Math 2270 Spring 2004
Homework 3 Solutions
Section 1.3) 1, 2, 6, 10, 17, 18, 22, 24, 25, 26, 28, 34, 50
(1a) This system has no solutions (inconsistent), as the bottom row of the augmented matrix
in reduced row-echelon form consists of coefficients equal to zero followed by a one.
(1b) There is one unique solution to this system, as rref (A) is equal to the identity matrix.
(1c) There are infinitely many solutions to this system as the first variable x1 is free. (It can
take on all real number values.)
(2) The original matrix is upper triangular, with all 1’s on the diagonal. Therefore, it reduces
to the identity matrix with rank 3.
(6) The system has no solution (it is inconsistent). The vectors v~1 and v~2 lie on the same line
and are scalar multiples of one another. Therefore, any linear combination of them will yield
a new vector that lies on the same line and is just a scalar multiple of either of them. The
vector v~3 does not lie on the line spanned by v~1 , so cannot be written as a scalar multiple of
either v~1 or v~2 .
(10)


 
1
1

 

−2
2
·
 = 1(1) + 2(−2) + 3(1) = 0
 

1
3
(17) This product is undefined since the matrix has three columns, but the vector only has
two rows.
(18)






5
1(1) + 2(2)
1 2 " #
1

 



=  3(1) + 4(2)  =  11 
 3 4 
2
17
5(1) + 6(2)
5 6
(22) A system with three equations and three unknowns can be written as the matrix equation
~y = A~x where A is 3x3. If the solution is unique, the rank of A must equal 3, and rref (A) =
I, where I is the 3x3 identity matrix (Fact 1.3.4).
(24) We are told that the system A~x = ~b has a unique solution, so rank(A)= 4 by Fact 1.3.4.
Therefore, also by Fact 1.3.4, the system ~c = A~x must have a unique solution as well.
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Math 2270 Spring 2004
Homework 3 Solutions
(25) If the system A~x = ~b is inconsistent, there is at least one non-leading variable (rank(A)<
4) and the reduced row-echelon form of A must have at least one row of zeros. Therefore,
the system ~c = A~x could also be inconsistent, or if it is consistent, it has infinitely many
solutions.
(26) We are told that the system A~x = ~b has a unique solution, so all variables must be
leading (rank(A)= 3) and the reduced row-echelon form of A has one row of zeros. Therefore,
the system A~x = ~c is either inconsistent, or, if it is consistent, it has a unique solution.
(28a) TRUE - If we omit one row of a matrix E in reduced row-echelon form, the resulting
matrix will still be in reduced row-echelon form. Removing a row will not introduce nonzero
entries in front of, or below the leading ones, and will also not alter the ordering of the
leading ones.


"
#
1 0 0 5
0 1 0 3


 0 1 0 3 ⇒
0 0 1 2
0 0 1 2
(28b) FALSE - For example, omit the third column from the following matrix.




1 0 5
1 0 0 5




 0 1 0 3 ⇒ 0 1 3 
0 0 2
0 0 1 2
(34a)




b


e
A~
e2 =  
h
a


A~
e1 =  d 
g
(34b) B e~1 = v~1 , B e~2 = v~2 , B e~3 = v~3


c


f
A~
e3 =  
k
(50) The rank of the augmented matrix is 4. Therefore,
h
i




rref ( A : ~b ) =
1
0
0
0
0
1
0
0
0
0
1
0
|
|
|
|
0
0
0
1





Looking at row four, we see that the system is inconsistent so there is no solution.
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