Math 2270, Fall 2014
Instructor: Thomas Goller
28 October 2014
Test #3
1
The Four Subspaces (10 points)
(a) Compute bases for the four fundamental subspaces for the matrix A =
are the dimensions of the four subspaces? (6 points)
✓
1
1
◆
T
✓
1
1
◆
Solution: N (A) = span
, C(A ) = span
, C(A) = span
✓ ◆
1
span
. Each of the four subspaces has dimension 1.
2

2
1
✓
2
1
2
. What
1
◆
, N (AT ) =
(b) Draw the four fundamental subspaces in (a) as pairs of orthogonal lines. Be sure to label
which line is which subspace! (4 points)
Idea of solution: You should draw N (A) and C(AT ) as orthogonal lines in one plane,
T
and C(A)
For example, N (A) =
✓ and
◆ N (A ) as orthogonal lines in another plane.

1
1
span
is the line through the origin with direction
, namely with slope 1.
1
1
Math 2270, Fall 2014
2
Instructor: Thomas Goller
28 October 2014
Inventions (10 points)
(a) Invent a matrix A such that det A = 3 and det(2A) = 12. (2 points)
Possible solution: A =

3 0
.
0 1
(b) Invent a permutation matrix P that is not the identity such that det P = 1. (2 points)
2
3
0 1 0
Possible solution: P = 40 0 15.
1 0 0
(c) Invent two matrices A and B so that det(A + B) 6= det A + det B. (2 points)


1 0
0 0
Possible solution: A =
,B=
.
0 0
0 1
(d) Invent three vectors (points) in R2 that do not lie on a common line. (2 points)
Possible solution:

0
,
1


1
2
,
.
1
1
(e) Invent a non-symmetric matrix A such that C(A) = C(AT ). (2 points)
Possible solution: A =

1 2
.
3 4
Math 2270, Fall 2014
3
Instructor: Thomas Goller
28 October 2014
Basis (10 points)
2
3
1 2 3
(a) Compute the determinant of the matrix A = 42 3 15. (4 points)
3 1 2
Solution:
1 2 3
3 1
2 3 1 =1
1 2
3 1 2
2
2 1
2 3
+3
=5
3 2
3 1
2
21 =
18.
(b) Is A invertible? (Hint: use part (a).) (1 point)
Solution: Yes because det A 6= 0.
(c) Are the columns of A independent? Explain your reasoning. (Hint: what is N (A)? ) (2
points)
Solution: Yes. Since A is invertible, N (A) = {~0}, so the columns of A are independent.
(d) Do the columns of A span R3 ? Explain your reasoning. (Hint: what is the dimension of
C(A)? ) (2 points)
Solution: Yes. Since A is invertible, it has rank 3, so C(A) is a 3-dimensional subspace
of R3 , which must be all of R3 .
(e) Are the columns of A a basis for R3 ? (Hint: combine (c) and (d).) (1 point)
Solution: Yes because the columns of A are independent and span R3 .
Math 2270, Fall 2014
4
Instructor: Thomas Goller
28 October 2014
Projection (10 points)
02 3 2 3 1
1
1
@
4
5
4
0 , 15A in R3 . (3 points)
(a) Find an orthonormal basis ~q1 , ~q2 for the plane S = span
0
1
2 3
2 3
1
1
~
4
5
4
Possible solution: Let ~q1 = 0 . Using Gram-Schmidt, B = 15
0
1
2 3
0
1 4 5
p
take ~q2 = 2 1 .
1
2 3 2 3
1
0
1 4 5
4
0 = 15, so
1
0
1
(b) Compute the projection matrix P that projects vectors in R3 orthogonally onto S. (Hint:
P = QQT when the columns of Q are an orthonormal basis of S.) (3 points)
2
1
4
Solution: Q = 0
0
0
3
2
1
p1 5, so P = 40
2
p1
0
2
0
3
p1 5
2
p1
2

1
0
0
0
p1
2
p1
2
2
3
1 0 0
= 40 12 12 5.
0 12 12
2 3
0
~
4
(c) Use your answer in (b) to compute the vector p~ in S that is closest to b = 15. Show
0
that the error vector ~e is orthogonal to S. (4 points)
2 3
0
~
4
Solution: p~ = P b = 12 5 and ~e = ~b
1
2
orthogonal to S = span(~q1 , ~q2 ).
2
p~ = 4
0
1
2
1
2
3
5. Since ~e · ~q1 = 0 and ~e · ~q2 = 0, ~e is
Math 2270, Fall 2014
5
Instructor: Thomas Goller
28 October 2014
Inverse (10 points)
2
3
1 2 3
(a) Compute the cofactor matrix C for the matrix A = 40 2 35. (Hint: the (i, j) entry
0 0 3
i+j
of C is ( 1) Mij .) (8 points)
Solution:
2
6
6
6
C=6
6
6
4
(b) Show that
1
CT
det A
2
0
2
0
2
2
3
3
3
3
3
3
0
0
1
0
1
0
3
3
3
3
3
3
0
0
1
0
1
0
2
0
2
0
2
2
3
7 2
7
6
7
7=4 6
7
7
0
5
3
0 0
3 05 .
3 2
is the inverse of A. (2 points)
Solution: Check that det1 A C T A = I,
2
32
6
6 0
1 2
1 4
5 40 2
0
3
3
6
0 0
2
0 0
where det A = 1 · 2 · 3 = 6:
3
2
3 2
3
3
6 0 0
1 0 0
35 = 61 40 6 05 = 40 1 05 .
3
0 0 6
0 0 1