Properties: For n = 1, the definition reduces to the multiplicative
inverse (ab = ba = 1).
If B is an inverse of A, then A is an inverse of B, i.e.,
A and B are inverses to each other.
Example:
⇒
Example: Some matrices have no inverse, like O ∈ Rn×n and
since
Proof Suppose B and C are both inverses of A. Then
Notation: The inverse of A, if exists, is denoted by A−1.
Symbolically, the inverse may be used to solve matrix equations:
Ax = b
However, this method is computationally inefficient.
1
Example:
⇒
⇒
then AT is also invertible, and (AT)−1 = (A−1)T.
Proof (a) A−1 and A are the inverse to each other.
(b) (AB)(B−1A−1) = A(BB−1)A−1 = AInA−1 = AA−1 = In;
similarly, (B−1A−1) (AB) = In.
(c) AT(A−1)T = (A−1A)T = (In)T = In ; similarly, (A−1)TAT = In .
Definition. An n×n matrix E is called an elementary matrix if E can
be obtained from In by a single elementary row operation.
1 0 0 
1 0 0
1 0 0




Example : E1 = 0 0 1, E2 = 0 − 4 0, E3 = 0 1 0.
2 0 1
0 0 1
0 1 0
1 2
1 0 0 1 2 1 2 


Example : A = 3 4 ⇒ E3 A = 0 1 0 3 4 = 3 4 .
5 6
2 0 1 5 6 7 10
2
0
0
 1 0 0
1
1 0 0




−1
−1
Example : E = 0 0 1, E2 = 0 − 1 / 4 0, E3 =  0 1 0.
− 2 0 1
0
0 1 0
0
1
−1
1
In general, if E ∈ Rn×n is an elementary matrix that corresponds to
some elementary row operation, then E−1 is the elementary matrix
that corresponds to the reverse elementary row operation.
Proof ∃ Ek , Ek−1 , …, E1 such that Ek Ek−1L E1 A = R.
P: invertible
−1 −1
(P−1 = E1−1L Ek−1
Ek )
Equivalence of Ax = b and Rx = c (the reduced row echelon form),
an analytical proof:
∃ an invertible P such that P[ A b ] = [ PA Pb ] = [ R c ]
⇒ PA = R, Pb = c or A = P−1R, b = P−1c
⇒ Av = b implies Rv = (PA)v = P(Av) = Pb = c
and Ru = c implies Au = (P−1R)u = P−1(Ru) = P−1c = b
Definition. For a matrix A, a linear relationship among the columns
of A is an equation in which each side of the equation is either 0 or
a linear combination of one or more columns of A.
Example:
or
or
3
Thus a linear relationship among the columns of A can be expressed
as Av = 0 for some vector v.
Linear Correspondence Property:
A certain linear relationship among the columns of A holds if and
only if the same linear relationship holds among the same columns
of R, the reduced row echelon form of A.
Proof Av = 0 ⇔ P−1Rv = 0 ⇔ Rv = 0.
Corollary:
pivot columns
4
Uniqueness of the reduced row echelon form of a matrix:
Let A ∈ Rm×n and R be a reduced row echelon form of A.
Claim I: Contents and positions of the pivot columns of R are
uniquely determined by A.
Proof By the Property (b) of R, the contents of the pivot columns of
R are fixed (e1, e2, … in Rm).
Suppose ri is a pivot column of R. Then Property (a) of R and
Corollaries 1, 3 of the Linear Correspondence Property imply
ai ≠ 0 and ai is not a linear combination of the preceding
columns of A. ⇒ Positions of the pivot columns of R are
uniquely determined by positions of A’s nonzero columns
that are not linear combinations of the preceding columns.
Claim II: The non-pivot columns of R are uniquely determined by A.
Proof Suppose rn = e1, rn = e2, …, rn = ek are the pivot columns of R,
and rj is a non-pivot column of R. ⇒
rn , rn , …, rn are L.I. and rj = c1rn + c2rn + L + ckrn for some
c1, c2, …, ck.
1
1
2
2
k
k
1
2
k
⇒ By the Linear Correspondence Property and its Corollary 2
an , an , …, an are L.I. and aj = c1an + c2an + L + ckan
Thus c1, c2, …, ck and rj are uniquely determined by A.
1
2
k
1
2
k
Proof With the above notations and by the definition of the pivot
columns of A, an , an , …, an are the pivot columns of A.
1
2
k
5