Let A be a square, nxn matrix. The determinant of A is defined by:
det(A) =
(1) a 


Sn
a
1 (1) 2 (2)
ai (i ) an ( n)
This means the sum is taken over all n! permutations of Sn . Each term in the sum has one element
from each row and one from each column. Let's look at the determinants of sizes 1, 2 and 3.
n = 1: There is only one permutation (which is even) and det(A) = a11
n = 2: There are two permutations: 1 2 (even) and 2 1 (odd) det(A) = a11a22  a12 a21
 a11 a12 
a
a22 
You can remember this by looking at the two diagonals of A:  21
The main diagonal (top-left to
bottom-right) is added and the anti-diagonal is subtracted.
n = 3: 6 permutations
3 even: 1 2 3
231
312
a11a22 a33  a12 a23a31  a13a21a32
det(A) =
and 3 odd:
213
132
321
a12 a21a33  a11a23a32  a13a22 a31
The three positive terms come from the top-left to bottom-right diagonals where the entries rotate
cyclically as necessary, while the negative terms come from the top-right to bottom-left diagonals.
 a11
a
 21
 a31
a13 
 a11 a12 a13  a11
a

a22 a23 
 21 a22 a23 
a32 a33  a31
a33  a31 a32 a33 
a13
n = 4: There are 24 permutations and not enough diagonals in a 4x4 matrix to work out mnemonic
device. Other ways of compute determinant are needed.
a12
Here are two propositions that follow directly from the definition of determinant.
Proposition: Let A have a row of zeros. Then det(A) = 0.
Proof: Suppose row i of A = 0. Then every term in det(A) has one factor from row i and hence is zero.
t t t
t
Proposition: If T = [tij] is triangular then det(T) = 11 22 33 nn the product of the diagonal
entries.
Proof: Assume for convenience that T is lower triangular. Each term in the determinant has one factor
from each row and one from each column. What permutations can possibly give a non-zero product?
In row 1, only t11 can be chosen. In row 2, only t22 can be chosen, since column one has been is used. In
each row only the diagonal element can be chosen for a non-zero product. This means all but one
permutation leads to a zero product, and this is the identity permutation 1 2 3 . . . n. Since this is an
even permutation, the result follows. A similar argument works if T is upper triangular.
Corollary det(I) = 1.
Proposition: Let A be nxn and let B = A with rows k and m interchanged. Then det(A) = -det(B)
Proof: Assume that k < m. We have bij = aij if i is not k or m; bkj = amj and bmj = akj
(1) b1 (1)b2 (2) bk ( k ) bm ( m ) bn ( n )

det(B) =  Sn
(1) a1 (1) a2 (2) am ( m ) ak ( k ) an ( n )

=  Sn
Switching am ( m ) and ak ( k ) puts the row indices in
the correct order, but this will change the permutation  Write  for withentries k and m swapped.
  (1) a1 (1) a2 (2) ak ( k ) am ( m ) an ( n )
Then det(B) =  Sn
= -det(A) since  and have opposite signs.
Corollary: Let A be an nxn matrix with two rows equal. Then det(A) = 0.
Proof: A = (A with two rows swapped), so by the last proposition det(A) = -det(A).
Proposition: Let A be an nxn matrix and let B be obtained from A by adding a multiple of row k
to row m. Then det(A) = det(B).
Proof: We have bij = aij if i is not m and bmj =amj+ cakj . Let D be matrix A with row k replacing row m.
(1) b1 (1)b2 (2) bm ( m ) bn ( n )

det(B) =  Sn
(1) a1 (1) a2 (2) (am ( m )  cak ( m ) ) an ( n )

=  Sn
( expand the sum)


(

1)
a
a
a
a

c
(

1)
a1 (1) a2 (2) ak ( k ) ak ( m ) an ( n )


1 (1) 2 (2)
m ( m )
n ( n )
 Sn
 Sn
=
= det(A) + c det(D)= det(A), since D has two identical rows and det(D) = 0.
Proposition: Let A be an nxn matrix and let B be obtained from A by multiplying row m by c.
Then det(A) = cdet(B).
Proof: We have bij = aij if i is not m and bmj = camj .
(1) b1 (1)b2 (2) bm ( m ) bn ( n )

det(B) =  Sn
(1) a1 (1) a2 (2) cam ( m ) an ( n )

=  Sn
c  (1) a1 (1) a2 (2) am ( m ) an ( n )
=  Sn
= c det(A)
Corollary: det(cA) = cndet(A), if A is nxn.
Proof: Apply the last result to each of the n rows of A.
Apply these last three results with A as the identity matrix.
If E is I with 2 rows swapped, det(E) = -1.
If E is I with a multiple of one row added to another, det(E) = 1
If E is I with a row multiplied by c, det(E) = c.
These last three propositions together show det(EA) = det(E) det(A), where E is any
elementary matrix. This is used to prove that det(AB) = det(A) det(B) for all matrices A and B.
Cofactor expansion:
Here is a sketch of the proof. Assume the expansion is on row 1. The key idea is to collect all
of the terms containing a11, all those with a12, etc. These means grouping all 's
by the value of (1). There are thus n terms in the sum :
(1) a1 (1) a2 (2) ai ( i ) an ( n )

det(A) =  Sn
a11  (1) a2 (2) ai ( i ) an ( n )  a12  (1) a1 (1) a3 (3) ai ( i ) an ( n ) 
 (1)  2
=  (1) 1

a1n  (1) a1 (1) a2 (2) ai ( i ) a( n 1) ( n 1)
 (1)  n
+
(1) a2 (2) ai ( i ) an ( n )

The sum multiplying a11  Sn ; (1) 1
has (n-1)! summands and each
one has one entry from each row and column except for row 1 and column 1. Each of the permutations
maps the set {2, . . . n} to itself and can be thought of as a permutation on Sn-1. This sum is a
determinant of the submatrix of A obtained by removing row 1 and column 1. This is called the (1,1)
minor of A, denoted M11. Each of similar terms in this expansion is a minor of A obtained by removing
row 1 and column j for each j. The only problem remaining is determining the signs of these restricted
permutations on Sn. It can be shown that they either all flip signs or all have the same sign as the
original permutation. The pattern is (-1)1+j for the jth term. This leads to the cofactors of A, denoted Cij:
Cij = (-1)i+ j Mij Then row 1 expansion becomes
n
 a1nC1n   a1 j C1 j
a11C11  a12C12  a13C13 
j 1
det(A) =
ai1Ci1  ai 2Ci 2  ai 3Ci 3 
and for any row i
n
 ainCin   aij Cij
j 1
det(A) =
Proposition: det(A) = det(AT)
Proof: Let B = AT, so bij = aji
det(B) =
=
(1) b 


b
bi ( i )
bn ( n )
(1) a


a
a ( i ) i
a ( n ) n
Sn
Sn
1 (1) 2 (2)
(1)1  (2)2
Reorder each product to put the row indices in the natural order. Then the column indices will have the
order of  the inverse of As runs through all permutations of Sn so does  Thus the two
expressions are equal since and have the same number of inversions.
Any of these results about rows of a matrix and determinants apply to columns as well, merely apply
the result to the transpose of the matrix.
For example, swapping columns negates a det(A) and if two columns of A are equal then det(A) = 0.