278 MATRICES AND DETERMINANTS(b) The procedure for solving linear simulta-
neous 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 25)(iv) multiply each side of (ii) by the inverse
matrix, and
(v) solve forx,yandzby equating the
corresponding elements.Problem 2. Use matrices to solve the simulta-
neous 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) Writing the equations 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 byA−^1 =adjA
|A|The adjoint ofAis the transpose of the matrix of
the cofactors of the elements (see Chapter 25).
The matrix of cofactors is
(
14 16 5
0 − 55
7 − 2 − 5)and the transpose of this matrix givesadjA=(
1407
16 − 5 − 2
55 − 5)The determinant ofA, i.e. the sum of the prod-
ucts 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)= 35
Hence the inverse ofA,A−^1 =1
35(
1407
16 − 5 − 2
55 − 5)(iv) Multiplying each side of (ii) by (iii), and
remembering thatA×A−^1 =I, the unit matrix,
gives
(
100
010
001)×(
x
y
z)=1
35(
1407
16 − 5 − 2
55 − 5)×(
4
33
2)(
x
y
z)=1
35×(
(14×4)+(0×33)+(7×2)
(16×4)+((−5)×33)+((−2)×2)
(5×4)+(5×33)+((−5)×2))=1
35(
70
− 105
175)=(
2
− 3
5)(v) By comparing corresponding elements,x= 2 ,
y=− 3 ,z = 5 , which can be checked in the
original equations.