matrix and ∗ denotes either a column or a row havinglength n− 1 and the 0 denotes either a column or a row of length n− 1 consisting entirely of 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 1.27. 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 = adet(A).
(k1,⋅⋅⋅,kn−1)n−1

To get the assertion in the situation of 1.26 use Corollary A.5.8 and 1.27 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 A.5.15Let A =

(aij)

be an n×n matrix. Thena new matrix called the cofactor matrix,cof

(A)

is defined bycof

(A )

=

(cij)

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

(n − 1)

×

(n − 1)

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

(− 1)

^{i+j}. To make the formulas 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 A.5.16Let A be an n × n matrix where n ≥ 2. Then

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 A.5.6, 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 A.5.14,