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=− 1Checking:
equation (1),3 × 4 + 5 ×(− 1 )− 7 = 0 =RHSequation (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 forma 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 ofA=⎛
⎝111
2 − 34
3 − 2 − 2⎞
⎠is given byA−^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 givesadjA=⎛
⎝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 is1∣
∣
∣
∣− 34
− 2 − 2∣
∣
∣
∣−^1∣
∣
∣
∣24
3 − 2∣
∣
∣
∣+^1∣
∣
∣
∣2 − 3
3 − 2∣
∣
∣
∣=( 1 × 14 )−( 1 ×(− 16 ))+( 1 × 5 )= 35Hence 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