Higher Engineering Mathematics

F

Matrices and Determinants

26

The solution of simultaneous

equations by matrices and

determinants

26.1 Solution of simultaneous

equations by matrices

(a) The procedure for solving linear simultaneous

equations intwo unknowns using matricesis:

(i) write the equations in the form

a 1 x+b 1 y=c 1

a 2 x+b 2 y=c 2

(ii) write the matrix equation corresponding to

these equations,

i.e.

(

a 1 b 1

a 2 b 2

)

×

(

x

y

)

=

(

c 1

c 2

)

(iii) determine the inverse matrix of

(

a 1 b 1

a 2 b 2

)

i.e.

1

a 1 b 2 −b 1 a 2

(

b 2 −b 1

−a 2 a 1

)

(from Chapter 25)

(iv) multiply each side of (ii) by the inverse

matrix, and

(v) solve forxandyby equating corresponding

elements.

Problem 1. Use matrices to solve the simulta-

neous equations:

3 x+ 5 y− 7 = 0 (1)

4 x− 3 y− 19 =0(2)

(i) Writing the equations in thea 1 x+b 1 y=cform

gives:

3 x+ 5 y= 7

4 x− 3 y= 19

(ii) The matrix equation is

(

35

4 − 3

)

×

(

x

y

)

=

(

7

19

)

(iii) The inverse of matrix

(

35

4 − 3

)

is

1

3 ×(−3)− 5 × 4

(

− 3 − 5

− 43

)

i.e.

⎛

⎜

⎝

3

29

5

29

4

29

− 3

29

⎞

⎟

⎠

(iv) Multiplying each side of (ii) by (iii) and remem-

bering thatA×A−^1 =I, the unit matrix, gives:

(

10

01

)(

x

y

)

=

⎛

⎜

⎝

3

29

5

29

4

29

− 3

29

⎞

⎟

⎠×

(

7

19

)

Thus

(

x

y

)

=

⎛

⎜

⎝

21

29

+

95

29

28

29

−

57

29

⎞

⎟

⎠

i.e.

(

x

y

)

=

(

4

− 1

)

(v) By comparing corresponding elements:

x= 4 and y=− 1

Checking:

equation (1),

3 × 4 + 5 ×(−1)− 7 = 0 =RHS

equation (2),

4 × 4 − 3 ×(−1)− 19 = 0 =RHS