ·

n∑r1=1

Bi1r1Ar1j1

n∑r2=1

Bi2r2Ar2j2 · · ·n∑

rm=1

BimrmArmjm

Now denote by Ik one of the subsets of {1, · · · , n} which has m elements. Thus there areC (n,m) of these.

C(n,m)∑k=1

∑{r1,··· ,rm}=Ik

∑(i1···im)

∑(j1···jm)

sgn (i1 · · · im) sgn (j1 · · · jm) ·

Bi1r1Ar1j1Bi2r2Ar2j2 · · ·BimrmArmjm

C(n,m)∑k=1

∑{r1,··· ,rm}=Ik

∑(i1···im)

sgn (i1 · · · im)Bi1r1Bi2r2 · · ·Bimrm ·

∑(j1···jm)

sgn (j1 · · · jm)Ar1j1Ar2j2 · · ·Armjm

C(n,m)∑k=1

∑{r1,··· ,rm}=Ik

m!sgn (r1 · · · rm)

2det (Bk) det (Ak) =

C(n,m)∑k=1

det (Bk) det (Ak)

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

3.3.5 Expansion Using Cofactors

Lemma 3.3.15 Suppose a matrix is of the form

M =

(A ∗0 a

)or

(A 0

∗ a

)(3.11)

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) .

3.3. THE MATHEMATICAL THEORY OF DETERMINANTS 95Theorem 3.3.14 Let A be annxm matriz with n >m and let B beamxn matria. Alsolet A;1=1,--- ,C'(n,m)be the m x m submatrices of A which are obtained by deleting n —m rows and let B; be themxm submatrices of B which are obtained by deleting corresponding n—m columns. ThenC(n,m)det(BA)= © det (By) det (Ax)k=1Proof: This follows from a computation. By Corollary 3.3.8 on Page 93, det (BA) => » sgn (1 — im) sgn (ji “ jm) (BA);,j, (BA). 55°" , (BA); im, (t1++4m ) (j1--dm)1 . . . .= De 88H (in ++i) 880 (Sam) +(t1-++im) (j1---Jm)n n nS> Bir Ari: S- Bigrz Ars jo a S> Binrm Armimry=1 rg=1 TmaHlNow denote by J; one of the subsets of {1,--- ,n} which has m elements. Thus there areC (n,m) of these.C(n,m)1 . . . .= DD de XL sen linen) 880 Gam)k=1 {ris + Pmp= Tx ‘ (i1-+im) (j1---jm)Bir, A B; AT1j1*?tiere44*rej2 "°° Bim ArmimC(n,m)= SY YS samls tn) Bar Bara Bink=1 {rij rm}=Ie (da-tm)S- sen (j1*** jm) Ary j, Aron ue Aon im(jai)— 1 C(n,m)» cn “Sen (ri Tm) det (By) det (Ay) = » det (B,) det (Ax)since there are m! ways of arranging the indices {r),--- ,7m}. Hi3.3.5 Expansion Using CofactorsLemma 3.3.15 Suppose a matrix is of the formu(r) o(4 eo) (3.11)where a is a number and A is an (n—1) x (n—1) matrix and * denotes either a columnor a row having length n — 1 and the O denotes either a column or a row of length n — 1consisting entirely of zeros. Then det (M) = adet (A).