Higher Engineering Mathematics, Sixth Edition

232 Higher Engineering Mathematics

(a) Adding the corresponding elements gives:

(

2 − 1

− 74

)

+

(

− 30

7 − 4

)

=

(

2 +(− 3 ) − 1 + 0

− 7 + 74 +(− 4 )

)

=

(

− 1 − 1

00

)

(b) Adding the corresponding elements gives:

⎛

⎝

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

⎞

⎠

=

⎛

⎝

58 − 9

24 1

77 1

⎞

⎠

(ii) Subtraction of matrices

IfAis a matrix andBis another matrix, then (A−B)

is a single matrix formed by subtracting the elements of

Bfrom the corresponding elements ofA.

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 ofBare

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, calledscalar

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.