22
Matrix Inverses
Let A be a square matrix of size n. A matrix B is called inverse of A if
AB = BA = I.
If A has an inverse, then we say that A is invertible. The inverse of A is unique and we
denote it by A−1 .
·
¸
·
¸
1 1
1 −1
−1
Example 20. Let A =
. Then A =
.
0 1
0 1
·
¸
·
¸
a b
d −b
1
−1
Example 21. If A =
and ad − bc 6= 0 then A = ad−bc
.
c d
−c a




1 4 −1
−3 −3 11
1 −3 .
Example 22. A =  2 7 1 . Then, A−1 = 12  1
1 3 0
−1 1 −1
Theorem 6. If a square matrix A, has a zero row (zero column), then A is not invertible.
Proof. Suppose that the i-th row of A is zero. For any square matrix C, the (i, i)-entry of
AC is equal to dot product of the i-th row of A with i-th column of C. Therefore, (i, i)-entry
of AC is 0. But, the (i, i)-entry of identity matrix is equal to 1. Thus,
AC 6= I
for any C.
Therefore, A is not invertible.



1 12 13 3

 0 0 0 0 

 and B = 
Example 23. A = 

 9 9 1 1 

0 9 3 2
1
2
1
4
4
0
0
0
0
0
5
9
8
2
6

9 11
5 1 

2 1 
 are not invertible.
1 1 
7 7
Theorem 7. If A and B are square matrices, then
• If A is invertible, then A−1 is invertible, and (A−1 )−1 = A.
• If A and B are invertible, then AB is invertible, and(AB)−1 = B −1 A−1 .
• If A is invertible, then AT is invertible, and(AT )−1 = (A−1 )T .
• If A is invertible, then 1c A is invertible for a number c 6= 0, and (cA)−1 = 1c A−1 .
Example 24. A,B, and C are invertible matrices. Simplify
C T B(AB)−1 [C −1 AT ]T .
23
Solution:
C T B(AB)−1 [C −1 AT ]T = C T B(B −1 A−1 )[(AT )T (C −1 )T ]
= C T IA−1 [A(C −1 )T ]
= C T I(C −1 )T
= I.
Example 25. Find the matrix A.
¸
·
¤T
£
−1
1 1
− 5A−1 = ( AT )−1 .
2
−2 3
4
Solution:
·
2
¸T
1 1
−2 3
Thus,
− 5(AT )−1 = −4(AT )−1
·
2
1 1
−2 3
Then,
1
A=
10
¸T
= (A−1 )T .
·
3 −1
2 1
¸
.
Theorem 8. Let A be a square matrix of size n. If A is invertible then the system
AX = B
has a unique solution.
Proof. Multiply both side of equation by A−1 . Then,
A−1 AX = A−1 B
IX = A−1 B
X = A−1 B
Therefore, the only solution for this system is X = A−1 B.


1 −2 2
Example 26. Let A =  2 1 1 .
1 0 1


1
2 −4
1. Show that A−1 =  −1 −1 3  .
−1 −2 5
24
2. Solve the following system of linear equations:
x +2y −4z = 3
−x −y 3z = 0 .
x
z
= −2






1 −2 2
1
2 −4
1
2 −4
1 −2 2
Solution. Since,  2 1 1   −1 −1 3  = I3 , and  −1 −1 3   2 1 1  =
−1 −2 5
1 0 1
 1 0 1
−1 −2 5
1
2 −4
I3 , then A−1 =  −1 −1 3  . By Theorem 8, the unique solution for the system is
−1 −2 5


 

1
2 −4
3
11
X = A−1 B =  −1 −1 3   0  =  −9  .
−1 −2 5
−2
−13
Theorem 9. If A is a square matrix of size n and there is a square matrix C such that
AC = I, then CA = I.
Proof. consider the homogeneous system CX = 0. Multiply both side with A, we get
X = 0. Let Ei be the column with the entry in i-th row and zero else where. Thus, the
system CX = Ei has a solution for each i. This exactly means that there is a matrix F
such that CF = I. Now,
AC
ACF
AI
A
=
=
=
=
I
IF
F
F.
Thus, CA = I.
Theorem 10. A matrix A is invertible if and only if its reduced row-echelon form is equal
to identity matrix.
Proof. Let A be invertible. If rankA = r < n, then the solutions for AX = 0 has n − r
parameters, and therefore there are infinitely many solutions. This is a contradiction with
the Theorem 8. Thus, rankA = n. This means that the reduced row-echelon form of A
is In . Conversely, if the reduced row-echelon form of A is identity, then AX = Ei has a
solution. Here, Ei is the column with the entry in i-th row is one and other entries are zero.
This exactly means that there is a matrix C such that AC = In . Thus, A is invertible.
Inversion Algorithm
Let A be an invertible matrix, then A can be changed to identity by a series of row operations. If we apply the same operations to identity we get A−1 . This method is written
shortly by
[A I] → [I A−1 ].
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In the following example, we see that why inversion algorithm works.
Example 27. Find the inverse of


2 7 1
A =  1 4 −1 
1 3 0
First, we find the

2
 1
1
reduced row-echelon form of A.





7 1
1 4 −1
1 4 −1
R1 ↔R2
1 ,R3 −R1
 0 −1 3 
 2 7 1  R2 −2R
4 −1  −→
−→
3 0
1 3 0
0 −1 1






1 0 11 −R , 1 R
1 0 7
1 0 0
2 2 3
R1 +4R2 ,R3 −R2
3 ,R1 −11R3
 0 −1 3  −→
 0 1 −3  R2 +3R−→
 0 1 0 .
−→
0 0 2
0 0 1
0 0 1
Since the reduced row echelon form is I3 ,

x1 x2
A−1 =  y1 y2
z1 z2


xi
Here, Ci =  yi , for i = 1, · · · 3. Then,
zi
then A−1 exists by Theorem 10. Let

x3
£
¤
y3  = C1 C2 C3 .
z3
AA−1 = [AC1 , AC2 , AC3 ] = In
Therefore, we have 3 linear system of linear of equations:
 
1

0  =: E1
AC1 =
0
 
0
AC2 =  1  =: E2
0
 
0
AC3 =  0  =: E3 .
1
To find A−1 , we need to find C1 , C2 , C3 . As usual, we need to change [A E1 ], [A E2 ], [A E3 ]
to reduced row-echelon form. In other words, to find Ci we have to apply the same row
26
operations(we applied to get identity from A) on Ei . To save the time we can change all
these three systems at the same time.
[A I] = [A E1 E2 E3 ] → [I C1 C2 C3 ] = [I A−1 ].
In this example,


2 7 1 1 0 0
1 0 0
 1 4 −1 0 1 0  −→  0 1 0
1 3 0 0 0 1
0 0 1

Thus

A−1 = 
−3
2
1
2
1
2
−3
2
1
2
−1
2
11
2
−3
2
−1
2
−3
2
1
2
1
2
−3
2
1
2
−1
2
11
2
−3
2
−1
2

.

.
Definition 8. Two statements P and Q are called equivalent if P ⇒ Q and Q ⇒ P . Thus,
P is true if and only if Q is true. For example, two statements P : x = 10 and Q : x = 12−2
are equivalent.
Theorem 11. The following conditions are equivalent for an n × n matrix A:
• A is invertible.
• The only solution for the homogeneous system AX = 0 is the trivial solution X = 0.
• The reduced row-echelon form of A is identity.
• For any column B, the system AX = B is consistent.
• There is an n × n matrix C, such that AC = I.
These five equivalent conditions enable us to solve problems related to linear systems by our
knowledge about inverse matrices and vise versa.
Example 28. Suppose that the only solution for the homogeneous systems AX = 0 and
BX = 0 is X = 0. Show that ABX = 0 does not have any non-trivial solution.
Solution. The only solution for the homogeneous systems AX = 0 and BX = 0 is
X = 0, thus A and B are invertible by Theorem 11. By Theorem 7, AB is invertible and
again Theorem 11 shows that X = 0 is the only solution for the ABX = 0.