242 Higher Engineering Mathematics
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
(b) The procedure for solving linear simultaneous
equations inthree unknowns using matricesis:
(i) write the equations in the form
a 1 x+b 1 y+c 1 z=d 1
a 2 x+b 2 y+c 2 z=d 2
a 3 x+b 3 y+c 3 z=d 3
(ii) write the matrix equation corresponding to
these equations, i.e.
⎛
⎝
a 1 b 1 c 1
a 2 b 2 c 2
a 3 b 3 c 3
⎞
⎠×
⎛
⎝
x
y
z
⎞
⎠=
⎛
⎝
d 1
d 2
d 3
⎞
⎠
(iii) determine the inverse matrix of
⎛
⎝
a 1 b 1 c 1
a 2 b 2 c 2
a 3 b 3 c 3
⎞
⎠(see Chapter 22)
(iv) multiplyeach side of (ii) by the inverse
matrix, and
(v) solve forx,yandzby equating the corre-
sponding elements.
Problem 2. Use matrices to solve the
simultaneous equations:
x+y+z− 4 =0(1)
2 x− 3 y+ 4 z− 33 =0(2)
3 x− 2 y− 2 z− 2 =0(3)
(i) Writingtheequations in thea 1 x+b 1 y+c 1 z=d 1
form gives:
x+y+z= 4
2 x− 3 y+ 4 z= 33
3 x− 2 y− 2 z= 2
(ii) The matrix equation is
⎛
⎝
111
2 − 34
3 − 2 − 2
⎞
⎠×
⎛
⎝
x
y
z
⎞
⎠=
⎛
⎝
4
33
2
⎞
⎠
(iii) The inverse matrix of
A=
⎛
⎝
111
2 − 34
3 − 2 − 2
⎞
⎠
is given by
A−^1 =
adjA
|A|
The adjoint ofAis the transpose of the matrix of
the cofactors of the elements (see Chapter 22). The
matrix of cofactors is
⎛
⎝
14 16 5
0 − 55
7 − 2 − 5
⎞
⎠
and the transpose of this matrix gives
adjA=
⎛
⎝
14 0 7
16 − 5 − 2
55 − 5
⎞
⎠
The determinant ofA, i.e. the sum of the products
of elements and their cofactors, using a first row
expansion is
1
∣
∣
∣
∣
− 34
− 2 − 2
∣
∣
∣
∣−^1
∣
∣
∣
∣
24
3 − 2
∣
∣
∣
∣+^1
∣
∣
∣
∣
2 − 3
3 − 2
∣
∣
∣
∣
=( 1 × 14 )−( 1 ×(− 16 ))+( 1 × 5 )= 35
Hence the inverse ofA,
A−^1 =
1
35
⎛
⎝
14 0 7
16 − 5 − 2
55 − 5
⎞
⎠
(iv) Multiplyingeach side of (ii) by (iii), and remem-
bering thatA×A−^1 =I, the unit matrix, gives