Page 356 - 35Linear Algebra

356 Sample Final Exam

eject the last two vectors and obtain as our basis

       
 1 4 1
 0 

2 3 0 1
 
       
,
,
,
        .
 3 2 0 0 
       
 
4 1 0 0
 
Of course, this answer is far from unique!
(c) The method is the same as above. Add the standard basis to {u, v}
to obtain the linearly dependent set {u, v, e 1 , . . . , e n }. Then put these
vectors as the columns of a matrix and row reduce. The standard
basis vectors in columns corresponding to the non-pivot variables can
be removed.
9. (a)
 1 
λ − −1
2

  1 1 1 λ 1 1
det  − 1 λ − 1 − 1  = λ (λ− λ− )+ − − − − +λ
 2 2 2 2 4 2 2 2 4
−1 − 1 λ
2
1 3 3
2
3
= λ − λ − λ = λ(λ + 1)(λ − ) .
2 2 2
3
Hence the eigenvalues are 0, −1, .
2
(b) When λ = 0 we must solve the homogenous system
 1   1   
0 1 0 1 0 0 1 0 −1 0
2 2
 1 1 1 0  ∼  0 1 1 0  ∼  0 1 2 0  .





 2 2 2 4 2
1 1 0 0 0 1 1 0 0 0 0 0
2 2
 
s
So we find the eigenvector  −2s  where s 6= 0 is arbitrary.
s
For λ = −1
 1   
1 1 0 1 0 1 0
2
 1 3 1   
 2 2 2 0  ∼  0 1 0 0  .
1 1 1 0 0 0 0 0
2
 
−s
So we find the eigenvector  0  where s 6= 0 is arbitrary.
s
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