Math 2270, Fall 2015
Instructor: Thomas Goller
29 September 2015
Test #2 Solutions
1
Inventions (10 points)
(a) Invent a 3 ⇥ 3 matrix A that is not invertible.(2 points)
Possible solution:
2
3
0 0 0
A = 4 0 0 05
0 0 0
(b) Invent 2 ⇥ 2 matrices A and B such that AB 6= BA. (2 points)
Possible solution:

0 0
A=
,
1 0
B=

1 0
0 0
(c) Invent a 2⇥3 matrix A with no zero entries such that the linear transformation T : R3 !
R2 defined by T (~x) = A~x is not onto. (2 points)
Possible solution:
A=

1
2
1
2
1
2
(d) Invent a 2 ⇥ 2 matrix A such that A 6= I2 and A2 = I2 . (2 points)
Possible solution:
A=

0 1
1 0
(e) Invent a set of linearly independent vectors in R3 that contains as many vectors as
possible. (2 points)
Possible solution:
8 2 3 2 3 2 39
0
0 =
< 1
405 , 415 , 405
:
;
0
0
1
Math 2270, Fall 2015
2
Instructor: Thomas Goller
29 September 2015
Inverse Matrix (10 points)
2
3
1 0 0
Let A = 4 0 1 15.
2 0 1
⇥
⇤
(a) Let B be the inverse of A. Compute B using row reduction on A I3 . (5 points)
Solution:
2
3
2
1 0 0 1 0 0
1 0 0 1 0
4 0 1 1 0 1 0 5 ! 40 1 1 0 1
2 0 1 0 0 1
0 0 1 2 0
2
1 0
B=4 2 1
2 0
3
2
0
1 0 0
05 ! 4 0 1 0
1
0 0 1
3
0
15
1
1 0
2 1
2 0
3
0
15
1
(b) Show B is the inverse of A by computing BA. (2 points)
Solution:
2
3
2
3 2
3 2
3
0
1 0 0
1
0
0
1 0 0
1 5 A = 4 0 1 15 = 4 2 + 2 1 1 1 5 = 4 0 1 0 5
1
2 0 1
2 2 0
1
0 0 1
1 0
4
2 1
BA =
2 0
2
3
1
(c) Use B to solve A~x = 4 35. (3 points)
4
Solution:
2
3 2
1
1 0
~x = B 4 35 = 4 2 1
4
2 0
32 3 2
3 2 3
0
1
1
1
15 4 35 = 4 2 3 45 = 4 95
1
4
2+4
6
Comment: You have to multiply both sides of the equation on the left by B to solve for
~x. Multiplying both sides of the equation on the right by B doesn’t make sense because
both sides of the equation are vectors in R3 . Recall that left versus right is an issue
because matrix multiplication doesn’t commute!
Math 2270, Fall 2015
3
Instructor: Thomas Goller
29 September 2015
Geometry (10 points)

2
and let T : R2 ! R2 be the linear transformation defined by T (~x) = A~x.
1
⇢
⇢
1
0
2
(a) In one copy of R , draw Span
and Span
. In another copy of R2 , draw
0
1
what you get when T acts on all points of those two lines. (4 points)
Let A =
1
0
Description
of solution:⇢In one copy of R2 , you should draw the axes and label them
⇢
1
0
Span
and Span
. In the other copy of R2 , you should draw the horizontal
0
1
⇢
1
axis (the image of T acting on Span
) as well as the line through the origin and
0

2
the vector
.
1
Comment: The question never asks you to draw the⇢
span of a set of two vectors, so you
1
0
shouldn’t be getting all of R2 . For instance, Span
,
would be all of R2 , but
0
1
the problem doesn’t ask for that.




0
1
0
1
(b) In one copy of R2 , draw the square with vertices
,
,
,
. In another copy of
0
0
1
1
R2 , draw what you get when T acts on all points of that square. (4 points)
Description of solution: In one copy of R2 , you should draw the square with the vertices
given in the problem.
copy of R2 , you should draw the parallelogram with

 In another

1
0
1
2
vertices
,
,
,
.
0
0
1
1
(c) What geometric shape would you get if T acts on all points of a line that does not
contain the origin? (1 point)
Solution: A line.
(d) What geometric shape would you get if T acts on all points of a parallelogram that is
far from the origin? (1 point)
Solution: A parallelogram.
Math 2270, Fall 2015
4
Instructor: Thomas Goller
29 September 2015
One-to-one (10 points)
2
1
4
2
Let A =
1
3
2
4 5 and let T : R2 ! R3 be the linear transformation defined by T (~x) = A~x.
2
(a) Write down a linear dependence relation among the columns of A. (2 points)
Possible solution:
2
3
2
4 4 5=
2
2
3
1
2 4 25
1
(b) Based on your answer in (a), is T one-to-one? (1 point)
Solution: No.
(c) Invent two distinct vectors ~x and ~y in R2 such that T (~x) = T (~y ). Hint: try to use your
answer in (a). (2 points)
Possible solution:
~x =

2
,
0
~y =

0
.
1
(d) Compute the echelon form of A. (3 points)
2
1
4 2
1
3
2
2
1
5
4
4 ! 0
2
1
3
2
2
1
5
4
0 ! 0
2
0
3
2
05
0
(e) Look at the pivots in your answer for (d). Is T one-to-one? Explain. (2 points)
Solution: T is not one-to-one because not every column of A has a pivot position.
Comment: Try to be precise with your explanation. Instead of just saying something
like “A has only one pivot”, point out that “there is a column of A that has no pivot”,
which emphasizes what is really causing T to not be one-to-one.
Math 2270, Fall 2015
5
Instructor: Thomas Goller
29 September 2015
Linear Transformations (10 points)
Let T : R3 ! R2 be a linear transformation such that T (~e1 ) =

02 3 1

0
1
2
@
4
5
A
1
and T
=
.
3
2
1
(a) Compute each of the following vectors if you can, or state that not enough information
is given: (3 points)
Solution:
02 3 1


0
1
0
(i) T @405A = T (0 ~e1 ) = 0 T (~e1 ) = 0
=
3
0
0
02 3 1
2 3
0
0
(ii) T @415A Not enough information is given: ~e2 is not in the span of ~e1 and 415.
0
1
02 3 1
02 3 2 31
02 3 1
02 3 1



1
1
0
1
0
1
2
1
@
4
5
A
@
4
5
4
5
A
@
4
5
A
@
4
5
A
1
0 + 1
0
1
(iii) T
=T
=T
+T
=
+
=
3
2
5
1
0
1
0
1
Now suppose you also know that T (~e3 ) =

0
.
5
(b) Compute all vectors you couldn’t compute in (a). (3 points)
Solution:
02 3 1
02 3
0
0
T @4 1 5 A = T @4 1 5
0
1
2 31
02 3 1
0
0
4 0 5 A = T @4 1 5 A
1
1
02 3 1

0
2
T @4 0 5 A =
2
1

0
=
5

2
.
7
(c) Compute a matrix A such that T (~x) = A~x for all ~x in R3 . (4 points)
Solution:
A=

1
3
2 0
7 5
Comment: You have already computed T (~e1 ), T (~e2 ), and T (~e3 ) in earlier parts of the
problem! The reason I wrote “Compute” and made this part worth 4 points is to allow
you to still get quite a few points by finding the matrix by another method (like trial
and error) if you were confused by the first parts of the problem. A number of you wrote
A = I3 , which I find ba✏ing. I am curious where I3 came from, so please enlighten me!