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Lecture 22
Andrei Antonenko
March 31, 2003
1 Properties of determinants
This lecture we will start studying a properties of determinants, and algorithms of computing
them. Let’s recall, that we defined a determinant by the following way:
det
a 11 a 12... a 1 n
a 21 a 22... a 2 n
an 1 an 2... ann
all permutations of
n elements σ
sgn(σ)a 1 σ(1)a 2 σ(2) · · · anσ(n). (1)
Now we’ll start with properties of determinants.
Theorem 1.1 (1st elementary row operation). If 2 rows of a matrix A are interchanged,
then the determinant changes its sign.
Proof. Suppose B arises from A by interchanging rows r and s of A, and suppose r < s. Then
we have that brj = asj and bsj = arj for any j, and aij = bij if i 6 = r, s. Now
det B =
all permutations of n elements σ
sgn(σ)b 1 σ(1) · · · brσ(r)... bsσ(s)... bnσ(n)
all permutations of n elements σ
sgn(σ)a 1 σ(1) · · · asσ(r)... arσ(s)... anσ(n)
all permutations of n elements σ
sgn(σ)a 1 σ(1) · · · arσ(s)... asσ(r)... anσ(n).
The permutation (σ(1)... σ(s)... σ(r)... σ(n)) is obtained from (σ(1)... σ(r)... σ(s)... σ(n))
by interchanging 2 numbers, so its sign is different, and det B = − det A.
Theorem 1.2 (Determinant of a matrix with 2 equal rows). If 2 rows of a matrix are
equal, then its determinant is equal to 0.
Proof. Suppose rows r and s of matrix A are equal. Interchange them to obtain matrix B.
Then det B = − det A. On the other hand, B = A, so det B = det A. So, det A = − det A, and
thus det A = 0.
Theorem 1.3 (2nd elementary row operation). If B is obtained from A by multiplying a
row of A by a real number c, then det B = c det A.
Proof. Suppose r-th row of A is multiplied by c to obtain B. Then brj = carj for any j and
bij = aij if i 6 = r. Thus
det B =
all permutations of n elements σ
sgn(σ)b 1 σ(1) · · · brσ(r)... bnσ(n)
all permutations of n elements σ
sgn(σ)a 1 σ(1) · · · (carσ(r))... anσ(n)
= c ·
all permutations of n elements σ
sgn(σ)a 1 σ(1) · · · arσ(r)... anσ(n)
= c det A.
Theorem 1.4. If a row of a matrix A consists entirely of zeros, then det A = 0.
Proof. Let’s multiply the zero row of a matrix A by a nonzero number c to obtain matrix
B. Then det B = c det A, But B = A, so det B = det A, and thus det A = c det A. So,
det A = 0.
Theorem 1.5 (Multilinearity by rows). If in matrix A row ar can be represented as sum
of rows b and c, i.e. arj = bj + cj , i.e.
a 11 a 12... a 1 n
ar 1 ar 2... arn
an 1 an 2... ann
a 11 a 12... a 1 n
b 1 + c 1 b 2 + c 2... bn + cn
an 1 an 2... ann
then
det
a 11 a 12... a 1 n
b 1 + c 1 b 2 + c 2... bn + cn
an 1 an 2... ann
= det
a 11 a 12... a 1 n
b 1 b 2... bn
an 1 an 2... ann
+det
a 11 a 12... a 1 n
c 1 c 2... cn
an 1 an 2... ann
Formally, this property tells us that the determinant is a multilinear function of rows of a
matrix.
Let a 1 k 1 a 2 k 2... ankn 6 = 0. Then
k 1 ≥ 1 , k 2 ≥ 2... , kn ≥ n
(otherwise the term is equal to 0). But (k 1 k 2... kn) is a permutation of numbers from 1 to n,
so
k 1 + k 2 + · · · + kn = 1 + 2 + · · · + n,
and it is possible only if
k 1 = 1, k 2 = 2,... , kn = n.
So, now we know what happens with the determinant after applying elementary row oper-
ations. So, we can now give the algorithm of computing the determinant.
Algorithm. Transform matrix A by elementary row operation to the triangular form keeping
track of how the determinant changes.
Example 1.8.
det
= 2 det
div. 3rd row by 2
= −2 det
interchange rows 1 and 3
= −2 det
mult. 1st row by 3 and sub. from 2nd
= −2 det
mult. 1st row by 4 and sub. from 3rd
= (−2)(4) det
div. 2nd row by 4
= (−2)(4)(5) det
div. 3rd row by 5
= (−2)(4)(5) det
3 2
mult. the 2nd row by 1/2 and sub. from 3rd
