Schaum's Outline of Discrete Mathematics, Third Edition (Schaum's Outlines)

412 VECTORS AND MATRICES [APP. A

EXAMPLE A.4 LetA=

[

1 − 23

045

]

andB=

[

468

1 − 3 − 7

]

. Then

A+B=

[

1 + 4 − 2 + 63 + 8

0 + 14 +(− 3 ) 5 +(− 7 )

]

=

[

54 11

11 − 2

]

3 A=

[

3 ( 1 ) 3 (− 2 ) 3 ( 3 )

3 ( 0 ) 3 ( 4 ) 3 ( 5 )

]

=

[

3 − 69

01215

]

2 A− 3 B=

[

2 − 46

0810

]

+

[

− 12 − 18 − 24

− 3921

]

=

[

− 10 − 22 − 18

− 31731

]

Matrices under matrix addition and scalar multiplication have the following properties.

Theorem A.1: LetA,B,Cbe matrices with the same size, and letkandk′be scalars. Then:

(i)(A+B)+C=A+(B+C) (v)k(A+B)=kA+kB

(ii)A+ 0 = 0 +A(vi)(k+k′)A=kA+k′A

(iii)A+(−A)=(−A)+ 0 =A (vii)(kk′)A=k(k′A)

(iv)A+B=B+A(viii) 1 A=A

Note first that the 0 in (ii) and (iii) refers to the zero matrix. Also, by (i) and (iv), any sum of matrices

A 1 +A 2 +···+An

requires no parentheses, and the sum does not depend on the order of the matrices. Furthermore, using (vi) and
(viii), we also have
A+A= 2 A, A+A+A= 3 A, ···

Lastly, sincen-component vectors may be identified with either 1×norn×1 matrices, Theorem A.1 also holds
for vectors under vector addition and scalar multiplication.
The proof of Theorem A.1 reduces to showing that theijentries on both sides of each matrix equation are
equal.

A.5Matrix Multiplication

The product of matricesAandB, writtenAB, is somewhat complicated. For this reason, we first begin with a
special case. (The reader is referred to Section 3.5 for a discussion of the summation symbol, the Greek capital
letter sigma.)
The productABof a row matrixA=[ai]and a column matrixB=[bi]with the same number of elements
is defined as follows:

AB=[a 1 ,a 2 ,...,an]

⎡

⎢

⎢

⎣

b 1

b 2

..

bn

⎤

⎥

⎥

⎦=a^1 b^1 +a^2 b^2 +···+anbn=

∑n

k= 1

akbk

That is,ABis obtained by multiplying corresponding entries inAandBand then adding all the products. We
emphasize thatABis a scalar (or a 1×1 matrix). The productABis not defined whenAandBhave different
numbers of elements.

EXAMPLE A.5

(a) [ 7 ,− 4 , 5 ]

⎡

⎣

3

2

− 1

⎤

⎦= 7 ( 3 )+(− 4 )( 2 )+ 5 (− 1 )= 21 − 8 − 5 = 8