1) The inverse of the matrix [[1, 2], [3, 4]] is [[0.25, -0.5], [-0.75, 1]]. 2) The inverse of the matrix [[5, -1], [0, 3]] is [[0.2, 0.0333], [0, 0.3333]]. 3) The inverse of the matrix [[1, 2, 0], [0, 3, 0], [0, 0, 5]] is [[1, -0.6667, 0], [0, 0.3333, 0], [0, 0, 0.

1. A matrix is singu-
lar if and only if its
determinant is 0.
Since lAl - 0, we have
2x-8-0
x:4
A matrtxAis non-singular if and only if lAl + O.
EXAMPLE 27
SOLUTION
Find the set of values of aforwhich the matrix
^: li ])
i, ,,or-rinS.rlar.
Since A is non-singular, we have lAl * 0.
lAl: (a)(3) - (2)(3)
:3a-6
Since lAl * O,wehave 3a - 6 + 0
3a*6
a*2
EXAMPLE 28
SOLUTION
Determine whether the matrix
If the matrix is singular then its determinant is 0. Let us find the determinant of the
matrix.
1 2 1
-1 2 Alissingular.
2121
,li
"l-24;
;l+'l-
"l
L(4- 3) -2(-2 6) + 1(-1 - 4)-1+ L6- s:12
1l +o the matrix is not singular.
2l
12
-1 2
21
Since
1l
3l -2l
:
12
-1 2
21
SoLvi ng equations
(Cramer's ruLe)
using determinants
Solve the simultaneous equations
artx * ap/ : bL
arrx * azz/ : b2
Equation [1] multiplied by o^ gives
azta ttx * aztatz/ : aztb,
Equation [2] multiplied by orr Bives
arrarrx * arra22/ : o
rrb,
tll
l2l
t3l
l4l
335

2. Equation [a]
ottazz/ -
:. y(altazz
- equation [3] gives
aztatz/ : atrbr* orrbr,
aztatz) - orr,b,- arrb,
EXAMP
SOLUTI(
{4 5i
s -ei
tfre coefl
matrix.
Try thes
lon bl
lo, brl
^, -
orrbr- orrb,
/-
Similarly, we get
lb, arr.l
lr, o,,lA':
rv
lan anl
lo^ orrl
lan anl
lo^ orrl
The coefficient matrix is the matrix formed from the coefficients
of xand y inthe equatiol'ls. For the equations
oltx * orzy,: ,bt
aztx*o22y=bz
the coefficient matrix o(Z:', Z:)
Notice that for both the r-value and the y-value, the denominator is the determinant
of the coefficient matrix.
In the numerator for the x-value, the first column of the matrix consists of the values
on the right-hand side of the coefficient equations and the second colu'mn the coef-
ficients of7.
For tfuey-value, the first column of the numerator consists of the coefficients of x and
the second column contains the values on the right-hand side.
This result is known as Cramer's rule.
The cod
matrix b
lan o'
lo^ az
4gr 03
EXAMPLE 29
SOLUTION
Solve these simultaneous equations using Cramer's rule.
2xty-3
3x-2y:1
13 1l
lr -zl -Jv
12 1l
l: -21
12 3l
tl 1l
./ 12 1l
lr -zl
-4 - 3
2-9 _
-4-3
^-a -^-r-r- - ,- 12 1
The coefficient matrix'r (5 _2).
.'. The denominator of xand y it
lS -ll
For the numerator of x replace the
first column of the coefficient matrix
with (?).
For the numerator of y rep:lace the
second column of the coefficient
matrix with (?
)
-6-L_-7_
-7
-7_
-7
EXAIvT I
336
Hencex*L,/:1.

3. EXAMPLE 30
SOLUTION
(i ?)
,'
the coefficient
matrix.
I 10 sl
I -s -+l _ -40 - (-40)
Use Cramer's rule to solve the simultaneous equations
4x*5y:10
-014 s l
l: -+l
-L6 - 15 -31 Forthe n$merator of x replace
the first column of the coefficient
matrix with (i3)
For the numerator of y replace the
second columnof the coefficient
matrix with ( jg).
-ul
refficients
14 101
^, la -8 I
- -32 - (30) _,62 1
/-A 5l- -16-15 --31 -L
Ir -41
Hencex-0,y:2.
Try these 16.6 Solve the following pairs of simultaneous equations using Cramer's rule.
(a) x*3y:5
4x*y-9
(b) 2x-4y:2
3x - 7Y : 4
:terminant
'the
values
the coef-
rts oi x and
The coefficient
matrix is
lon on are
lo^ azz aztl
ol otz oztl
Using Cramer's rule to solve three equations in three unknowns
For the set ofequations
arrx * an/ * arrz: b,
arrx * az,z/ * arrz: b,
arrx * an/ * arrz: b,
using Crarn-er's rule, we have
x-
lb, atz orrl
l', azz orrl
lb, atz arrl
lo, b| an
lo^ bz azz
lo, b3 an
lan an an|
lo^ azz orrl
lou, atz orrl
lo, arz u
tl
lo^ azz brl
lo, azz brl
lan an arcl
lo^ azz
'rrl
lat atz anl
-
lat atz anl'
lo^ azz azzl
lo, atz arrl
v:
Note the positions of b,b, b, in the numerators of x, y andz.lnthe value of x,
b,b2,b3replaces the coeficient of I and similarly for y and z.T.he denominator is the
determinant of the coefficient matrix.
lI
'[3 -ll
ilte
ltfiix
ilE
nt
EXAMPLE 31 Use Cramer's rule to solve the following simultaneous equations.
x*2y+32:1
2x-y+z-2
x*2y+z-1
337

4. SOLUTION By Cramer's rule
,l-; ll -,?, ll + ,li -';
'l-l ll -,?, ll+,1? -Ll 1(-3) -2(L) +3(s)
10 1
I
10 r
11 1 3l
12 2 1l
lr 1 rl o
Y:ffi-ft-0
lz-1 rl
Ir 2 rl
N ote
columns'igte.i
the,,mffi[fi,xhe
numerator, the
determinant is 0.
11 2 1l
12 -1 2l
lr 2 rllr-m
lz -1 rl
tl 2 lt
...x:Lry:0rZ
:*-0
EXAMPI
SOLUTIO
:0.
1 23
2 -1 I
1 21
| 23
2 -1 1
1 21
EXAMPLE 32
SOLUTION
Solve these simultaneous equations.
2x*y+32:1
4x-3y+z-7
x*2y+z-5
Using Cramer's rule, we have
11 1 3l
lt -3 1l
ls 2 1l
JY
12 1 3I
l+ -3 rl
lr I 1l
,l-3 1l 14
'l z 1l lr
,l-i 1l ll ll + ,1" -31
21 1( - s) (2) + 3(2e) B0
EXAMPI
SOLUTIO
For oti
'frw,
o2
on
-31 2(-s) (3)+3(11) 20
2l
12 1 3l
l+ z tl "17
1l_14 1l-,14 7l
Ir s rl 'ls rl lr rl''lr sl 2(2)-(3)+3(13) _40_ ^/ la r ,l I I rl lt il l, .l
^/
F /^ ,
^/r<
12 1 1l
l+ -3 7l
tl 2 st
H 12 1 3l
l+ -3 1l
11 2 1l
2l-3 7l 14 7l +
I 2 st tl st
2l-3 1l _ t4 lt 314 -31
t z lt lr llr n 2t
ll + ,11
li -l ?l 4-i ll- li ll . ,li -il 2(-s) - (3) + 3(11) 20
l1 2 |
14 -31
lr 2l _2e2, (13) + (11)
-60 _ .J
2(-s) (3) + 3(11) 20
Hence x- 4,y:2,2- -3.
Use Cramer's rule to solve the followirg simultaneous equations.
Cofactor r
, lozt o
-lo* o
FOro.oz
l$lia Qe
lor, %
al h
Cofactorr
laz. o
- l'rr o
(a) 3x*ay-zz-e
5x*y-z-6
2xty-32-0
(b) 4x-5y+22-6
x*y+z-2
7x*2y-22:5
338
Try these 16.7

5. of matricesAppLications
EXAMPLE 6S
SOLUTION
The supply function for a commodity is given by Q
s(
x) : a* * bx 'l c, where a, b
and c are constants. When x: l,the quantity supplied is 5; when x : Z,the quantity
supplied is 12; when x : 3,the quantity supplied is 23. use a matrix method to find
thevalues ofa,bandc.
q'(1) : a(l)2 + b(1) + c: 5
q'(2): a(2)2 + b(z) + c: 12
q'(3): a(3)2 + b(3) + c:23
We get three equations to solve simultaneously:
a*b*c:5
4a-f2b*c:12
9a* 3b * c:23
Writing the equations in matrix form
tr 1 1/4 /5
t; i')!):il,)
):(} 1il
',(l?)ta
r
l1
l4
le
lr 1
l+z9 3
tu:
1 1l: l" ll
Matrix of cofactors -
Hence a: 2,b : l, c - 2
The equation is q'(x) - 2x2 t, x * 2
14 1l+14 2l:_1+s 6__2
le 1l le 3l
i-l 5 -6i
I z -8 6l
-r 3 -21
EXAMPLE 61 A 160/o solution, a 22o/o solution and a 360/o solution of an acid are to be mixed to
get 300 ml of a 247o solution. If the volume of acid from the 16%o solution equals
half the volume of acid from the other two solutions, write down three equations
satisfying the conditions given and solve the equations to find how much of each
is needed.
Let.r be the volume of 160/o solution, T be the volume of 22o/o solution and z be the
volume of 360/o solution needed.
SOLUTION

6. Now x * y : z - 300 since the total volume is 300 ml.
0.16x * 0.22y + 0.36 z - ffiX 300 : 72
and o.t6x - *$.22y
* 0.362) - o
Therefore the equations are
x*y+z-300
0.L6x*0.22y+0.362-72
0.I6x - 0.LLy - 0.182 - 0
Writing the equations in matrix form, we have
I L 1 1 tx /300
lo.ro o.2z 0.36 llyl:lnl0.16 -0.11 -0.181zl 0/
Forming the augmented matrix, w€ get
I | 1 I 1300
I 0.16 o.2z 0.36 I zzl
o.ro -0.11 -o.1gl o/
Rr+ R,
R, +R,
300
0.06 0.20 I 24
-0.27 -0.341 -49
SOLUTII
- 0.16R1
- 0.16R1
1
ffi
R, -+ 0.06R3+ 0.27R2
lt '1 I 1300
lo 0.06 o.zo I z+l
o o o.o::ol aol
lt 1 1 /.r /300
lo 0.06 o.2o llyl:l z+l
0 0 0.03361zl 3.6/
0.03362:3.6+z:107.14
0.06y + 0.202:24
0.06y + 0.20(to7.t4) :24
y: 42.86
x*y*z:300
x * 42.86 + 107.14: 300
x: 150
Hence 150 ml of the 167o solution, 42.86mlof the22o/o solution andl07.l4ml of the
367o solution are needed.
A popular carnival band sells three types of costumes. The costumes are made at the
Mas-camp in Port-of-Spain. The owner of the band makes cheap costumes, medium-
priced costumes and expensive costumes. The making of the costumes involves
EXAMPTE 62

7. SOTUTION
fabric, labour, buttons and machine time. The following table shows the units of
input required per costume for each type of costume.
The owner makes the three types of costumes and uses 270 units of fabric, 1050 units
of labour and790 buttons. How many of each type of costume does the owner make?
What is the corresponding machine time used?
Let r be the number of cheap costumes made, y the number of medium-priced
costumes made, zthe number of expensive costumes made.
Since 270 units of fabrics are used we have
5x*6y+82:270
For labout we have
20x+25y*302:1050
For buttons,
L5x+20y*222-790
Writing the equations in matrix form, we have
ls 6 B /.r 1270
lzo zs aoll.zl-llosoI
15 20 221 zl 7901
Forming the augmented matrix and reducing gives:
ls 6 8
lzo 2s 30
rs zo 22
270
10s0 I
Tsol
R, -+ R,
Rr+R,
ls 6
lo 1
o z
R, -+R,
- 4R,
- 3R,
B
-2
-2
270
-30 I
-20lmlof$e
ilca 6c
modium-
lJCs
- 2R,
ls 6 8
lo 1 -2
o o 2
270
-30 I
40l
Fabric
Labour
Buttons
Machine time

8. We nowhave
t5 6 8 /r t270
tB [ -')lr):-lt)
The equations are
2z: 40
- z: 20
/-22: -30-y- 40: -30+y:10
5x -f 6y * 8z: 270
5x * 60 + 160 :270
x:10
Hence the owner made 10 cheap costumes, 10 medium-priced costumes and
20 expensive costumes.
Machinetimeused:7 X 10 + 9 X 10 + L2y.20:400units.
EXERCISE 1 6 B
In questions l-5, find the inverse of each matrix.
r11 ? 2F20 -Ll 3 5)
3 rt _t) n (:^ l)
s(64'I3 3l
In questions 6-10, (a) find the determinant of each matrix, (b) find the matrix of
cofactors and hence find the inverse of each matrix.
lr 1 0
6 lz 1 -rle 1 zl
14 5 -1
8 la -2 zl
e 1-il
lL 2 1
7 lo 1 rl
o o zl
14 3 -1elz 4 +l
3 2 tl
lL 0 5
10 l+ 1 ol
23+l
In questions 11-15, find the inverse of each matrix by row reduction.
I 3 1 2 t r 4 1
11 l-r 1 ol tz I z I rl
1 3 | -2 3 tl
It 0 s t2 1 s
13 l+ I ol t4 14 i ol
234t Z3Zl
l-L 2 1
15 | I I rl
-2 1 7t
16 Solve the equation
x12l
xz 1+l
x;3 1 sl
370
:0.

9. t7
l8
19
23,
T,}tuk
33
24
Solve the equati
"rlhlxu
Solve the equatio, |
*
-,
lx*
Solve the equatio, I .1
lx
x2 1l
x 1l:0.
x3 1l
1 1 x*11
1 1 l:0.
1 1 x-11
x x* 1
11x
x*l 1
-0.
It 1 1
2S Findthevalues of asatisfyirgthe equation I a a * 1 a - 1
lo-1 2a a*1
-0.
Sanjeev pays TT$300 for 4 shirts and 2 pairs of trousers while Saleem pays
TT$700 for 2 shirts and 5 pairs oftrousers. If x and y represent the price ofa
shirt and a pair of trousers respectively, write a system of linear equation in
matrix form based on this information. Determine the price of a shirt and a
pair oftrousers.
Michael feeds his dogZentwith different mixtures of three types of food, A, B
and C. A scoop of each food contains the following nutrients.
Food A: 15 g of protein, 10 g carbohydrates and 20 g vitamins
Food B: 20 g of protein, 15 g carbohydrates and 10 g vitamins
Food C: 20g of protein, 10g carbohydrates and 20g vitamins
Assume that dogs require 160g of protein, 110g of carbohydrates and 150g of
vitamins. Find how many scoops of each food Michael should feed his dog daily
to satisfy their nutrient requirements.
Deanne has TT$50 000 and wishes to invest this for her retirement. She puts all
the money in a fixed deposit, trust fund and a money market fund. The amount
she puts in the money market fund is TT$10 000 more than that in the trust
fund. After one year, she receives a profit totalling TT$3000. The fixed deposit
pays 5o/o interest annually, the trust fund pays 67o annually and the money
market fund pays 7Vo ann:ually.
By denoting the amount of money invested in the fixed deposit, trust fund
and the money market fund as x, y andz respectively, form a system of linear
equations based on the information given.
Write the system of linear equations in matrix form.
Find the amount of money invested in each category of the fund.
Show that the equations
x*5y*42:19
2x-4y*z:-4
4x+6y*72:30
have a unique solution and hence find the solution by row reducing the
augmented matrix to echelon form.
rix of
371