Higher Engineering Mathematics

268 MATRICES AND DETERMINANTS

Problem 2. Subtract

(a)

(

− 30

7 − 4

)

from

(

2 − 1

− 74

)

and

(b)

(

27 − 5

−21 0

63 4

)

from

(

31 − 4

43 1

14 − 3

)

To find matrixAminus matrixB, the elements of
Bare taken from the corresponding elements ofA.
Thus:

(a)

(

2 − 1

− 74

)

−

(

− 30

7 − 4

)

=

(

2 −(−3) − 1 − 0

− 7 − 74 −(−4)

)

=

(

5 − 1

− 14 8

)

(b)

(

31 − 4

43 1

14 − 3

)

−

(

27 − 5

−21 0

63 4

)

=

(

3 − 21 − 7 − 4 −(−5)

4 −(−2) 3− 11 − 0

1 − 64 − 3 − 3 − 4

)

=

(

1 − 61

621

− 51 − 7

)

Problem 3. If

A=

(

− 30

7 − 4

)

,B=

(

2 − 1

− 74

)

and

C=

(

10

− 2 − 4

)

findA+B−C.

A+B=

(

− 1 − 1

00

)

(from Problem 1)

Hence,A+B−C=

(

− 1 − 1

00

)

−

(

10

− 2 − 4

)

=

(

− 1 − 1 − 1 − 0

0 −(−2) 0 −(−4)

)

=

(

− 2 − 1

24

)

AlternativelyA+B−C

=

(

− 30

7 − 4

)

+

(

2 − 1

− 74

)

−

(

10

− 2 − 4

)

=

(

− 3 + 2 − 10 +(−1)− 0

7 +(−7)−(−2) − 4 + 4 −(−4)

)

=

(

− 2 − 1

24

)

as obtained previously

(iii) Multiplication

When a matrix is multiplied by a number, called

scalar multiplication, a single matrix results in

which each element of the original matrix has been

multiplied by the number.

Problem 4. IfA=

(

− 30

7 − 4

)

,

B=

(

2 − 1

− 74

)

andC=

(

10

− 2 − 4

)

find

2 A− 3 B+ 4 C.

For scalar multiplication, each element is multiplied

by the scalar quantity, hence

2 A= 2

(

− 30

7 − 4

)

=

(

− 60

14 − 8

)

3 B= 3

(

2 − 1

− 74

)

=

(

6 − 3

− 21 12

)

and 4 C= 4

(

10

− 2 − 4

)

=

(

40

− 8 − 16

)

Hence 2A− 3 B+ 4 C

=

(

− 60

14 − 8

)

−

(

6 − 3

−21 12

)

+

(

40

− 8 − 16

)

=

(

− 6 − 6 + 40 −(−3)+ 0

14 −(−21)+(−8) − 8 − 12 +(−16)

)

=

(

− 83

27 − 36

)

When a matrixAis multiplied by another matrixB,a

single matrix results in which elements are obtained

from the sum of the products of the corresponding

rows ofAand the corresponding columns ofB.

Two matricesAandBmay be multiplied together,

provided the number of elements in the rows of

matrixAare equal to the number of elements in the

columns of matrixB. In general terms, when multi-

plying a matrix of dimensions (mbyn) by a matrix of

dimensions (nbyr), the resulting matrix has dimen-

sions (mbyr).Thusa2by3matrix multiplied by a

3 by 1 matrix gives a matrix of dimensions 2 by 1.