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Inverse Matrices and Inverse
Mappings

3
...
We will make
many connections with the material we have covered so far and provide useful tools
for the material contained in the rest of the book
...


If there exists an

n x n matrix B such that AB

=

I

=

BA, then

A is said to be invertible, and B is called the inverse of A (and A is the inverse of B)
...


the

inverse of A
...
is depends on the easily

Theorem 1

Let A be a square matrix and suppose that BA = AB = I and CA = AC = I
...


Proof: We have B=BI=B(AC)=(BA)C=IC=C
...
Sometimes we say that if BA = I, then B is a
"left inverse" of A
...
The proof
shows that for a square matrix, any left inverse must be equal to any right inverse
...
We will now show that for square matrices, a right inverse is
automatically a left inverse
...
Then BA = I, so that

B =A-1• Moreover, B and A have rank

n
...
Suppose that B has rank
less than n
...
2
...
But this means that for some non-zero x, AB1 = A(Bx) = AO = 0
...
Hence, B must have
rank

n
...
11 by Theorem 2
...
2
...
11, so BA=I by Theorem 3
...
4
...

Theorem 2 makes it very easy to prove some useful properties of the matrix in­
verse
...


Theorem 3

Suppose that A and B are invertible matrices and that t is a non-zero real number
...
Fortunately, one procedure answers both questions
...
If a solution X can be found, then X = A- by Theorem 2
...

To keep it simple, the procedure is examined in the case where A is 3 x 3, but it

should be clear that it can be applied to any square matrix
...
Note

that each system has a different standard basis vector as its right-hand side, but all
have the same coefficient matrix
...
Therefore, write

and row reduce to reduced row echelon form to solve
...
It follows that the first column of the
�

=

desired matrix X is b1, the second column of X is b2, and so on
...
Hence, A is not
invertible, since Theorem 2 tells us that if A is invertible, then rank(A) = n
...


Algorithm 1
Finding A-1

To find the inverse of a square matrix A,

(1) Row reduce the multi-augmented matrix
reduced row echelon form
...

(3) If the reduced row echelon form of A is not/, then A is not invertible
...


3

II

A

J and row reduce:

l
ol
[1
1
1
1 1 1 -1 1
[l
l
[
1
[
1
-1
j]
[-l : j]
...


EXAMPLE3

Determine whether A =

[ � �]

Solution: Write the matrix

[

A

is invertible, and if it is, determine its inverse
...


EXERCISE 1

Determine whether A =

[� �]

is invertible, and if it is, determine its inverse
...
It is worth stating
them clearly as a theorem
...


Theorem 4

[Invertible Matrix Theorem]
Suppose that A is an

n xn

matrix
...


Theorem 4
(continued)

(1) A is invertible
...

(3) The reduced row echelon form of A is/
...
n, the system Ax= b is consistent and has a unique solution
...

(6) The columnspace of A is lln
...
)

(1)
(2)

=>
=>

nx n
...
" It is common

(4) :: (5)

::

(6)

::

(1),

(2): This is the second part of Theorem 2
...
2
...


(5): Assume that the only solution to Ax = 0 is the trivial solution
...
a,, , then 0 = Ax = X1 il1 +
...
=
Xn =0
...
'an of A are linearly independent
...
Thus, Ax = b is consistent for all

b

E IR
...


(6)

::

(1): If Col(A) = IR
...
Thus, A is

invertible
...
However, the way we proved
the theorem does not immediately tell us how to find the unique solution
...

Let A be an invertible square matrix and consider the system Ax=

b
...

_,

_,

EXAMPLE4

Let A=

[� n

Solution: By Example 1, A-1 =

EXERCISE 2

Let A=

[

_

1
2

[�]
[ � -n

Find the solution of Ax=

_


...

and use 1t to find the solution of Ax=b
...

7
1

]

_,

[ ]

·

·

_,

It likely seems very inefficient to solve Exercise 2 by the method described
...
However, observe that if we wanted to solve many systems of equations
with the same coefficient matrix A, we would need to compute A-1 once, and then
each system can be solved by the problem of solving the system to simple matrix
multiplication
...
Of course, with some thought, one realizes that the elementary row operations
are "contained" inside the inverse of the matrix (which we obtained by row reducing)
...


Inverse Linear Mappings
It is useful to introduce the inverse of a linear mapping here because many geometrical
transformations provide nice examples of inverses
...
Recall that the identity transformation Id is the linear mapping
defined by ld( x)= x for all x
...
-- JR
...
11
JR
...


Theorem 5

n
Suppose that L : JR
...
11 is a linear mapping with standard matrix [L] = A and
n
that M : JR
...
11 is a linear mapping with standard matrix [M] = B
...


Proof: By Theorem 3
...
5, [Mo L] = [M][L]
...


•

For many of the geometrical transformations of Section 3
...


EXAMPLES

For each of the following geometrical transformations, determine the inverse transfor­
mation
...

(a) The rotation Re of the plane
(b) In the plane, a stretch T by a factor oft in the x1-direction

Solution: (a) The inverse transformation is to just rotate by -e
...
Hence,

�

c se

sine

- sm e

cose

]

cos2 e + sin2e

cose sine - cose sine

- sine cose+sine cose

cos2e+sin2e

1
(b) The inverse transformation r- is a stretch by a factor of

EXERCISE 3

=

1

] [ ]
=

O

0
1

? in the Xi-direction:

For each of the following geometrical transformations, determine the inverse transfor­
mation
...

(a) A reflection over the line x

=

2

x1 in the plane

(b) A shear in the plane by a factor oft in the x1 -direction

Observe that if

y

E

JR
...
Therefore, it follows that if L has an inverse, the range of L
must be all of the codomain JR
...
11, there exists a unique x E JR" such that
y
...


E

Theorem 6

[Invertible Matrix Theorem, cont
...

Then, the following statements are equivalent to each other and to the statements of
Theorem 4
...

(8) Range(L)=]Rn
...


Proof: (We use P <:=> Q to mean "P if and only if Q
...

(4) <:=> (8): Since L(x)=Ax, we know that for every b E JRll there exists a x E JR11 such
n
that L(x)=bif and only if Ax=bis consistent for every b E IR
...
Hence, there is a linear mapping L-1 such that
L-1(L(x)) = x
...
But, this
gives
x=L-1(0) =0
_,

_,

since L-1 is linear
...

(9) � (3): Assume that Null(L) = {O}
...

Thus, rank(A)=n, by Theorem 2
...
2
...
You should be able to start with any one of the nine statements and show
that it implies any of the other eight
...

Solution: By definition, projil(x) = tV, for some t E R Hence, any vector y
that is not a scalar multiple of v is not in the range of prok Thus, Range(projil)
hence projil is not invertible, by Theorem 6
...

Solution: Assume that x is in the nullspace of L
...
Thus, Null(L)= {0} and hence L
is invertible, by Theorem 6
...
Here is an example that illustrates for

linear mappings that the domain and codomain of L must be the same if it is to have
an inverse
...
4 � JR
...
� JR
...


EXAMPLE 8

It is easy to see that Poinj = Id but that injoP 1- Id
...
Notice that P satisfies the condition that its range is all of its codomain, but it fails
the condition that its nullspace is trivial
...


PROBLEMS 3
...
Check
by multiplication
...

(a) B

X

=

-

(b)

(c)

[� ! il
[i � � ]
[� : l

(e)

(f)

1
2
2
6

3

1
0
0
0
0

0
1
0
0
0

1
0

0

0
0

1
0

A2 Let B =

3

(c) B

X

=

[ � �]

and B =

(a) Find A-1 andB-1•

1
0
2
0

7

=
X

A3 Let A =

0

(d)

(b) B

[;]
nl
[[ : ]{�]]

(b) Calculate

and (ABt1
(ABt1 = s-1A-1•

1
0

AB

and check that

(c) Calculate(3At1 and checkthat it equals

t A-1•

(d) Calculate(A7t1 and checkthat A7(A7t1 =I
...

(a) The matrix of the rotation Rn:;6 in the plane
...

(a) Determine the matrix of the shear S by a factor
of 2 in the x2-direction and the matrix of S -1•
(b) Determine the matrix of the reflection R in the
line x
x2
0 and the matrix of R-1
...


A6 Suppose that L
that

M

11
: JR - JR" is a linear mapping and

: JR11 - JR11 is a function (not assumed to be

linear) such that x
Show that

=

M(J) if and only if y

=

L(x)
...


=

Homework Problems
Bl For each of the following matrices, either show that
the matrix is not invertible or find its inverse
...

(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

U)

[-� �]

[� : �l
[� 3 �]
[� i -�1
[f -� �]
-

1
0
2
1

-1

0
0
1
3

2
1
3
6

2
0

1
0
0
0

0
0
0
1
0

0
0

0
0
0
0

0
1
3
1

-2
0

0
0
0
0

0

1
0
o
0

B2 Let A

0
1
o
0
1

=

0
0
1
0

[�

-

0
0
o
1

� 1-

�

-1
(a) Find A-1
...

(AB)-1

=

=

B4 By geometrical arguments, determine the inverse of
each of the following matrices
...

(b)

[� �]

(c)

(d)

[�

- 1/

[-�

�]

-

�

0

0

B6 For each of the following pairs of linear mappings

�i

from JR
...

(a) Determine the matrix of the stretch

S

-t

2
JR
...

(b) Determine the matrix of the reflection
line x1 + x2

=

0 and the matrix of

R

R-1•

3

3
JR
...

(a) R is the rotation about the x1 -axis through angle
7r /2, and S is the stretch by a factor of 0
...


(b)

R is the reflection in the plane x1 - x3 0, and
S is a shear by a factor of 0
...


1

in the

(c) Determine the matrix of (Ros)-1 and the matrix

(S o R)-1 (without
of R o S and S o R)
...
23
2
...
11
2
...
11
-2
...
01

0
...


3
...
04

(b) Use computer software to calculate the inverse

0
...
11

-1
...
05


...
56
1
...
Explain your answer
...
Find an expression for A-1 in terms
of A
...
)
Suppose that B satisfies B5 + B3 + B
I
...

=

D3 Prove that if A and B are square matrices such that

l�

AB is invertible, then A and B are invertible
...


(b) Show that there cannot exist a matrix C such
that

CA =

I
...


DS Prove that the following are equivalent for an n
matrix

A
...

{O}
...


is invertible

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