applied linear algebra, homework help

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applied linear algebra

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Explanation & Answer

Attached.

Problem 1:

a

Consider a linear transformation T = 
c

b
. Since T maps e1 to −e2 and e2 to e1 + 3e2 , we have
d 

 e   a b   e1   ae1 + be2   −e2 
T  1 = 
=

  = 
e2   c d  e2   ce1 + de2  e1 + 3e2 
which gives a = 0, b = −1, c = 1, d = 3 .

0 −1
.
3 

Hence, the linear transformation is T = 
1
Problem 2:

1 1  s  1 s + 1 0   s 
  = 
 =   for s  , it implies that the transformation maps the x-axis
0 1 0 0  s + 1 0 0

Since A = 

to the x-axis itself.

 t 

1 1 0 1 0 + 1 t  t 
=
=   for t  , and the set    : t   is the line y = x in




0 1  t  0  0 + 1 t  t 
 t 

2
Hence, the transformation maps the y-axis to the line y = x in
.
Since A = 

Problem 3:
The unit square is described as below:

Problem 4:

2

.

We compute

0   2
A1   = 
0  1
1   2
A1   = 
0  1

0  0 0 
0   2 0  0  0 
=   , A1   = 



  =  ,
1  0 0 
1   1 1  1  1 
0  1   2 
1  2 0 1  2
=   , A1   = 



  =  
1  0 1 
1 1 1  1  2

Hence, A1 maps the unit square to (0,0), (0,1), (2,1) and (2,2), as sketched below

We compute

0   0
A2   = 
0   −1
1   0
A2   = 
0   −1

1  0  0 
0   0
=   , A2   = 



0  0  0 
1   −1
1  1   0 
1  0
=   , A2   = 



0  0   −1
1  −1

1   0  1 
=
,
0  1  0 
1  1  1 
=
0  1  −1

Hence, A2 maps the unit square to (0,0), (1,0), (0,-1) and (1,-1), as sketched below

We compute

0 �...


Anonymous
Awesome! Perfect study aid.

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