1001 Algebra Problems.PDF

(Marvins-Underground-K-12) #1

  1. b.Identify the following determinants to be


used in the application of Cramer’s rule for the

matrix equation :

So, from Cramer’s rule, we have:

Thus, the solution is x= 2,y= 3.


  1. d.Identify the following determinants to be
    used in the application of Cramer’s rule for the


matrix equation :

Since applying Cramer’s rule requires that we
divide by Din order to determine x and y,we
can only conclude that the system has either
zero or infinitely many solutions. We must
consider the equations directly and manipu-
late them to determine which is the case. To
this end, as in Problem 925, subtracting the
second equation from the first equation yields
the false statement 0 = –3. From this, we con-
clude that the two lines must be parallel
(which can also be checked by graphing
them). Hence, the system has no solution.


  1. d.Identify the following determinants to be
    used in the application of Cramer’s rule for the


matrix equation :

Since applying Cramer’s rule requires that we
divide by Din order to determine x and y,we
can conclude only that the system has either
zero or infinitely many solutions. We must
consider the equations directly and manipu-
late them to determine which is the case. To
this end, as in Problem 926, the second equa-
tion in the system is obtained by multiplying
both sides of the first equation by 3. The two
lines are identical, so the system has infinitely
many solutions.


  1. b.Identify the following determinants to be
    used in the application of Cramer’s rule for


the matrix equation :

So, from Cramer’s rule, we have:

Thus, the solution is x= –1,y= 2.

y===DDy –– 12 2

x===DDx –^11 – 1

D ()()()()


1


1


1


1 11 1 1^2





y==–––––=

Dx==–– 11 01 ( )() ()( )––– 10 1 1 1=

D==–– 11 – 01 ()()()()––––– 10 1 1= 1


x
y

1


1


1


0


1


1




>>>––HH H=


D ()( ) ()()


3


9


3


y== 12 312 – 93 9=

Dx== 124 – –^26 ()()()()46 122 0–– –=

D==^39 – –^26 ()( ) ()( )36 92 0–– –=


x
y

3


9


2


6


4


12



>>>– HH H=


D^3 ()() ()( )


3


2


1


y==–^31 ––3 2 9=

Dx==– 12 22 ( )() ()()–– 22 12 =– 6

D==^3322 ()() ()() 32 – 23 0=


x
y

3


3


2


2


2


1


>>>HH H= –


y===DDy 13 3

x===DDx 12 2

D ( )() ()( )


1


2


2


1 11 2 2 3


––


y==–––=

Dx==– 12 –^01 ()()()()–––21 10 2=

D==– 21 –^01 ()()()()–––11 20 1=


x
y

1


2


0


1


2


1




>>>HH H= –


ANSWERS & EXPLANATIONS–
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