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MATH 331 QUIZ # 2 SOLUTIONS
Hi kids! Here are the solutions to the recent Quiz # 2. Enjoy!
(1) (a) By preforming row operations, we find the that (unique) reduced row-echelon form of
A is


1 2 0 10 0
A ∼ 0 0 1 −1 0 .
0 0 0 0 1
Thus, the solution (x1 , x2 , x3 , x4 , x5 ) to the system Ax = 0 has free variables x2 and x4 ,
and satisfies (reading from the bottom up) x5 = 0, x3 = x4 , and x1 = −2x2 − 10x4 . So, an
arbitrary solution to Ax = 0 can be written in parametric form as

  


 
x1
−10
−2x2 − 10x4
−2
x2  
 0 

1
x2

  


 

x3  = 




x4
  
 = x2  0  + x4  1  .
x4  




1 
0
x4
0
x5
0
0
Thus, the vectors (−2, 1, 0, 0, 0) and (−10, 0, 1, 1, 0) make up a basis for the null space of A.
(b) Since there are three columns with pivots, the dimension of the column space of A,
also known as the Rank of A, is 3. Another way to find this is through the rank-nullity
theorem: since the dimension of the nullity of A is 2 by the above (there are two elements in
that basis), and there are 5 columns of A, the rank must be 3, since Rank(A)+Nullity(A) =
the number of columns of A.
(2) By expanding about the 2nd row, and not including the zeros in our calculations, we have
the determinant as (−1)2+3 4(3 − 16) = −4 ∗ (−13) = 52.
(3) We use Cramer’s rule. For the given A and b = (7, 3, 4), the matrix


1 7 1
A2 (b) = 0 3 1 .
0 4 3
We expand about the first column to compute the determinant of A to be 5 and the
determinant of A2 (b) to also be 5, therefore, applying Cramer’s rule, we see that x2 = 55 = 1.
(4) (a) This is NOT a vector space, consider the zero vector in this space–it must be the zero
polynomial, and the zero polynomial is NOT of the form a + t2 for any real number a. This
also fails to be closed under addition and scaling by real numbers.
(b) This fails to be a vector space for the same reasons as above: there is no zero vector
in this space, and it is not closed under addition of vectors or scaling by real numbers. We
exhibit the failure of closure of addition: Let x6 + 1 and −x6 be polynomials of degree
exactly 6. Clearly, (x6 + 1) + (−x6 ) = 1 is not a polynomial of degree exactly 6.
(c) This IS a vector space, consult the class notes.
(d) This is NOT a vector space, it is not closed under scaling by real numbers. The
element (1, 2) is in this set, although (−5) · (1, 2) = (−5, −10) is not in this set.
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