Matrix Inverses
€
Suppose A is an m × n matrix. We have learned
that matrix addition with A is defined for any other
m × n matrix B; that the m × n zero matrix (all
entries equal to 0) is the identity element for this
€
operation; and that the matrix −1 ⋅ A = − A is the
additive inverse of A, that is,
€
A + (− A ) = 0 = − A + A.
€
This allows us to define matrix subtraction in the
natural way: B − A = B + (− A ).
€
We have also seen that matrix multiplication is
defined only if a particular compatibility condition
€ amongst the matrices being multiplied and
is met
their product: if AB makes sense, and A is m × n
and B is p × q, then necessarily n must equal p and
the product AB is m × q. For this and a number of
other reasons (including for instance
€ that this
operation is not commutative), matrix
€
multiplication
is a more complex operation than is
€
multiplication of real numbers. However, we know
that there is a multiplicative identity element: if I n
is the n × n identity matrix (1’s along the diagonal,
0’s elsewhere), then AI n = A; similarly, if I m is the
m × m identity matrix, then I m A = A. Does
€ A have
€ a multiplicative inverse akin to its additive inverse
–A?
€
€
€
€
The results of Exercises 23 and 24, p. 117, show
that
• if there were a matrix C that satisfied CA = I n ,
then m must equal n; and
• if there were a matrix D that satisfied AD = I m ,
then m must equal n.
€
That is, only square matrices can have
multiplicative inverses. Still, even we restrict our
€
attention only to matrices with square
size,
multiplicative inverses are not guaranteed for
every matrix, as we shall see shortly.
€
€
The n × n matrix A is invertible if it has a
multiplicative inverse, that is, if there exists a
matrix C such that CA = I n and a matrix D such
that AD = I n . However, if these matrices exist,
then by the observation that
€
C = CI n = C ( AD) = ( CA ) D = I n D = D,
these two matrices are identical. In other words, A
can have only one associated matrix that acts as a
€multiplicative inverse. We typically denote the
mulitplicative inverse of A by A −1 .
It is instructive to look at the 2 × 2 case in detail. If


€ then its multiplicative
A =  a b  is invertible,
 c d
€
€
inverse is defined; denote it A
−1
w x
=
. Then
y
z


 1 0   a b   w x   aw + by ax + bz 
,

=
 ⋅
=
0 1   c d  € y z   cw + dy cx + dz 
€
leading to a pair of systems of linear equations, one
in w and y, the other in x and z:
aw + by = 1
cw + dy = 0
and
ax + bz = 0
cx + dz = 1
Solving the two systems produces
€
d
−c
−b
a
w=
,y=
and x =
,z =
ad − bc
ad − bc
ad − bc
ad − bc
thereby proving the
€
a b 
Theorem If A = 
, then A is invertible, with
c
d


inverse
€
A −1 =
1
 d −b 

,
ad − bc  − c a 
if and only if ad − bc ≠ 0 . //
€
€
The most important feature of this theorem is its
assertion that not all square matrices have
inverses. (See Practice Problem 1(c), p. 125.) Such
matrices are called singular. (You will also see
noninvertible as a synonym for singular, as well
as nonsingular as a synonym for invertible.)
€
€
Theorem Suppose A and B are invertible n × n
matrices. Then
1. A −1 is invertible, with inverse ( A −1 ) −1 = A;
2. AB is invertible, with inverse ( €
AB ) −1 = B −1 A −1 ;
3. A T is invertible, with inverse ( A T ) −1 = ( A −1 ) T .
€
Proof In each case, just€multiply the given matrix
with its supposed inverse and check that the
€
product is the identity matrix. //
How then do we tell whether an n × n matrix with
n > 2 is invertible? The answer can be formulated
once again in terms of row reduction.
€
€
The first step in this effort is to relate the
elementary row operations to matrix
multiplication.
As Example 5, p. 122, indicates, performing an
elementary row operation on an n × n matrix A can
be represented in terms of matrix multiplication: if
we perform a certain row operation to the n × n
identity I n and obtain the
€ matrix E (called an
elementary matrix), then performing the same
row operation on A produces the matrix
EA.
€
€
€
In particular, this means that elementary matrices
are invertible, for we know that row operations are
reversible: if we perform a row operation to I n and
obtain the matrix E, then there is a second row
operation that can be performed on E to return to
I n ; since this second operation can be represented
€
by multiplication by some elementary matrix F, we
have FE = I n . These row operations are reversible
in the other order as well, so EF = I n , showing that
E (and F) are invertible matrices. This idea is
€ central to the following
€
Theorem The n × n matrix A is invertible if and
only if A is row equivalent to I n . Furthermore, the
same sequence of row operations that carries A into
I n will€carry I n into A −1 .
€
Proof If A is invertible then any system of linear
equations with
€ coefficient matrix A, having matrix
€
form Ax = b, will have a solution obtained by
€
€
multiplying through by A −1 : x = A −1 b. Also, this is
the only possible solution, for if y were another
solution, then Ay = b ⇒ y = A −1 b = x . So the
system has a unique
€
€solution, meaning that in the
row reduction process on the augmented matrix
A b , every column and every row of A contains
€
a pivot entry. Thus, A will row reduce to I n .
Conversely, if A is row equivalent to I n , we can
represent the row operations in the reduction of A
to I n in terms of elementary matrices:
€
[
€
]
€
A ~ E1 A ~ E 2 ( E1 A ) ~ L ~ Ek ( Ek − 1 L E1 A ) = I n
€
€
Since each of these elementary matrices are
invertible, so is their product E = E k E k − 1 L E 1,
therefore EA = I n ⇒ E −1 EA = E −1 ⇒ A = E −1 , so A
is invertible with inverse E.
In the process of €
this argument, we have
−1
determined
that
A
= E = E k E k − 1 L E 1 is the
€
matrix obtained from the identity I n by applying
the same sequence of row operations that were used
to reduce€A to I n . //
€
It follows that to find A we can augment A by I n
to obtain
A I n , then apply row operations to
€
−1
[
]
reduce the left side to I n ; the matrix appearing on
€
the right will€have to be A −1 .
€
€
€
Bringing together a number of notions in our
discussion so far, we can now state
The Invertible Matrix Theorem Suppose that A
is an n × n matrix. Then the statements listed
below are logically equivalent (i.e., all are
simultaneously true or simultaneously false):
• A is invertible.
€
• There is an n × n matrix C satisfying CA = I n .
• There is an n × n matrix D satisfying AD = I n .
• A is row equivalent to I n .
• A has
€ n pivot entries.
€
• The€matrix equation Ax = 0 €
has only the trivial
solution x = 0.
€ A form a set of linearly
• The columns of
independent vectors.
€
n
• The
columns
of
A
span
R
.
€
• The linear transformation T:R n → R n defined by
the formula T( x ) = Ax is one-to-one.
• The linear transformation
T:R n → R n defined by
€
the formula T( x ) =
€ Ax is onto.
• For €
every vector b in R n , the matrix equation
Ax = b has at least
€ one solution.
• For €
every vector b in R n , the matrix equation
Ax = b has exactly
one solution. //
€
€
€
Corollary A is a noninvertible n × n matrix if and
€
only if every one of the statements above is false. //
€
Corollary The linear transformation T:R n → R n
with standard n × n matrix A is invertible if and
only if A is invertible. Further, the standard
matrix for the inverse function€is A −1 . //
€
€