Page 330 - 35Linear Algebra

330 Sample First Midterm

   
1 0 0 0 1 1 −1 2
 1 1 0 0   0 2 3 0 
=    
−1 −1 1 0 0 0 −2 8
   
0 2 0 1 0 0 1 −2
   
1 0 0 0 1 1 −1 2
1 1 0 0 0 2 3 0
   
=     .
 −1 −1 1 0   0 0 −2 8 
0 2 − 1 1 0 0 0 2
2
To solve MX = V using M = LU we first solve LW = V whose augmented
matrix reads

   
1 0 0 0 7 1 0 0 0 7
1 1 0 0 6 0 1 0 0 −1
   
  ∼  
 −1 −1 1 0 12   0 0 1 0 18 
0 2 − 1 1 −7 0 2 − 1 1 −7
2 2
 
1 0 0 0 7
0 1 0 0 −1
 
∼   ,
 0 0 1 0 18 
0 0 0 1 4
from which we can read off W. Now we compute X by solving UX = W
with the augmented matrix

   
1 1 −1 2 7 1 1 −1 2 7
0 2 3 0 −1 0 2 3 0 −1
   
  ∼  
 0 0 −2 8 18   0 0 −2 0 2 
0 0 0 2 4 0 0 0 1 2
   
1 1 −1 2 7 1 0 0 0 1
0 2 0 0 2 0 1 0 0 1
   
∼   ∼   .
 0 0 1 0 −1   0 0 1 0 −1 
0 0 0 1 2 0 0 0 1 2
So x = 1, y = 1, z = −1 and w = 2.

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