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− 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,

∑
det(M ) ≡ sgn(k1,⋅⋅⋅,kn)m1k1 ⋅⋅⋅mnkn
(k1,⋅⋅⋅,kn) n

Letting θ denote the position of n in the ordered list,

Now suppose 1.11. 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)m1k ⋅⋅⋅m(n−1)k = a det(A ).
(k1,⋅⋅⋅,kn−1)n−1 1 n−1

To get the assertion in the situation of 1.10 use Corollary A.2.5 and 1.11 to
write

(( ) )
det(M ) = det(M T) = det AT 0 = a det(AT) = adet(A).■
∗ 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.2.12Let A =

(aij)

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

(A)

is defined bycof

(A )

=

(cij)

where toobtain c_{ij}delete the i^{th}row and the j^{th}column of A, take the determinant of the

(n − 1)

×

(n − 1)

matrix which results, (This is called the ij^{th}minor of A. ) andthen 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.2.13Let 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

( ij)
A

. At this point, recall
that from Proposition A.2.3, 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.2.11,