Solution Set
Bretscher 1.3 - 4, 8, 14, 47, 48, 49, 57
2/6/16
1.3 / 4
7 
1 4 7
1 4
1 4 7  4( II ) 1 0  1
2 5 8  2( I )  0  3  6   3  0 1 2
 0 1 2 






3 6 9  3( I )
0  6  12   2( II ) 0 0 0
0 0 0 
Rank = # of leading 1’s = 2
1.3 / 8
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There are infinitely many solutions to the system xv1  yv2  zv3  v4 . Notice that any vector in the plane can be
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written as a linear combination of v1 and v2 . Thus, by varying z, we would form an infinite number of vectors
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v4  zv3 that can be written as a linear combination of v1 and v2 .
1.3 / 14
 1
1 2 3  
1 
 2
3 6
2 3 4   2    12  23  14  8

 1
 
 
   
 
 1
1 2 3    1 1  22  31 6
2 3 4   2   2 1  32  41  8

 1 
  
 
1.3 / 47
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a. All homogeneous systems Ax  0 have x  0 as a solution, thus are consistent.
b. Since a linear system with fewer equations than unknown has either no solutions or infinitely many solutions
(Fact 1.3.3) and a homogeneous system has at least one solution (a), we can conclude that a homogeneous
system has infinitely many solutions.
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c. Since x1 and x2 are solutions of the homogeneous system Ax  0 , Ax1  x2   Ax1   Ax2   0  0  0 .
 
Therefore, x1  x2 is also a solution.
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d. Since x is a solution of the homogeneous system Ax  0 , Akx   k  Ax   k  0  0 . Therefore, kx is also a
solution.
1.3 / 48
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a. If xh is a solution to Ax  0 , then Ax1  xh   Ax1   Axh   b  0  b .
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Thus, x1  xh is a solution to Ax  b .
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b. If x2 is also a solution to Ax  b , then Ax2  x1   Ax2   A x1 
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 Ax2   Ax1   b  b  0 . Thus, x2  x1 is a solution to Ax  0 .
c. We know from part a that adding the solution of the homogeneous equation
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yield another solution to Ax  b . Thus, we have:
Solutions of
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Ax  b
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x1
0
Solutions of
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Ax  0
1.3 / 49
a. The system can have zero or infinitely many solutions. It can have no solutions if one of the equations is 0 =
1. Otherwise can have infinite solutions because of 4 – 2 = 2 free variables.
b. The system has zero or one solution. Again, it can have no solution if one of the equations is 0 = 1. It cannot
have infinite solutions because there are 3 – 3 = 0 variables.
c. The system has zero solution. Since the rank of the 4x4 augmented matrix is 4, the last row must be
0 0 0 1 . This corresponds to 0 = 1, an inconsistency.
d. The system has infinitely many solutions. It cannot be inconsistent because the coefficient side of the
augmented matrix is never 0. It has infinitely many solutions because it has 4 – 3 = 1 free variable.
1.3 / 57
1 
2
A vector on the line y  3x is   . A vector on the line y  x 2 is   . Thus, we can express the given vector
3
 
1 
7
1
 2   a  2b 
7
1
2
as:    a    b    
. Solving this linear system, we have a  3, b  2 . Thus,    3     2    .

11
3
1  3a  b 
11
3
1 