1001 Algebra Problems.PDF

(Marvins-Underground-K-12) #1
the given system can be written as the equivalent matrix equation. The solution is

therefore given by. Using the calculation for the inverse from Problem 904 yields

the following solution:

So, the solution of the system is x= –^4265 ,y=  215 .


  1. b.The solution to the matrix equation , where a, b, c, d, e,and fare real numbers, is


given by , provided that the inverse matrix on the right side exists. From Problem 889,

the given system can be written as the equivalent matrix equation. The solution is

therefore given by. Using the calculation for the inverse from Problem 905 yields

the following solution:

So, the solution of the system is x= –^179 ,y= ^272 .


  1. c.The solution to the matrix equation , where a, b, c, d, e,and fare real numbers, is


given by , provided that the inverse matrix on the right side exists. From Problem 890,

the given system can be written as the equivalent matrix equation. The solution is,

therefore, given by. Using the calculation for the inverse from Problem 906 yields

the following solution:

So, the solution of the system is x= –1,y= –^136 .

x
y 2

0 2


4


1



– –



21
31 163

>> > >HHHH==


x
y

2


12


0


3


2


– 4


= –^1 –


>> >HHH


x
y

2


12


0


3


2


– 4


>>>HH H= –


x
y

a
c

b
d

e
f


  • 1



    HHH=







a
c

b
d

x
y

e
>>>HH H= f

x
y

5


9





– –


52 –


51

51
53

51

(^225)


>> > >HHHH==


x
y

3


1


1


2


5


9



––



  • 1



    HHH=







x
y

3


1


1


2


5


9



>>>––HH H=


x
y

a
c

b
d

e
f


  • 1



    HHH=







a
c

b
d

x
y

e
>>>HH H= f

x
y

1


0


2


1


254 – –


251

2546
251

>> >HHH==> H


x
y

1


0


4


25


2


1


= –––^1


>> >HHH


x
y

1


0


4


25


2


1


––=


>>>HH H


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