In this section, we introduce the concept of invertible matrix and examine special invertible matrices that are intimately associated with elementary row operations, the elementary matrices.
For any real number a =0, there is a unique real number b, called the multi- plicative inverse of a, with the property that ab =ba =1. For example, if a =2, then b =1/2. In the context of matrices, the identity matrix In is a multiplicative identity; so it is natural to ask for what matricesA does there exist a matrixB such thatAB =BA=In. Notice that this last equation is possible only if bothAandB are n×n matrices. This discussion motivates the following definitions:
Definitions Ann×n matrixAis calledinvertibleif there exists ann×n matrixB such thatAB =BA=In. In this case,B is called aninverseof A.
IfA is an invertible matrix, then its inverse is unique. For if both B andC are inverses ofA, then AB =BA=In andAC =CA=In. Hence
B =BIn =B(AC)=(BA)C =InC =C.
When A is invertible, we denote the unique inverse of A by A−1, so that AA−1= A−1A=In. Notice the similarity of this statement and 2ã2−1=2−1ã2=1, where 2−1 is the multiplicative inverse of the real number 2.
Example 1 LetA=1 2 3 5
andB =
−5 2 3 −1
. Then AB =
1 2
3 5 −5 2
3 −1
= 1 0
0 1
=I2,
and
BA=
−5 2
3 −1 1 2
3 5
= 1 0
0 1
=I2. SoAis invertible, and B is the inverse ofA; that is,A−1=B.
Practice Problem 1 䉴 IfA=
−1 0 1
1 2 −2
2 −1 −1
andB =
4 1 2 3 1 1 5 1 2
, isB =A−1? 䉴
Because the roles of the matricesAandBare the same in the preceding definition, it follows that if B is the inverse of A, then A is also the inverse of B. Thus, in Example 1, we also have
B−1=A= 1 2
3 5
.
Just as the real number 0 has no multiplicative inverse, then×n zero matrixO has no inverse becauseOB =O=In for anyn×n matrixB. But there are also other square matrices that are not invertible; for example,A=
1 1 2 2
. For ifB = a b
c d
is any 2×2 matrix, then AB =
1 1 2 2
a b c d
=
a+c b+d 2a+2c 2b+2d
.
Since the second row of the matrix on the right equals twice its first row, it cannot be the identity matrix
1 0 0 1
. SoB cannot be the inverse of A, and henceA is not invertible.
In the next section, we learn which matrices are invertible and how to compute their inverses. In this section, we discuss some elementary properties of invertible matrices.
The inverse of a real number can be used to solve certain equations. For example, the equation 2x =14 can be solved by multiplying both sides of the equation by the inverse of 2:
2−1(2x)=2−1(14) (2−12)x =7
1x =7 x =7
In a similar manner, ifAis an invertiblen×n matrix, then we can useA−1 to solve matrix equations in which an unknown matrix is multiplied by A. For example, ifA is invertible, then we can solve the matrix equationAx=bas follows:2
Ax=b A−1(Ax)=A−1b
2Although matrix inverses can be used to solve systems whose coefficient matrices are invertible, the method of solving systems presented in Chapter 1 is far more efficient.
(A−1A)x=A−1b Inx=A−1b x=A−1b
IfAis an invertiblen×n matrix, then for everybinRn,Ax=bhas the unique solution A−1b.
In solving a system of linear equations by using the inverse of a matrix A, we observe thatA−1“reverses” the action ofA; that is, if Ais an invertiblen×n matrix andu is a vector inRn, thenA−1(Au)=u. (See Figure 2.7.)
A1(Au) u Au
multiply by A1 multiply
by A
Rn Rn
Figure 2.7 Multiplication by a matrix and its inverse
Example 2 Use a matrix inverse to solve the system of linear equations x1+2x2=4
3x1+5x2=7.
Solution This system is the same as the matrix equationAx=b, where A=
1 2 3 5
, x= x1
x2
, and b= 4
7
.
We saw in Example 1 thatAis invertible. Hence we can solve this equation forx by multiplying both sides of the equation on the left by
A−1=
−5 2 3 −1 as follows:
x1
x2
=x=A−1b=
−5 2 3 −1
4 7
= −6
5
Thereforex1 = −6 andx2=5 is the unique solution of the system.
Practice Problem 2 䉴 Use the answer to Practice Problem 1 to solve the following system of linear equations:
−x1 + x3= 1 x1+2x2−2x3= 2
2x1− x2− x3= −1 䉴
Example 3 Recall the rotation matrix
Aθ =
cosθ −sinθ sinθ cosθ
considered in Section 1.2 and Example 4 of Section 2.1. Notice that for θ=0◦, Aθ =I2. Furthermore, for any angleα,
AαA−α=Aα+(−α)=A0◦ =I2.
Similarly, A−αAα=I2. Hence Aα satisfies the definition of an invertible matrix with inverseA−α. Therefore (Aα)−1=A−α.
Another way of viewingA−αAα is that it represents a rotation byα, followed by a rotation by−α, which results in a net rotation of 0◦. This is the same as multiplying by the identity matrix.
The following theorem states some useful properties of matrix inverses:
THEOREM 2.2
LetAandB ben×n matrices.
(a) If Ais invertible, thenA−1 is invertible and (A−1)−1=A.
(b) IfA andB are invertible, thenAB is invertible and (AB)−1 =B−1A−1. (c) If Ais invertible, thenAT is invertible and (AT)−1=(A−1)T.
PROOF The proof of (a) is a simple consequence of the definition of matrix inverse.
(b) Suppose thatAandB are invertible. Then
(AB)(B−1A−1)=A(BB−1)A−1=AInA−1=AA−1=In.
Similarly, (B−1A−1)(AB)=In. HenceAB satisfies the definition of an invertible matrix with inverseB−1A−1; that is, (AB)−1=B−1A−1.
(c) Suppose thatAis invertible. ThenA−1A=In. Using Theorem 2.1(g), we obtain
AT(A−1)T =(A−1A)T =InT =In.
Similarly, (A−1)TAT =In. HenceAT satisfies the definition of an invertible matrix with the inverse (A−1)T; that is, (AT)−1 =(A−1)T.
Part (b) of Theorem 2.2 can be easily extended to products of more than two matrices.
LetA1,A2,. . .,Ak be nìn invertible matrices. Then the productA1A2ã ã ãAk is invertible, and
(A1A2ã ã ãAk)−1 =(Ak)−1(Ak−1)−1ã ã ã(A1)−1.