422 VECTORS AND MATRICES [APP. A
EXAMPLE A.12
Find the inverse ofA=
⎡
⎣102
2 − 13
418⎤
⎦.Form the matrixM=(A, I )and reduceMto echelon form:M=[
102 ... 100
2 − 13 .... 010
418 .... 001]
∼[
102 ... 100
0 − 1 − 1 .... 210
010 ....− 401]
∼[
102 ... 100
0 − 1 − 1 ....− 210
00 − 1 ....− 611]In echelon form, the left half ofMis in triangular form; henceAis invertible. Further row reduceMto row
canonical form:
M∼[
100 ...− 1122
0 − 10 .... 40 − 1
001 .... 6 − 1 − 1]
∼[
100 ...− 1122
010 ....− 401
001 .... 6 − 1 − 1]The identity matrix is in the left half of the final matrix; hence the right half isA−^1. In other words,
A−^1 =⎡
⎣− 1122
− 401
6 − 1 − 1⎤
⎦A.11Boolean (Zero-One) Matrices
Thebinary digitsorbitsare the symbols 0 and 1. Consider the following operations on these digits:+ 01
0 01
1 11× 01
0 00
1 01Viewing these bits as logical values (0 representing FALSE and 1 representing TRUE), the above operations
correspond, respectively, to the logical operations of OR(∨)and AND(∧); that is,
∨ FT
F FT
T TT∧ FT
F FF
T FT(The above operations on 0 and 1 are calledBoolean operationssince they also correspond to the operations of
a Boolean algebra discussed in Chapter 15.)
Now letA=[
aij]
be a matrix whose entries are the bits 0 and 1 subject to the above Boolean operations.
ThenAis called aBoolean matrix. TheBoolean productof two such matrices is the usual product except that
now we use the Boolean operations of addition and multiplication. For example, if
A=[
11
10]
andB=[
01
01]
,thenAB=[
0 + 01 + 1
0 + 01 + 0]
=[
01
01]One can easily show that ifAandBare Boolean matrices, then the Boolean productABcan be obtained by finding
the usual product ofAandBand then replacing any nonzero digit by 1.