Notes on Determinants and Matrix Inverse
University of British Columbia, Vancouver
Yue-Xian Li
March, 2014
1
1
Definition of determinant
Determinant is a scalar that measures the “magnitude” or “size” of a square
matrix.
Notice that conclusions presented below are focused on rows and row expansions. Similar results apply when the row-focus is changed to column-focus.
Definition (verbal): For each n × n (square) matrix A, det A is the sum of
n! terms, each is the product of n entries in which one and only one is taken
from each row and column. The sum exhausts all permutations of integers
{1, 2, · · · , n}. Terms with even permutations carry a “+” sign and those with
odd permutations a “-” sign.
Permutation: any rearrangement of an ordered list of numbers, say S, is a
permutation of S.
Example 1.1: For S = {1, 2, 3}, there are a total of 3! = 6 permutations:
{123, 231, 312, 132, 213, 321}.
Definition: For each permutation (j1 j2 · · · jn ) of the ordered list S = {1, 2, · · · , n},
σ(j1 j2 · · · jn ) = min number of pairwise exchanges to recover order (12 · · · n).
Even/odd permutations: a permutation (j1 j2 · · · jn ) is even/odd if the
value of σ(j1 j2 · · · jn ) is even/odd.
Example 1.2: For S = {1, 2, 3}:
σ(123) = 0,
σ(231) = 2,
σ(312) = 2 are even;
σ(132) = 1,
σ(213) = 1,
σ(321) = 1 are odd.
2
Example 1.3: Show that the determinant of diagonal and triangular matrices
is equal to the product of diagonal entries.
Answer: In each one of these matrices, there is only one way of picking one
entry from each and only one row and column that is not a zero.
a11 0 · · · 0
0 a22 · · · 0
..
.. . .
..
.
.
.
.
0
0 · · · ann
=
a11 0 · · · 0
∗ a22 · · · 0
..
.. . .
..
.
.
.
.
∗
∗ · · · ann
3
=
a11 ∗ · · · ∗
0 a22 · · · ∗
..
.. . .
..
.
.
.
.
0
0 · · · ann
= a11 a22 · · · ann .
Definition (formula): ∀ n × n matrix A = [aij ] (1 ≤ i, j ≤ n)
a11 a12 · · · a1n a21 a22 · · · a2n X
det A = ..
=
(−1)σ(j1 j2 ···jn ) a1j1 a2j2 · · · anjn ,
..
.
.
..
.. .
.
(j1 j2 ···jn )
an1 an2 · · · ann where the sum is taken over all possible permutations of integers (12 · · · n) (a
total of n!) and that
+1, if σ(j1 j2 · · · jn ) is even,
σ(j1 j2 ···jn )
(−1)
=
−1, if σ(j1 j2 · · · jn ) is odd.
Example 1.4: For
a11 a12
det A = a21 a22
a31 an2
a 3 × 3 matrix A = [aij ] (1 ≤ i, j ≤ 3),
a13 X
a23 =
(−1)σ(j1 j2 j3 ) a1j1 a2j2 a3j3
a33 (j1 j2 j3 )
= (−1)σ(123)=0 a11 a22 a33 + (−1)σ(231)=2 a12 a23 a31 + (−1)σ(312)=2 a13 a21 a32
+(−1)σ(132)=1 a11 a23 a32 + (−1)σ(213)=1 a12 a21 a33 + (−1)σ(321)=1 a13 a22 a31
= a11 a22 a33 + a12 a23 a31 + a13 a21 a32 − a11 a23 a32 − a12 a21 a33 − a13 a22 a31 .
Remark: For matrices larger than 3×3, this formula is not practical to use by
hand. So, it shall be mainly employed in demonstrating important properties
of determinants.
4
2
Determinants calculated by row expansions
For all n × n matrix A = [aij ] (1 ≤ i, j ≤ n), expansion in the ith row yields
det A =
n
X
i+j
aij (−1)
det Aij =
j=1
n
X
aij cij ,
j=1
or in expanded form
det A = ai1 (−1)i+1 det Ai1 + ai2 (−1)i+2 det Ai2 + · · · + ain (−1)i+n det Ain
= ai1 ci1 + ai2 ci2 + · · · + ain cin
where
• Aij = (n − 1) × (n − 1) matrix obtained by deleting the ith row and j th
column of A;
• det Aij = (i, j)th −minor of A;
• cij = (−1)i+j det Aij = (i, j)th −cofactor of A.
Formula for expansion in the j th column is obtained by changing the summation index from j to i:
det A =
n
X
aij (−1)
i+j
i=1
det Aij =
n
X
aij Cij .
i=1
Remark: The determinants for 2 × 2 and 3 × 3 matrices are easy to calculate.
Most students are familiar with the expansion in the first row (i.e., i = 1).
5
Example 2.1:
det A = 2
0
2
3
0
0
2
0
2
3
4
6
4
2
4
2
=?
Ans: Using expansion in the first row, we obtain
0 3 2
det A = 2 2 4 4
0 6 2
0 0 2
+ 2 2 2 4
3 0 2
0 0 3
− 4 2 2 4
3 0 6
.
Using expansion in the second row, we obtain
2 0 4
2+3 det A = (−1) 3 2 4 4
3 0 2
2 0 2
+(−1)2+4 2 2 2 4
3 0 6
2 0 4
= −3 2 4 4
3 0 2
2 0 2
+2 2 2 4
3 0 6
But the simplest is to expand in the second column,
2 2 4
3+2 det A = (−1) 2 0 3 2
3 6 2
= (−2)(12 + 12 − 36 − 24) = 72.
6
.
Derivation/proof of the expansion formula: (not required!)
det A =
n
X
i+j
aij (−1)
det Aij =
j=1
n
X
aij cij .
j=1
Proof: Based on definition:
det A =
X
(−1)σ(l1 ···li ···ln ) [a1l1 · · · aili · · · anln ]
(l1 ···li ···ln )
=
X
σ(li l1 ···l(i−1) l(i+1) ···ln )−(i−1)
(−1)
h
aili a1l1 · · · a(i−1)l(i−1) a(i+1)l(i+1) · · · anln
i
(l1 ···li ···ln )
=
X
i
h
(−1)σ(jl1 ···l(i−1) l(i+1) ···ln )−(i−1)−(j−1) aij a1l1 · · · a(i−1)l(i−1) a(i+1)l(i+1) · · · anln
(l1 ···li ···ln )
=
n
X
j=1
=
n
X

lk 6=j for all k
X
aij (−1)i+j 
h
(−1)σ(l1 ···l(i−1) l(i+1) ···ln ) a1l1 · · · a(i−1)l(i−1) a(i+1)l(i+1)
(l1 ···l(i−1) l(i+1) ···ln )
aij (−1)i+j det Aij .
j=1
In the derivation above, we used the following identities:
(i) σ(l1 · · · li · · · ln ) = σ(li l1 · · · l(i−1) l(i+1) · · · ln ) − (i − 1),
because it takes (i−1) pairwise exchanges to turn (li l1 · · · l(i−1) l(i+1) · · · ln )
into (l1 · · · li · · · ln ).
(ii) σ(li l1 · · · l(i−1) l(i+1) · · · ln ) = σ(jl1 · · · l(i−1) l(i+1) · · · ln ) − (j − 1),
because it takes (j − 1) pairwise exchanges to turn (1 · · · j · · · n) into
(j1 · · · (j − 1)(j + 1) · · · ln ).
(iii) (−1)−(i+j)−2 = (−1)−(i+j) = (−1)i+j .
7

i
· · · anln 
3
3.1
Important properties of determinants
det A = det AT
Because the “row-focused” and “column-focused” formulas are identical.
det A =
X
(−1)σ(j1 j2 ···jn ) a1j1 a2j2 · · · anjn
(j1 j2 ···jn )
=
X
(−1)σ(i1 i2 ···in ) ai1 1 ai2 2 · · · ain n = det AT .
(i1 i2 ···in )
Example 3.1:
1 2
3 4
=4−6= 1 3
2 4
8
.
det Ak↔l = − det A
3.2
Suppose that matrix A0 = Ak↔l is obtained by swapping the k th and j th rows
of A. Thus, a0ij = aij except that a0kj = alj and a0lj = akj for all 1 ≤ j ≤ n.
Then,
det A0 =
=
X
X
(−1)σ(j1 j2 ···jk ···jl ···jn ) a01j1 a02j2 · · · a0kjk · · · a0ljl · · · a0njn
(j1 j2 ···jn )
σ(j1 j2 ···jk ···jl ···jn )
(−1)
a1j1 a2j2 · · · aljk · · · akjl · · · anjn
(j1 j2 ···jn )
=
X
(−1)σ(j1 j2 ···jk ···jl ···jn ) a1j1 a2j2 · · · akjl · · · aljk · · · anjn
(j1 j2 ···jn )
=
X
(−1)σ(j1 j2 ···jl ···jk ···jn )+1 a1j1 a2j2 · · · akjl · · · aljk · · · anjn = − det A.
(j1 j2 ···jn )
Note that σ(j1 j2 · · · jk · · · jl · · · jn ) = σ(j1 j2 · · · jl · · · jk · · · jn ) + 1 because the
two differ by one pairwise swap.
Example 3.2:
1 2
3 4
= 4 − 6 = −2,
3 4
1 2
but
9
2 1
=
4 3
= 6 − 4 = 2.
3.3
det Ak=l = 0
If two rows (columns) of a matrix A are identical (i.e. A = Ak↔l ), then
det A = det Ak↔l = − det A = 0.
Alternatively, if two rows are identical or are constant multiples of each other,
then the rows in A are not linearly independent. Thus, A is not invertible
which implies det A = 0.
Example 3.3:
1 2
1 2
= 2 − 2 = 0.
10
3.4
Determinant is linear in each row/column
Suppose a row vector is a linear combination of two row vectors, say the k th
row ak = sb + tc (k = 1, 2, · · · , n), where s, t are scalars. Then,


 
a1
a1


 a2 
 a2 


.
.


.
.


.
.
det 
=
s
det
+
t
det

 
 .
sb + tc
b
c
 . 
.
.
 .. 
 .. 
 .. 
an
an
an
a1
a2
..
.


Equivalently, we can also express this property in the following form
a11
a12
a
a22
21
..
..
.
.
sb1 + tc1 sb2 + tc2
..
..
.
.
an1
an2
···
···
..
.
···
..
.
···
= s
sbn + tcn ..
.
ann
a1n
a2n
..
.
a11 a12 · · ·
a21 a22 · · ·
..
..
..
.
.
.
b1 b2 · · ·
..
..
..
.
.
.
an1 an2 · · ·
+t bn .. . ann a1n
a2n
..
.
a11 a12 · · ·
a21 a22 · · ·
..
..
..
.
.
.
c1 c2 · · ·
..
..
..
.
.
.
an1 an2 · · ·
.
cn .. . ann a1n
a2n
..
.
Proof:
det A =
X
(−1)σ(j1 j2 ···jn ) a1j1 a2j2 · · · akjk · · · anjn
X (j1 j2 ···jn )
(−1)σ(j1 j2 ···jn ) a1j1 a2j2 · · · (sbjk + tcjk ) · · · anjn
=
(j1 j2 ···jn )
=s
X
(−1)σ(j1 j2 ···jn ) a1j1 a2j2 · · · bjk · · · anjn +t
(j1 j2 ···jn )
X
(j1 j2 ···jn )
11
(−1)σ(j1 j2 ···jn ) a1j1 a2j2 · · · cjk · · · anjn .
This property can also be expressed in form of two related properties:
(1) ···
···
..
.
a1n
a2n
..
.
sak1 sak2 · · ·
..
..
..
.
.
.
an1 an2 · · ·
sakn
..
.
a11
a21
..
.
a12
a22
..
.
a11
a12
a21
a22
..
..
.
.
(2) b1 + c 1 b2 + c 2
..
..
.
.
a
a
n1
n2
ann
···
···
..
.
···
..
.
···
= s
a11 a12 · · ·
a21 a22 · · ·
..
..
..
.
.
.
ak1 ak2 · · ·
..
..
..
.
.
.
an1 an2 · · ·
=
bn + c n ..
.
ann a1n
a2n
..
.
,
akn .. . ann a1n
a2n
..
.
a11 a12 · · ·
a21 a22 · · ·
..
..
..
.
.
.
b1 b2 · · ·
..
..
..
.
.
.
an1 an2 · · ·
+
bn .. . ann a1n
a2n
..
.
a11 a12 · · ·
a21 a22 · · ·
..
..
..
.
.
.
c1 c2 · · ·
..
..
..
.
.
.
an1 an2 · · ·
Example 3.4:
a b
3 5
= 5a − 3b,
a+c b+d
3
5
2a 2b 3 5 = 10a − 6b = 2(5a − 3b).
= 5(a+c)−3(b+d) = (5a−3b)+(5c−3d) = a b
3 5
12
c d
+
3 5
.
.
cn .. . ann a1n
a2n
..
.
3.5
Adding a scalar multiple of one row to another
Let matrix A0 be identical to A except that k th row multiplied by scalar s
(6= 0) is added to lth row (1 ≤ k < l ≤ n).
Thus, a0ij = aij for all 1 ≤ i, j ≤ n except that a0lj = alj + sakj . Then,
det A0 = det A.
Proof:
X
det A0 =
=
X
(−1)σ(j1 j2 ···jn ) a01j1 a02j2 · · · a0ljk · · · a0njn
(j1 j2 ···jn )
σ(j1 j2 ···jn )
(−1)
a1j1 · · · akjk · · · (aljl + sakjl ) · · · anjn
(j1 j2 ···jn )
=
X
(−1)σ(j1 j2 ···jn ) a1j1 · · · akjk · · · aljl · · · anjn +s
(j1 j2 ···jn )
X
(−1)σ(j1 j2 ···jn ) a1j1 · · · akjk · · · akjl · · · anjn
(j1 j2 ···jn )
= det A + 0 = det A.
Example 3.5:
a b
c d
= ad − bc.
a
b
c + 2a d + 2b
= a(d + 2b) − b(c + 2a) = ad + 2ab − bc − 2ba = ad − bc.
13
3.6
det AB = det A det B.
Example 3.6:
2 1
det A = 3 4
= 8 − 3 = 5,
−2 5
det AB = −13 15
1 1 = 3 − (−4) = 7.
det B = −4 3 = −30 − (−65) = 35 = det A det B.
14
Proof(1): (not required!):
Case I: If at least one of A and B is noninvertible, i.e. either det A = 0 or
det B = 0 or both, then AB as the composite transformation of the two must
also be noninvertible. Thus,
det(AB) = 0 = det A det B.
Case II: If both A and B are invertible, then based on the results on elementary matrices:
A = A1 A2 · · · Ak ,
B = B1 B2 · · · Bl ,
where A1 , · · · , Ak and B1 , · · · , Bl are all elementary matrices. Therefore,
det(AB) = det(A1 A2 · · · Ak B1 B2 · · · Bl ) = (det A1 · · · det Ak )(det B1 · · · det Bl ) = det A det B.
15
Proof(2): (not required!):
n
X
Note that AB = [(AB)ij ], where (AB)ij =
ail blj .
l=1
X
det AB =
(−1)σ(j1 j2 ···jn ) (AB)1j1 (AB)2j2 · · · (AB)njn
(j1 j2 ···jn )
=
X
n
X
(−1)σ(j1 j2 ···jn )
(j1 j2 ···jn )
!
n
X
a1l1 bl1 j1
!
a2l2 bl2 j2
···
n
X
!
anln bln jn
l1 =1
l2 =1
ln =1
All non−selfavoiding terms cancel (i.e. l1 ,l2 ,··· ,ln must be distinct)
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−→
because an identical term with opposite sign occurs when sumed over all permutations of (j1 j2 ···jn )
X
X
=
(−1)σ(j1 j2 ···jn )
(a1l1 bl1 j1 )(a2l2 bl2 j2 ) · · · (anln bln jn )
(j1 j2 ···jn )
=
=
X
(l1 l2 ···ln )
σ(j1 j2 ···jn )
(−1)
(a1l1 a2l2 · · · anln )(bl1 j1 bl2 j2 · · · bln jn )
(j1 j2 ···jn )
(l1 l2 ···ln )
X
X
(−1)σ(j1 j2 ···jn )
(j1 j2 ···jn )
=
X
X
(a1l1 a2l2 · · · anln )(−1)σ(l1 l2 ···ln ) (b1j1 b2j2 · · · bnjn )
(l1 l2 ···ln )
σ(j1 j2 ···jn )
(−1)
X
(b1j1 b2j2 · · · bnjn )
(j1 j2 ···jn )
(−1)σ(l1 l2 ···ln ) (a1l1 a2l2 · · · anln )
(l1 l2 ···ln )
= det B det A.
The key step in the proof above is to eliminate all the non-selfavoiding terms.
Here, a “non-selfavoiding term” refers to those terms in which li = lj (i 6= j)
happens for at least one pair of i, j (1 ≤ i, j ≤ n). To see this more clearly,
we show the case when only two subscripts exists for 2×2 matrices. In this case,
(12) or (21)
X
det AB =
(−1)σ(j1 j2 ) (AB)1j1 (AB)2j2
(j1 j2 )
(12) or (21)
=
X
σ(j1 j2 )
(−1)
(j1 j2 )
(12) or (21)
=
X
2
X
!
a1l1 bl1 j1
l1 =1
2
X
!
a2l2 bl2 j2
l2 =1
(−1)σ(j1 j2 ) (a11 b1j1 + a12 b2j1 ) (a21 b1j2 + a22 b2j2 )
(j1 j2 )
= (a11 b11 + a12 b21 ) (a21 b12 + a22 b22 ) − (a11 b12 + a12 b22 ) (a21 b11 + a22 b21 )
Terms with identical color cancel each other.
−−−−−−−−−−−−−−−−−−−−−−−−−−→
Self−avoiding colors!
16
= a11 b11 a22 b22 + a12 b21 a21 b12 − a11 b12 a22 b21 − a12 b22 a21 b11
2
X
=
=
=
(a1l1 bl1 1 a2l2 bl2 2 − a1l1 bl1 2 a2l2 bl2 1 )
l1 ,l2 =1, (l1 6=l2 )
2
X
X
(−1)σ(j1 j2 ) a1l1 bl1 j1 a2l2 bl2 j2 =
l1 ,l2 =1, (l1 6=l2 ) (j1 j2 )
X X
σ(j1 j2 )
(−1)
=
(−1)σ(j1 j2 ) a1l1 bl1 j1 a2l2 bl2 j2
1 l2 ) (j1 j2 )
X (lX
(a1l1 a2l2 )(bl1 j1 bl2 j2 ) =
(−1)σ(j1 j2 ) (a1l1 a2l2 )(−1)σ(l1 l2 ) (b1j1 b2j2 )
(l1 l2 ) (j1 j2 )
X
X X
(l1 l2 ) (j1 j2 )
σ(l1 l2 )
(−1)
(l1 l2 )
(a1l1 a2l2 )
X
(−1)
σ(j1 j2 )
(b1j1 b2j2 )
(j1 j2 )
= det A det B.
Here, non-selfavoiding terms are terms like a11 b11 a21 b12 and a12 b21 a22 b22 in
which entries from the same column or row of a matrix occur twice in the same
product. They all cancel out in the summation. Therefore, “self-avoiding”
means in each term of the expression, there must be one and only one entries
from each row and each column of each matrix. The same conclusion works
in a similar way when the matrix is bigger. For any term that contains at
least one pair of non-selfavoiding entries, we can use the same argument for
this particular pair of entries in exactly the same way as we treat the 2×2 case.
17
4
Formula for the inverse of n × n matrices
Definition of minor and cofactor:
∀ n × n matrices A = [aij ], the (i, j)th −minor and (i, j)th −cofactor are
mij = det Aij ,
cij = (−1)i+j mij ,
where Aij is the (n − 1) × (n − 1) matrix obtained by deleting the ith row and
j th column of A.
Definition of cofactor matrix:
∀ n × n matrices A = [aij ], its cofactor

c11
 c21

C = [cij ] =  ..
 .
cn1
matrix is defined as

c12 · · · c1n
c22 · · · c2n 

.. . .
..  ,
.
.
. 
cn2 · · · cnn
where cij = (−1)i+j det Aij is the (i, j)th −cofactor of matrix A.
Formula for A−1 :
∀ invertible n × n matrices A with a cofactor matrix C:
1
−1
A =
CT .
det A
18


0
2 1
Example 4.1: Consider the matrix A =  −1 1 0 .
2 −5 2
(a) Show that it is invertible using the row expansion formula;
(b) Find A−1 using the cofactor formula.
Ans:
(a) First calculate the cofactors for the expansion in row 1:
−1 0 −1 1
1 0 = 2; c12 = − = 2; c13 = + c11 = + −5 2
2 2
2 −5
=⇒
= 3.
det A = (0)c11 + (2)c12 + (1)c13 = 4 + 3 = 7 6= 0!
(b) Then, calculate all the other cofactors:
0 1
2 1 =
−9;
c
=
+
c21 = − 22
2 2
−5 2 2 1 = −1; c32 = − 0 1
c31 = + −1 0
1 0
= −2;
= −1;
c23
c33
0 2
= − 2 −5
0 2
= + −1 1
= 4.
= 2.
Therefore,


2
2 3
C =  −9 −2 4 
−1 −1 2
⇒
A−1 =
1
det A


2 −9 −1
1
C T =  2 −2 −1  .
7
3 4
2
Verify,




 

2 −9 −1
0
2 1
7 0 0
1 0 0
1
1
A−1 A =  2 −2 −1   −1 1 0  =  0 7 0  =  0 1 0  = I3 .
7
7
3 4
2
2 −5 2
0 0 7
0 0 1
19
Derivation/proof of the cofactor formula for inverse: (not required!)
−1
A
=
1
det A
CT .
Proof: Based on the definition of det A:
det A =
n
X
aij cij
n
X
=⇒
j=1
aij
j=1
cij
= 1.
det A
Let A0 = [a0ij ] be a matrix obtained by replacing the ith row of A by its lth
row. Then,
0
det A =
n
X
a0ij cij
=
j=1
n
X
alj cij .
j=1
If l = i (no change), then A0 = A
0
det A =
n
X
a0ij cij
j=1
=
n
X
alj cij =
j=1
n
X
aij cij = det A
=⇒
j=1
n
X
j=1
alj
cij
= 1 ( for l = i).
det A
If l 6= i, then A0 has two identical rows i and l, thus
det A0 = 0 =
n
X
j=1
a0ij cij =
n
X
alj cij
=⇒
j=1
n
X
alj
j=1
Therefore,
n
X
cij
alj
=
det A
j=1
20
1, if l = i,
0, if l 6= i.
cij
= 0 ( for l 6= i).
det A
Introduce a new matrix B = [bij ] such that
bij =
cji
.
det A
Now, let D = AB = [dij ]. Based on definition of matrix multiplication
n
X
n
X
cij
dli =
alj bji =
alj
=
det
A
j=1
j=1
1, if l = i,
0, if l 6= i.
Therefore, D = I = AB ⇒
A
−1
cji
]=
= B = [bij ] = [
det A
21
1
det A
CT .