Page 365 - 35Linear Algebra

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(b) The rows of M span (kerL) ⊥
(c) First we put M in RREF:

   
1 2 1 3 1 2 1 3
2 1 −1 2 0 −3 −3 −4
   
M =   ∼  
1 0 0 −1 0 −2 −1 −4
   
4 1 −1 0 0 −7 −5 −12
1 0 −1 1 0 0 −1
 1   
3
0 1 1 0 1 0
 4  8
 .
 
3  3 
∼  4  ∼  4
0 0 0 0 1 − 
3 3
1 − 
0 0 2 − 8 0 0 0 0
3
Hence
8 4
ker L = span{v 1 − v 2 + v 3 + v 4 }
3 3
and
imL = span{v 1 + 2v 2 + v 3 + 4v 4 , 2v 1 + v 2 + v 4 , v 1 − v 2 − v 4 } .
Thus dim ker L = 1 and dim imL = 3 so
dim ker L + dim imL = 1 + 3 = 4 = dim V .

17. (a)

 5 = 4a − 2b + c

2 = a − b + c

 0 = a + b + c


3 = 4a + 2b + c .
(b,c,d)
     
4 −2 1 5 1 1 1 0 1 0 1 −1
1 −1 1 2 0 −6 −3 5 0 1 0 1
     
  ∼   ∼  
 1 1 1 0   0 −2 0 2   0 0 −3 11 
4 2 1 3 0 −2 −3 3 0 0 −3 3
The system has no solutions because c = −1 and c = − 11 is impossible.
3
(e) Let
   
4 −2 1 5
1 −1 1 2
   
M =   and V =   .
1 1 1 0
   
4 2 1 3
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