Math 1502 Quiz 5
March 21, 2000


Name: ____________________________

Section: Circle One:
Section A1:Lindsay Bates.Classroom:Skiles 202
Section A2:Eric Forgoston.Classroom:Skiles 246
Section A3:Mohammed Sinnokrot.Classroom:Skilkes 256
Section A4:Kasso Okoudjou.Classroom:Skiles
Section A5:Marcus Sammer. Skiles 140

Open Book and Notes.  Carefully explain your proceedures and answers. Calculators allowed, but answers mush be exact.

Problem #1 (   points)

Let:
                         
                  A =     
[Graphics:Images/index_gr_1.gif]


a)(5 points)  Find the column space of this matrx.

b) (5 points) Find the null space of this  matrix


c)(5 points)  Find all solutions to Ax =
[Graphics:Images/index_gr_2.gif].



d) (3 points)  Express the result of c) as a translate   of the null space of A.

Ans
[Graphics:Images/index_gr_3.gif]
[Graphics:Images/index_gr_4.gif]
[Graphics:Images/index_gr_5.gif]
[Graphics:Images/index_gr_6.gif]
[Graphics:Images/index_gr_7.gif]
[Graphics:Images/index_gr_8.gif]
[Graphics:Images/index_gr_9.gif]
[Graphics:Images/index_gr_10.gif]

Columm space is 2b1 - 3 b2 + b3 = 0.
    

[Graphics:Images/index_gr_11.gif]
[Graphics:Images/index_gr_12.gif]
[Graphics:Images/index_gr_13.gif]

Or : z = t , y = t, x = -t.  Or the line  t [Graphics:Images/index_gr_14.gif].
    

c)

[Graphics:Images/index_gr_15.gif]
[Graphics:Images/index_gr_16.gif]
[Graphics:Images/index_gr_17.gif]
[Graphics:Images/index_gr_18.gif]
[Graphics:Images/index_gr_19.gif]
[Graphics:Images/index_gr_20.gif]
[Graphics:Images/index_gr_21.gif]

Or  [Graphics:Images/index_gr_22.gif]   =   [Graphics:Images/index_gr_23.gif] +    t [Graphics:Images/index_gr_24.gif],

which also answers d.


[Graphics:Images/index_gr_25.gif]
[Graphics:Images/index_gr_26.gif]

Problem 2 (14 points)

a:    (5 points)   Find the  null  space of the matrixc:


[Graphics:Images/index_gr_27.gif]

b:  (3 points)   Find the dimension of the null space of A

c:  (3 points) Using consertvation of dimension, find the dimension  k of the column space of A.
d (3 points) Find   k linearly inde[pendent coumns of A. (Here k is the dimensoin of the column space of A>)

Ans
[Graphics:Images/index_gr_28.gif]
[Graphics:Images/index_gr_29.gif]

x  + 2 y + 3 z + 4 w = 0
x +  3 y + 4 z + 5 w =  0
x +  5 y + 6 z + 7 x  =  0


Subract multiples of the 1st from the 2nd and third:
  x  + 2 y + 3 z + 4 w = 0
            y +    z  +   w = 0
          3y + 3 z  + 3 w  = 0
          
          
   And then:
   
   x  + 2 y + 3 z + 4 w = 0
            y +    z  +   w = 0
                              0  = 0
                              
                              
Set w = t, z = s, solve for the rest of the variables:

y = -s - t,
x + 2 (-s -t) + 3 s + 4 t = 0  or


x = -s -2t


Its two dimensional

Therefore the range (column space) is 4-2 dimensional

The 1st two colums of A are linearly independent.


Converted by Mathematica      March 21, 2000