APPM 3310: Matrix Methods — Exam #1.001 Solutions — October 1, 2010
Problem 1. (30 points) For this problem, let A=
1−2 1 −1
2−3 2 −3
3−5 3 −4
−1 1 −1 2
.
(a) (7 pts) Find the determinant of A, justify your answer.
(b) (7 pts) Is the matrix A regular? Why or why not?
(c) (4 pts) What is the rank of A?
(d) (12 pts) Find all solutions to the system Ax =b, where b=
1
1
2
0
Solution:
(a) det(A) = 0 since Ahas two identical columns.
(b) Note that reducing A to upper triangular form yields U=
1−2 1 −1
0 1 0 −1
0 0 0 0
0 0 0 0
and so A is not regular
since it does not have non-zero pivots.
(c) rank(A) = 2
(d) Using Gaussian elimination we have,
1−2 1 −1 1
2−3 2 −3 1
3−5 3 −4 2
−1 1 −1 2 0
→
1−2 1 −1 1
0 1 0 −1−1
0 1 0 −1−1
0−1 0 1 1
→
1−2 1 −1 1
0 1 0 −1−1
0 0 0 0 0
0 0 0 0 0
so we have
x−2y+z−w= 1
y−w=−1
0 = 0
0 = 0
→
x=−1−z+ 3w
y=−1 + w
z=z
w=w
→
x
y
z
w
=
−1
−1
0
0
+
−1
0
1
0
z+
3
1
0
1
w
thus the solution of the system is
−1
−1
0
0
+
−1
0
1
0
s+
3
1
0
1
t
s, t ∈R
.
Problem 2. (40 points) Let S M2×2denote the set of all symmetric 2 ×2 matrices, that is,
SM2×2= a b
b c :a, b, c ∈R
(a) (12 points)Prove SM2×2is a subspace of M2×2(the space of all 2 ×2 matrices).
(b) (8 points)Find the dimension of SM2×2by exhibiting a basis.
(c) (10 points) Given the following matrices
M1=1 1
1 2 , M2=−2 0
0−2, M3=3 2
2 5
is {M1, M2, M3}a basis of SM2×2? Justify your answer.
