MATH 3191 Midterm Exam Solutions
Timothy Vis
May 3, 2010
1. Solve the following linear system completely.
x1 + 2x2 2x3 + 5x4 =
4
3x3 2x4 =
4
2x1 4x2 + 4x3 6x4 = 4
Set up and reduce the augmented matrix for this system.
1
1
2 2
5
4
0
0
3 2
4 0
0
2 4
4
MATH 3191. Homework 6
7 solution key.
4.3
SOLUTIONS
1. Consider the matrix whose columns are the given set of vectors. This 3 3 matrix is in echelon form,
and has 3 pivot positions. Thus by the Invertible Matrix Theorem, its columns are linearly
independe
MATH 3191. Homework 4 solution key.
2.1
SOLUTIONS
2
1. 2 A = ( 2)
4
1 4
=
2 8
0
5
7
B 2A =
1
5
4
2
. Next, use B 2A = B + (2A):
4
0
110
1 4
+
3 8
2 3
=
4 7
0
10
5
6
3
7
The product AC is not defined becauuse the number of columns of A does not match t
MATH 3191. Homework 7
8 solution key.
4.6
SOLUTIONS
1. The matrix B is in echelon form. There are two pivot columns, so the dimension of Col A is 2. There
are two pivot rows, so the dimension of Row A is 2. There are two columns without pivots, so the
equ
MATH 3191. Homework 10 solution key.
6.2
SOLUTIONS
1 3
1. Since 4 4 = 2 0, the set is not orthogonal.
3 7
6 3
3. Since 3 1 = 30 0, the set is not orthogonal.
9 1
3 1 3 3 1 3
2 3 2 8 3 8
5. Since = = = 0, the set is orthogonal.
1 3 1 7 3 7
3
MATH 3191-001. Exam 3. April 29, 2015. Solution key.
1. Find the eigenvalues of a given matrix.
1
2 3
0 1
2 3
= 2 3 + 2 = ( 1)( 2)
1 = 1 , 2 = 2 .
2. Determine whether a given vector is an eigenvector of a given matrix.
1
32 9
v=
, A=
3
108 31
32
MATH 3191-001. Exam 2. March 18, 2015. Solution key.
1. If the null space of a 9 7 matrix is 4-dimensional, find Rank A , dim Row A , and
dim Col A .
By the Rank theorem, Rank A = 7 dim Nul A = 3 , dim Row A = Rank A = 3 .
Also, dim Col A = Rank A = 3 .
2
MATH
3191.
Homework
11
key. key.
10 solution
MATH
3191.
Homework
10
9 solution
5.5
SOLUTIONS_
1
1. A =
1
2
1
, A I =
3
1
2
3
det( A I ) = (1 )(3 ) (2) = 2 4 + 5
Use the quadratic formula to find the eigenvalues: = 4 16 20 = 2 i. Example 2 gives a
2
Homework 3 solution key.
1.7
SOLUTIONS
1. Use an augmented matrix to study the solution set of x1u + x2v + x3w = 0 (*), where u, v, and w are
7
9 0 5 7 9 0
5
2
4 0 ~ 0 2 4 0 , there are no free variables. So
the three given vectors. Since 0
0 6 8 0 0 0 4
MATH 3191. Homework 11 solution key.
6.5
SOLUTIONS
1. To find the normal equations and to find x , compute
1
A A=
2
T
1
A b=
2
T
2
3
2
3
1
1
2
3
1
1
3
2
6
3 =
11
3
11
22
4
4
1 = 11
2
6
a. The normal equations are ( AT A)x = AT b :
11
b.
MATH 3191. Homework 8
9 solution key.
5.1
SOLUTIONS
1. The number 2 is an eigenvalue of A if and only if the equation Ax = 2x has a nontrivial solution.
s
This equation is equivalent to ( A 2 I )x = 0. Compute
3 2 2 0 1 2
A 2I =
=
3 8 0 2 3 6
The colum