A matrix is singu-lar if and only if itsdeterminant is 0.

Since lAl - 0, we have

2x-8-0x: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*6a*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-\

"lL(4- 3) -2(-2 6) + 1(-1 - 4)-1+ L6- s:12

1l +o the matrix is not singular.2l

12-1 221

Since

1l3l -2l

:

12-1 221

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

Equation [a]

ottazz/ -:. y(altazz

- equation [3] gives

aztatz/ : atrbr* orrbr,

aztatz) - orr,b,- arrb,

EXAMP

SOLUTI(

{4 5i\s -eitfre coeflmatrix.

Try thes

lon bllo, 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 coefficientsof 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 determinantof the coefficient matrix.

In the numerator for the x-value, the first column of the matrix consists of the valueson 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 andthe second column contains the values on the right-hand side.

This result is known as Cramer's rule.

The codmatrix b

lan o'

lo^ az

\4gr 03

EXAMPLE 29

SOLUTION

Solve these simultaneous equations using Cramer's rule.

2xty-33x-2y:1

13 1llr -zl -Jv 12 1ll: -21

12 3l

tl 1l./ 12 1llr -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 -llFor the numerator of x replace thefirst column of the coefficient matrixwith (?).

For the numerator of y rep:lace thesecond column of the coefficient

matrix with (? )

-6-L_-7_-7

-7_-7

EXAIvT I

336

Hencex*L,/:1.

EXAMPLE 30

SOLUTION

(i ?) ,'

the coefficientmatrix.

I 10 slI -s -+l _ -40 - (-40)

Use Cramer's rule to solve the simultaneous equations

4x*5y:10

-014 s ll: -+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 thesecond columnof the coefficient

matrix with ( jg).\ -ul

refficients

14 101

^, la -8 I - -32 - (30) _,62 1/-A 5l- -16-15 --31 -L

Ir -41Hencex-0,y:2.

Try these 16.6 Solve the following pairs of simultaneous equations using Cramer's rule.

(a) x*3y:54x*y-9

(b) 2x-4y:23x - 7Y : 4

:terminant

'the values

the coef-

rts oi x and

The coefficientmatrix 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 tllo^ azz brl

lo, azz brl

lan an arcl

lo^ azz 'rrllat 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 -llilteltfiix

ilEnt

EXAMPLE 31 Use Cramer's rule to solve the following simultaneous equations.

x*2y+32:12x-y+z-2x*2y+z-1

337

SOLUTION By Cramer's rule

,l-; ll -,?, ll + ,li -';'l-l ll -,?, ll+,1? -Ll 1(-3) -2(L) +3(s)

10 1I

10 r

11 1 3l12 2 1llr 1 rl oY:ffi-ft-0lz-1 rlIr 2 rl

N ote

columns'igte.ithe,,mffi[fi,xhenumerator, thedeterminant is 0.

11 2 1l12 -1 2llr 2 rllr-m

lz -1 rltl 2 lt

...x:Lry:0rZ

:*-0EXAMPI

SOLUTIO

:0.

1 232 -1 I1 21| 232 -1 1

1 21

EXAMPLE 32

SOLUTION

Solve these simultaneous equations.

2x*y+32:14x-3y+z-7x*2y+z-5

Using Cramer's rule, we have

11 1 3llt -3 1lls 2 1lJY 12 1 3Il+ -3 rllr I 1l

,l-3 1l 14'l z 1l lr

,l-i 1l ll ll + ,1" -3121 1( - s) (2) + 3(2e) B0

EXAMPI

SOLUTIO

For oti'frw,o2on

-31 2(-s) (3)+3(11) 202l

12 1 3ll+ z tl "17

1l_14 1l-,14 7lIr 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 7ltl 2 stH 12 1 3ll+ -3 1l11 2 1l

2l-3 7l 14 7l +I 2 st tl st

2l-3 1l _ t4 lt 314 -31t 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 -31lr 2l _2e2, (13) + (11) -60 _ .J2(-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 Qelor, %\al hCofactorr

laz. o- l'rr o

(a) 3x*ay-zz-e5x*y-z-62xty-32-0

(b) 4x-5y+22-6x*y+z-27x*2y-22:5

338

Try these 16.7

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:54a-f2b*c:129a* 3b * c:23Writing the equations in matrix form

tr 1 1\/4\ /5\t; i')\!):\il,)

):(} 1il ',(l?)ta

\r

l1l4le

lr 1

l+z\9 3

tu:

1 1l: l" llMatrix 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__2le 1l le 3l

i-l 5 -6iI 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

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-3000.L6x*0.22y+0.362-720.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:lnl\0.16 -0.11 -0.181\zl \ 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 0.03361\zl \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:300x * 42.86 + 107.14: 300

x: 150

Hence 150 ml of the 167o solution, 42.86mlof the22o/o solution andl07.l4ml of the367o solution are needed.

A popular carnival band sells three types of costumes. The costumes are made at theMas-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

SOTUTION

fabric, labour, buttons and machine time. The following table shows the units ofinput required per costume for each type of costume.

The owner makes the three types of costumes and uses 270 units of fabric, 1050 unitsof 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-pricedcostumes 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 8lzo 2s 30\rs zo 22

270\10s0 I

Tsol

R, -+ R,

Rr+R,

ls 6lo 1

\o z

R, -+R,

- 4R,

- 3R,

B

-2-2

270\-30 I

-20lmlof$e

ilca 6cmodium-lJCs

- 2R,

ls 6 8lo 1 -2\o o 2

270\-30 I

40l

Fabric

Labour

Buttons

Machine time

We nowhave

t5 6 8\ /r\ t270\

tB [ -')lr):\-lt)The equations are

2z: 40 - z: 20

/-22: -30-y- 40: -30+y:105x -f 6y * 8z: 270

5x * 60 + 160 :270

x:10Hence the owner made 10 cheap costumes, 10 medium-priced costumes and20 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 ?\ 2F2\\0 -Ll \3 5)

3 rt _t) n (:^ l)s(64'I\3 3l

In questions 6-10, (a) find the determinant of each matrix, (b) find the matrix ofcofactors and hence find the inverse of each matrix.

lr 1 0\6 lz 1 -rl\e 1 zl14 5 -1\8 la -2 zl\e 1-il

lL 2 1\7 lo 1 rl\o o zl14 3 -1\elz 4 +l\3 2 tl

lL 0 5\10 l+ 1 ol\23+lIn 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 tlIt 0 s\ t2 1 s\13 l+ I ol t4 14 i ol\234t \Z3Zll-L 2 1\

15 | I I rl\-2 1 7t

16 Solve the equationx12lxz 1+lx;3 1 sl

370

:0.

t7

l8

19

23,

T,}tuk

33

24

Solve the equati "rlhlxu

Solve the equatio, |

* -,lx*

Solve the equatio, I .1lx

x2 1lx 1l:0.x3 1l

1 1 x*111 1 l:0.

1 1 x-11

x x* 1

11xx*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 ofashirt and a pair of trousers respectively, write a system of linear equation inmatrix 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, Band 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 ofvitamins. Find how many scoops of each food Michael should feed his dog dailyto 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 trustfund. 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 fundand 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*42:192x-4y*z:-44x+6y*72:30have a unique solution and hence find the solution by row reducing theaugmented matrix to echelon form.

rix of

371