matrix and ∗ denotes either a column or a rowhaving length n− 1 and the 0 denotes either a column or a row of length n− 1 consisting entirelyof zeros. Then det

(M )

= adet

(A)

.

Proof:Denote M by

(mij)

. Thus in the first case, m_{nn} = a and m_{ni} = 0 if i≠n while in the
second case, m_{nn} = a and m_{in} = 0 if i≠n. From the definition of the determinant,

Now suppose the second case. Then if k_{n}≠n, the term involving m_{nkn} in the above
expression equals zero. Therefore, the only terms which survive are those for which θ = n
or in other words, those for which k_{n} = n. Therefore, the above expression reduces
to

∑
a sgn(k1,⋅⋅⋅kn− 1)m1k1 ⋅⋅⋅m (n−1)kn−1 = a det(A ).
(k1,⋅⋅⋅,kn−1)n−1

To get the assertion in the first case, use Corollary 3.3.8 to write

( ( ))
( T ) AT 0 ( T)
det(M ) = det M = det ∗ a = adet A = a det(A ).■

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

Definition 3.3.16Let A =

(aij)

be an n×n matrix. Thena new matrix called the cofactormatrixcof

(A)

is defined bycof

(A)

=

(cij)

where to obtain c_{ij}delete the i^{th}row and thej^{th}column of A, take the determinant of the

(n − 1)

×

(n− 1)

matrix which results, (Thisis called the ij^{th}minor of A. ) and then multiply this number by

(− 1)

^{i+j}. To make theformulas easier to remember,cof

(A )

_{ij}will denote the ij^{th}entry of the cofactor matrix.

The following is the main result. Earlier this was given as a definition and the outrageous
totally unjustified assertion was made that the same number would be obtained by
expanding the determinant along any row or column. The following theorem proves this
assertion.

Theorem 3.3.17Let A be an n × n matrix where n ≥ 2. Then

( a b c ) ( a b c ) ( a b c )
| | | | | |
B1 = ( d 0 0 ) ,B2 = ( 0 e 0 ) ,B3 = ( 0 0 f )
h i j h i j h i j

Denote by A^{ij} the

(n − 1)

×

(n− 1)

matrix obtained by deleting the i^{th} row and the j^{th}
column of A. Thus cof

(A)

_{ij}≡

(− 1)

^{i+j} det

(Aij)

. At this point, recall that from Proposition
3.3.6, when two rows or two columns in a matrix M, are switched, this results in multiplying the
determinant of the old matrix by −1 to get the determinant of the new matrix. Therefore, by
Lemma 3.3.15,

(( ))
n− j n−i Aij ∗
det (Bj ) = (− 1) (− 1) det 0 aij
( ( ) )
i+j Aij ∗
= (− 1) det 0 aij = aijcof (A )ij.

Therefore,

∑n
det(A ) = aijcof (A)ij
j=1

which is the formula for expanding det

(A)

along the i^{th} row. Also,

( ) ∑n ( ) ∑n
det(A) = det AT = aTijcof AT ij = ajicof (A )ji
j=1 j=1