Download Math 194 Clicker Questions

Math 194
Clicker Questions
Section 1.2: Row Reduction and Echelon Forms
1. Clicker Question: Suppose the 3 × 5 augmented matrix for a system of linear equations has three
pivot columns. True or False: The system must be consistent. Answer: False. If the rightmost
column of the augmented matrix is a pivot column (that is, if the third pivot column is the rightmost
column), then the bottom row of the augmented matrix would have the following form:
0 0 0 0 b ,
where b is some number not equal to 0. In this case, the corresponding linear equation would be 0 = b,
which indicates that the system is inconsistent.
2. Clicker Question: Suppose the coefficient matrix of a 3 × 3 SLE has a pivot in each column. True
or False: The system has exactly one solution. Answer: True. This implies that the augmented
matrix cannot have a bottom row like the one in the previous question, which means that the system
is consistent. Furthermore, there cannot be any free variables and so there cannot be infinitely many
solutions.
3. Clicker Question: Suppose you row-reduce the coefficient matrix for a linear system and end up
with a row of all zeros. How many solutions does the system have? Mark all that are possible.
(a) No solutions
(b) Exactly one solution
(c) Infinitely many solutions
Answer: All three are possible. Consider the following augmented matrices.


∗ ∗
0 ∗
0 0 

∗ ∗
 0 ∗
0 0 0


∗ ∗ ∗
 0 0 ∗
0 0 0 0
4. Clicker Question: We find that a system of three linear equations in three variables has an infinite
number of solutions. How could this happen? Mark all that are possible.
(a) We have three equations for the same plane.
(b) At least two of the equations represent the same plane.
(c) The three planes intersect along a line.
(d) The planes represented are parallel.
Answer: All four are possible, given the wording of the answer choices. Responses (a) and (c) are
relatively straightforward. If at least two of the equations represent the same plane and the third plane
intersects this plane or is the same as this plane, then you’ll have infinitely many solutions. So it’s
possible that response (b) could be true. As for response (d), if the planes are parallel and happen to
all be the same plane, then you’re back to response (a). If, however, the planes are parallel and one of
them is not the same as another, then there would be no solution.
Section 1.3: Vector Equations
1. Clicker Question:
Every vector in R2 can be written as a linear combination of the
True or False:
−1
2
vectors v1 =
and v2 =
.
1
1
(a) True
(b) False
Answer: (a). We have a few options for showing this to be true: solving a system of linear equations,
showing that the system is consistent using row reduction, showing that the system is consistent by
noting that the lines described by the system must intersect, or noting that the vectors are not parallel
(allowing us to use the head-to-tail method to get anywhere in the plane).
2. Clicker Question: Suppose v1 , v2 , and v3 are vectors in R3 . Consider the set of all linear combinations of v1 , v2 , and v3 . If x and y are two vectors in this set, is any linear combination of x and y
also in the set?
(a) Yes
(b) No
Answer: (a). Given that x and y are in the span of v1 , v2 , and v3 , it must be the case that
x = a1 v1 + a2 v2 + a3 v3
y = b1 v1 + b2 v2 + b3 v3
for some scalars a1 , a2 , a3 , b1 , b2 , and b3 . Now take an arbitrary linear combination of x and y:
αx + βy = α(a1 v1 + a2 v2 + a3 v3 ) + β(b1 v1 + b2 v2 + b3 v3 )
= αa1 v1 + αa2 v2 + αa3 v3 + βb1 v1 + βb2 v2 + βb3 v3
= (αa1 + βb1 )v1 + (αa2 + βb2 )v2 + (αa3 + βb3 )v3
which is also a linear combination of v1 , v2 , and v3 .




 
6
3
1
3. Clicker Question: Let v1 =  1  , v2 =  0  , v3 =  0  . Which of the following vectors is
−2
−1
2
not in the span of {v1 , v2 , v3 }?
 
1
(a) 0
0
 
4
(b) 1
1
 
3
(c) 3
6
(d) All of these are in the subspace of R3 spanned by {v1 , v2 , v3 }.
Answer: (a). The only way to make
 the
 second component zero is to only use
 vectors v2 and v3 .
1
1
However, we note that v3 = 2v2 and 0 is not a multiple of v2 , so the vector 0 cannot be written
0
0
 
 
4
3
as a linear combination of the vectors. Also note that 1 = v1 + v2 and 3 = 3v1 , so both of these
1
6
vectors are in the span of {v1 , v2 , v3 }.






1
3
6
4. Clicker Question: Let v1 =  1  , v2 =  0  , v3 =  0  . Geometrically, what is the
2
−1
−2
subspace of R3 spanned by {v1 , v2 , v3 }?
(a) A point
(b) A line
(c) A plane
(d) All of R3
Answer: (c). The vectors v2 and v3 are multiples, thus our set contains only two linearly independent
vectors, and so the subspace spanned by these three vectors is two-dimensional.






1
3
6
5. Clicker Question: Let v1 =  1  , v2 =  0  , v3 =  0  . Which of the following sets has
2
−1
−2
the same span as the set of all three vectors {v1 , v2 , v3 }? Mark all that apply.
(a) {v1 , v2 }
(b) {v2 , v3 }
(c) {v1 , v3 }
(d) None of the above
Answer: Since v2 and v3 are multiples of each other, either of (a) or (c) will work.
 
 
 
−1
3
1
6. Clicker Question: Let v1 =1, v2 = 0 , and v3 = 2 . Geometrically, what is the subspace
5
−1
2
spanned by {v1 , v2 , v3 }?
(a) A point
(b) A line
(c) A plane
(d) All of R3
Answer: (c). Since v1 and v2 are not parallel, they generate a plane, which eliminates options (a)
and (b). The question now is, Is v3 a linear combination of v1 and v2 ?
In this case, one can show that v3 = 2v1 − v1 . (You’ll need to do some row reduction here and show
how we can find those coefficients.) Since v3 can be written as a linear combination of v1 and v2 , it
lies in that plane and thus doesn’t add anything to the span of the vectors. If v3 could not be written
as a linear combination of the other two vectors, then it would project from the plane and allow the
three vectors to generate all of R3 .
 
 
 
1
3
−1
7. Clicker Question: Let v1 =1, v2 = 0 , and v3 = 2 . Which of the following sets has the
2
−1
5
same span as the set of all three vectors {v1 , v2 , v3 }? Mark all that apply.
(a) {v1 , v2 }
(b) {v2 , v3 }
(c) {v1 , v3 }
(d) None of the above
Answer: Since v3 = 2v1 − v2 , it follows that v2 = 2v1 − v3 and v1 = 21 v2 + 12 v3 . Thus we can remove
any one of the three vectors from the set without changing the span of the set and so (a), (b), and (c)
are all correct.
Section 1.4: Matrix Equations
1. Clicker Question: Note that x1 = 11, x2 = 2, x3 = 1 is a solution to the following system of linear
equations.
x1 + 4x2 + 2x3 = 21
2x1 − x2 − 5x3 = 15
Which of the following is not true?
 
11
(a) The vector x =  2  is a solution to the matrix equation
1
1 4
2
21
x=
2 −1 −5
15
.
(b) b1 = 21, b2 = 15 is a solution to the vector equation
   
 
2
11
1
b1 4 + b2 −1 =  2 
−5
1
2
.
(c) x1 = 11, x2 = 2, x3 = 1 is a solution to the vector equation
1
4
2
21
x1
+ x2
+ x3
=
2
−1
−5
15
.
Answer: (b). Given the row-vector rule for computing Ax found on page 45, we have that
 
11
(1)(11) + (4)(2) + (2)(1)
21
1 4
2  
2 =
=
.
(2)(11) + (−1)(2) + (−5)(1)
15
2 −1 −5
1
Thus choice (a) is true.
Using the definition of the product Ax, we have that
 
11
21
1 4
2  
1
4
2
21
2 = 11
=
+2
+1
=
.
15
2 −1 −5
2
−1
−5
15
1
Thus choice (c) is true.
The easiest way to show that choice (b) is not true is to consider the first row of the equation given in
(b) and note that
(b1 )(1) + (b2 )(2) = (21)(1) + (15)(2) 6= 1(1)
2. Clicker Question: Note that x1 = 11, x2 = 2, x3 = 1 is a solution to the following system of linear
equations.
x1 + 4x2 + 2x3 = 21
2x1 − x2 − 5x3 = 15
Which of the following is not a direct implication of this fact?
1 4
2
(a) The matrix equation Ax = b, where A =
, has a solution for any b in R2 .
2 −1 −5
21
1 4
2
(b) The vector
is a linear combination of the columns of the matrix
.
15
2 −1 −5
21
1
4
2
2
(c) The vector
is in the subset of R spanned by the vectors
,
, and
.
15
2
−1
−5
Answer: (a). As noted in the previous question,
21
1
4
2
21
= 11
+2
+1
=
.
15
2
−1
−5
15
Thus choice (b) is true.
1
4
2
Since the the subset of R2 spanned by the vectors
,
, and
consists of all possible linear
2
−1
−5
21
combinations of those vectors, and the vector
is a linear combination of those vectors, then choice
15
(c) is true.
21
As for choice (a), even though the matrix equation Ax = b is consistent for b =
, that does not
15
imply that it is consistent for any choice of b. Further analysis would be needed to verify this fact.
(It is, in fact, true.)


 
1
5
b1
3. Clicker Question: Let A = −2 −13 and b = b2 . Suppose that the matrix equation Ax = b
3
−3
b3
is consistent. Which of the following is not true?
(a) The entries of b must satisfy the equation
15b1 + 6b2 − b3 = 0.
(b) The vector b lies in the plane consisting of all linear combinations of the columns of A.
(c) The echelon form of A has a row of zeros.
(d) The vector b can be any vector in R3 .
Answer: (d). The matrix equation Ax = b is consistent if and only if the echelon form of the
augmented matrix


1
5
b1
−2 −13 b2 
3
−3 b3
does not contain a row of the form 0 0 c where c is any nonzero number. Using the row reduction
algorithm, we have

 
 

1
5
b1
1
5
b1
1 5
b1
−2 −13 b2  ∼ 0 −3
.
2b1 + b2  ∼ 0 −3
2b1 + b2
3
−3 b3
0 −18 −3b1 + b3
0 0 −15b1 − 6b2 + b3
Thus, the matrix equation Ax = b is consistent if and only if
−15b1 − 6b2 + b3 = 0.
Thus, choice (a) is true.
As for choice (2), the matrix equation Ax = b is consistent if and only if b is a linear combination of
the columns of A. Note that the columns of A are members of R3 and are not parallel vectors. Thus
their span is a plane in R3 . Thus, choice (b) is true.
As verified for choice (a), the echelon form of A does indeed have a row of zeros. Thus choice (c) is
true.
As verified for choice (a), the elements of b must satisfy is certain linear equation. Since not all vectors
in R3 satisfy this equation, choice (d) is not true.
Note that we can’t have both (a) and (d) true!
Section 1.5: Solution Sets of Linear Systems
1. Clicker Question: Suppose the matrix equation Ax = 0 has only the trivial solution x = 0. True or
False: The matrix equation Ax = b is consistent for any vector b of the appropriate size.
(a) True - High Confidence
(b) True - Low Confidence
(c) False - Low Confidence
(d) False - High Confidence
Answer: False, but this one is tricky. Since Ax = 0 has only the trivial solution, it follows that the
corresponding system of linear equations has no free variables. That means that A has a pivot position
in every column. If A is square, then this implies that A has a pivot position in every row and so by
Theorem 4, Ax = b has a solution for any b. However, if A is not square, what could happen?
It can’t be the case that A has more columns than rows. Consider a 3 × 4 matrix: It can only have 3
pivot positions, but A needs to have a pivot position in each of its 4 columns. So if A is not square, it
must have more rows than columns. Consider a 4 × 3 matrix: If it has a pivot position in every column,
it will have to have a row of all zeros. This means that for some vectors b, the augmented matrix
could have a row of all zeros except for the last column entry, which would imply that the system is
inconsistent.
Whew! So if A is square, then this statement is true. However, if A has more rows than columns, the
statement is false.
Section 1.7: Linear Independence
1. Clicker Question: Which of the following is a set of linearly independent vectors?
     
2
−2
0
(a)  1 ,  4 , 0
−1
7
0
       
2
−2
5
1
(b)  1 ,  4 ,  2 ,  1 
−1
7
−1
10
     
2
−2
2
(c)  1 ,  4 , −2
−1
7
6
    

2
−2
6
(d)  1 ,  4 , −12
−1
7
−21
(e) None of these are linearly independent sets of vectors.
Answer: (c) Note that set 1 contains the zero vector, and so the set must be linearly dependent.
Note that set (b) contains more vectors (four) than there are entries in each vector (three), and so the
matrix equation Ax = 0, where the columns of A are the vectors in question, must have a free variable,
implying that there are nontrivial linear combinations of the columns that equal the zero vector.


 
6
−2
Note that in set (d), the third vector −12 is equal to −3 times the second vector  4 . Since one
−21
7
vector in the set is a linear combination of the prior vectors in the set (in this case, one vector is a
scalar multiple of a prior vector), it follows that the set is linearly independent.
This leaves only set (c). To confirm that this is a linear independent set, we must show that the vector
equation
 
 
 
2
−2
2
x1  1  + x2  4  + x3 −2 = 0
−1
7
6
has only the trivial solution. This vector equation is equivalent to the system of linear equations with
augmented matrix


2 −2 2 0
1
4 −2 0
−1 7
6 0
which is row-equivalent to the augmented matrix

1 −1 1


0 1 − 53

0 0
1
0



0 .

0
Since this matrix has a pivot position in all but its rightmost columns, the corresponding system of
linear equations has no free variables. Thus, the system can only have a unique solution–the trivial
solution x1 = x2 = x3 = 0. Thus set (c) is a linearly independent set.
2. Clicker Question: Which subsets of the set of vectors shown are linearly dependent?
(a) u, w
(b) t, w
(c) t, v
(d) t, u, v
(e) None of these sets are linearly dependent.
(f) More than one of these sets is linearly dependent.
Answer: (f) Note that t and w are multiples of each other and so (b) is linearly dependent. Also,
you can’t have three linearly independent vectors in R2 , so (d) is also linearly dependent.
3. Clicker Question: To determine whether a set of n vectors from Rn is independent, we can form a
matrix A whose columns are the vectors in the set and then put that matrix in reduced row echelon
form. If the vectors are linearly independent, what will we see in the reduced row echelon form?
(a) A row of all zeros.
(b) A row that has all zeros except in the last position.
(c) A column of all zeros.
(d) An identity matrix.
Answer: (d) If the vectors are linearly independent, then the matrix equation Ax = 0 will have a
unique solution. Thus there must be a pivot in every column. Since A is a square matrix in this case,
that implies that A must row-reduce to the identity.
4. Clicker Question: If v1 , v2 and v3 are linearly dependent, then which of the following statements
must be true? Mark all that apply.
(a) There is a nontrivial solution to the vector equation c1 v1 + c2 v2 + c3 v3 = 0.
(b) There are infinitely many solutions to the vector equation c1 v1 + c2 v2 + c3 v3 = 0.
(c) The vector v3 can be written as a linear combination of v1 and v2 .
(d) None of the three vectors is a scalar multiple of another.
Answer: All four are true.
5. Clicker Question: Suppose you wish to determine whether
a set of vectors {v1 , v2 , v3 , v4 } is linearly
v1 v2 v3 v4 , and you calculate its reduced row echelon
independent.
You
form
the
matrix
A
=


1 0 2 1
form, R =  0 1 3 1  . You now decide to write v2 as a linear combination of v1 , v3 , and v4 .
0 0 0 0
Which is a correct linear combination?
(a) v2 = 3v3 + v4
(b) v2 = −3v3 − v4
(c) v2 = v4 − 3v3
(d) v2 = −v1 + v4
(e) v2 cannot be written as a linear combination of v1 , v3 , and v4 .
Answer: (d). The long way to this answer is to see that the parametric form of the general solution
to Ax = 0 must be
 
 
−2
−1
−3
−1

 
x = x3 
 1  + x4  0  .
0
1
 
−1
−1

Choose x3 = 0 and x1 = 1, we get the solution x =
 0  and so
1
−v1 − v2 + v4 = 0
which implies that v2 = −v1 + v4 . The short way is to see that r2 = −r1 + r4 and so
−r1 − r2 + r4 = 0
 
−1
−1

and so x =
 0  is a solution to Rx = 0. Since R is the reduced row echelon form of A, Rx = 0 has
1
 
−1
−1

is a solution to Ax = 0 and
the same solutions as Ax = 0. This implies that x = 
0
1
−v1 − v2 + v4 = 0
which implies that v2 = −v1 + v4 .
Section 1.8: Linear Transformations
1. Clicker Question: Which of the following is not a linear transformation?
x1
(a) L(x) =
x2 + 1
x − 2x2
(b) L(x) = 1
x1
4x2
(c) L(x) =
x1 − 2x2
x
(d) L(x) = 1
0
(e) All are linear transformations
(f) None are linear transformations
1
Answer: (a). Consider the vector x =
and the scalar α = 2.
3
1
2
2
L(αx) = L 2
=L
=
,
3
6
7
but
αL(x) = 2L
1
1
2
=2
=
,
3
4
8
so αL(x) 6= L(αx) and thus L is not linear. Options (b), (c), and (d) all satisfy the linearity conditions.
Section 1.9: The Matrix of a Linear Transformation
1. Clicker

1
A =0
0
Question: Consider the linear transformation L : R4 ⇒ R3 defined by L(x) = Ax where
−4 8 1
2 −1 3. Which of the following is true?
0
0 3
(a) L is onto, but not one-to-one.
(b) L is one-to-one, but not onto.
(c) L is both onto and one-to-one.
(d) L is neither onto nor one-to-one.
Answer: (a). First, consider the matrix equation Ax = b. This equation will always have a solution
since the reduced-row echelon form of A does not have a row of all zeroes. Thus, L is onto. Second,
note that the equation Ax = b will have infinitely many solutions because A has a free variable. Thus
L is not one-to-one.
Another way to see this is to note that L maps all of R4 into R3 . You can’t map a four-dimensional
space into a three-dimensional space in a one-to-one manner. Dimensional analysis shows us that L is
not one-to-one, but it doesn’t verify that L is onto. A linear transformation from R4 to R3 could map
R4 into a one- or two-dimensional subspace of R3 .
2. Clicker
Question:
Consider the linear transformation L : R2 ⇒ R3 defined by L(x) = Ax where


3 1
A =5 7. Which of the following is true?
1 3
(a) L is onto, but not one-to-one.
(b) L is one-to-one, but not onto.
(c) L is both onto and one-to-one.
(d) L is neither onto nor one-to-one.
Answer: (b). In this case, row reducing the matrix equation Ax = b will necessarily result in a row
of zeroes, and so there’s no guarantee of a solution for every b. Thus, L is not onto. Consider also the
dimensional analysis: There aren’t enough vectors in R2 to map onto everything in R3 .
As for checking that L is one-to-one, note that there will be no free variables in the reduced-row echelon
form of L and thus any solution to Ax = b will have to be unique. Dimensional analysis doesn’t help
us here: It’s possible for a linear transformation to map R2 into a one-dimensional subspace of R3 ,
which would not be a one-to-one mapping.