MATH 323.501
Exam 1 Solutions
October 10, 2013
1.
Consider the system of equations
3x − 3z = 9
4x − 2y − 8z + w = 21
2y + 4z = −4
(a) Write down the augmented matrix for this system of equations with the variables ordered
x, y, z, w.
(b) Carry out the row reduction of the augmented matrix in (a) and transform it into
reduced row echelon form. Show all of the steps you are taking.
(c) Determine the complete set of solutions of this system of equations.
Solution: (a) Reading off the coefficients of the variables, the augmented matrix is


3 0 −3 0 9
 4 −2 −8 1 21  .
0 2
4 0 −4
(b) We perform the following row operations: (there are several ways to do do this correctly)




3 0 −3 0 9
1 0 −1 0 3
(i)
 4 −2 −8 1 21  −→
 4 −2 −8 1 21 
0 2
4 0 −4
0 2
4 0 −4




1 0 −1 0 3
1 0 −1 0 3
(ii)
(iii)
4 0 −4 
−→  0 −2 −4 1 9  −→  0 2
0 2
4 0 −4
0 −2 −4 1 9




1 0 −1 0 3
1 0 −1 0 3
(v)
(iv)
2 0 −2  −→  0 1 2 0 −2  .
−→  0 1
0 −2 −4 1 9
0 0 0 1 5
The row operations performed were (i) divided Row 1 by 3; (ii) replaced Row 2 with its sum
with −4 times Row 1; (iii) swapped Rows 2 and 3; (iv) divided Row 2 by 2; (v) replaced
Row 3 with its sum with 2 times Row 2.
(c) Using the matrix in reduced row echelon form from (b), we have the equivalent system
of equations,
x−z =3
y + 2z = −2
w = 5.
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We see that z is a free variable, and x, y, and w are all leading variables. The solution set
is then



3+α






 −2
−
2α

α∈R .
 α







5
2.
Consider the matrix


0 1 1
A = −1 1 0 .
2 0 1
Find the inverse matrix of A, and show all of the steps that you take.
Solution: We aim to put the matrix


0 1 1 1 0 0
 −1 1 0 0 1 0 
2 0 1 0 0 1
in reduced row echelon form. We perform the following row operations (and again there are
several ways to do this correctly):




0 1 1 1 0 0
1 −1 0 0 −1 0
(i)
 −1 1 0 0 1 0  −→
 0 1 1 1 0 0 
2 0 1 0 0 1
2 0 1 0 0 1




1 0 1
1 −1 0 0 −1 0
1 −1 0
(iii)
(ii)
1
0 0 
−→  0 1 1 1 0 0  −→  0 1 1
0 2 1 0 2 1
0 0 −1 −2 2 1




1 0 1 1 −1 0
1 0 0 −1 1
1
(iv)
(v)
0  −→  0 1 0 −1 2
1 .
−→  0 1 1 1 0
0 0 1 2 −2 −1
0 0 1 2 −2 −1
From this we see that
A−1
3.


−1 1
1
1 .
= −1 2
2 −2 −1
For each statement below, write down whether it is true or false.
(a) A system of 4 linear equations in 5 variables must have more than one solution.
(b) For invertible n × n matrices A and B, (AB)−1 = A−1 B −1 .
(c) For an n × n matrix A of determinant 5, if B is the matrix obtained by multiplying one
of the rows of A by 3 and then interchanging two other rows, then the determinant of B
must be 15.
(d) For an elementary matrix E, the matrix E T must be an elementary matrix.
(e) For a positive integer d, the set of all polynomials of degree d with real coefficients is a
vector space.
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Solution: Credit was given for the correct answers. The explanations were not necessary,
but the ones below may answer some of your questions about these problems.
(a) False : Even though the system is underdetermined, it is possible that it could be
inconsistent.
(b) False : In general, (AB)−1 = B −1 A−1 .
(c) False : Multiplying a row by 3 will increase the determinant by a factor of 3; interchanging two rows will change the sign of the determinant. So the determinant will be
−15.
(d) True : For each of the three types of elementary matrices, it is easily seen that the
transpose takes an elementary matrix to an elementary matrix.
(e) False : The set of all polynomials of degree d with real coefficients is not closed under
addition, so it cannot be a vector space For example, if d = 2, then (2x2 + x + 1) + (−2x2 +
3x − 4) = 4x − 3. Since the degree of the sum is actually 1 and not 2, 4x − 3 is not an
element of the set of all polynomials of degree 2.
4.
Find all values of x such that the matrix
6
x+3
A=
x−2
x
is singular. Explain.
Solution: A is singular if and only if its determinant is 0, so we want to solve det(A) = 0
for x. Now
det(A) = 6x − (x − 2)(x + 3) = 6x − (x2 + x − 6) = −x2 + 5x + 6.
So we solve x2 − 5x − 6 = 0, which factors as (x − 6)(x + 1) = 0. Thus A is singular precisely
when x = 6, −1 .
5.
Let B be the matrix

1 −3 0 2
B = 0 0 1 −1 ,
0 0 0 0

and let N (B) be its nullspace.
(a) N (B) is a subspace of Rn for what value of n? Explain.
(b) Find two vectors a, b so that N (B) = Span(a, b).
Solution: (a) By definition of nullspace, N (B) = {v ∈ Rn | Bv = 0}. In order for the
product Bx to make sense, we need that the number of columns of B match the number of
entries of v, which is n. Thus we need n = 4 .
(b) Since B is already in reduced row echelon form, we can read off the nullspace as



3α − 2β 






α
 α, β ∈ R .
N (B) = 
 β 






β
3
Since we can rewrite
 

 

−2
3
3α − 2β
0
1
 α 
 

 

 β  = α 0 + β  1  ,
1
0
β
we see that we can take
 
3
1

a=
0 ,
0
6.
 
−2
0

b=
1.
1
Consider the matrices


3 2 1
A = 1 0 2  ,
1 1 −2


1 0 5
B = 4 2 3  .
1 1 −2
Find elementary matrices E1 and E2 so that B = E1 E2 A. (Hint: You can take both E1
and E2 to be of type III.)
Solution: We can get to B

3 2
1 0
1 1
from A by performing the following elementary row operations:





1 0 5
3 2 1
1
(i)
(ii)
2  −→ 4 2 3  −→ 4 2 3  ,
1 1 −2
1 1 −2
−2
where in (ii) we replace Row 2 with its sum with Row 1, and in (i) we then replace Row 1
with its sum with −2 times Row 3. These two operations correspond to the matrices


1 0 0
E2 = 1 1 0 ,
0 0 1


1 0 −2
E1 = 0 1 0  .
0 0 1
Note that this gives E1 E2 A = B, as desired, since in the product we first multiply E2
against A, and then E1 .
7.
Let V be a vector space, and suppose that W1 and W2 are both subspaces of V . Let
W = {w1 + 3w2 | w1 ∈ W1 , w2 ∈ W2 }.
Prove that W is a subspace of V .
Solution: By the definition of subspace, in order to show that W is a subspace of V , we
need to verify that it is nonempty and that it is closed under both scalar multiplication and
addition. It is certainly nonempty because if we take any w1 ∈ W1 and any w2 ∈ W2 , then
w1 + 3w2 ∈ W .
W is closed under scalar multiplication: Let w ∈ W and c ∈ R be arbitrary. Then w =
w1 + 3w2 for some w1 ∈ W1 and some w2 ∈ W2 . We scale by c:
cw = c(w1 + 3w2 ) = cw1 + 3(cw2 ).
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Now since cw1 ∈ W1 and cw2 ∈ W2 (as W1 and W2 are themselves closed under scalar
multiplication), it follows that cw is an element of W , and so W is closed under scalar
multiplication.
W is closed under addition: Let w, v ∈ W . Then we can find w1 , v1 ∈ W1 and w2 ,
v2 ∈ W2 so that w = w1 + 3w2 and v = v1 + 3v2 . Then
w + v = (w1 + 3w2 ) + (v1 + 3v2 ) = (w1 + v1 ) + 3(w2 + v2 ).
Since w1 +v1 ∈ W1 and w2 +v2 ∈ W2 (as W1 and W2 are themselves closed under addition),
it follows that w + v is an element of W , and so W is closed under addition.
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