MATH 1105, PARTIAL EXAM 3
PRACTICE PROBLEMS
MAY 7, 2011
JOHN GOODRICK
These are solutions to some of the practice problems from the textbook for
the partial exam on Wednesday, May 11.
Section 4.1
9. The determinant is 9.
13. a b = 6i + 3j + 5k = (6, 3, 5)
MATEMATICAS 1105, PARTIAL 1
SOLUTIONS
JOHN GOODRICK
1. For each pair of vectors, say whether they are (i) parallel, (ii) orthogonal,
or (iii) neither parallel nor orthogonal.
(a) [1, 2, 3, 1] and [1, 1, 0, 1]
If these vectors were parallel, then we would
MATEMATICAS 1105, QUIZ 1
SOLUTIONS
JOHN GOODRICK
1. Write down three dierent vectors in R3 which are in the span of [1, 0, 1]
and [0, 1, 0].
There are innitely many possible vectors in this span, for example:
1 [1, 0, 1] + 0 [0, 1, 0] = [1, 0, 1] ,
0 [1,
MATEMATICAS 1105, QUIZ 1
SOLUTIONS
JOHN GOODRICK
1. Write down three dierent vectors in R4 which are in the span of [1, 0, 1, 1]
and [0, 1, 0, 1].
SOLUTION: There are innitely many possible solutions, for instance:
1 [1, 0, 1, 1] + 0 [0, 1, 0, 1] = [1, 0
MATH 1105, PARTIAL EXAM 1
FEBRUARY 23, 2011
SOLUTIONS
JOHN GOODRICK
Problem 1. For each of the matrices below, put it into reduced row echelon
form using elementary row operations.
1 2 1 1
A= 0 0 0 0
1 1 4 5
1 0 2 1 0
B= 0 1 3 3 0
0 0 0 0 2
0 1 0
1
3
C
MATH 1105, PARTIAL EXAM 1
SOLUTIONS TO PRACTICE PROBLEMS
JOHN GOODRICK
Problem 1. For each of the matrices below, determine whether or not it is
in row echelon form. If it is not, put it into row echelon form using a single
elementary row operation.
1 2 3
MATEMATICAS 1105, QUIZ 3
PRACTICE VERSION, SOLUTIONS
JOHN GOODRICK
1. For each matrix below, nd its determinant.
1 2 1
A= 2 1 2
1 2 1
Solution: The row replacement operation of subtracting Row 3 from Row 1
does not change the determinant, so
1 2
2 1
1 2
MATEMATICAS 1105, QUIZ 1
SOLUTIONS TO PRACTICE VERSION
JOHN GOODRICK
All the topics for this quiz are explained in sections 1.4 (solving systems of
linear equations) and 1.5 (inverse matrices) of the textbook.
1. (a) For each augmented matrix below, expl
MATEMATICAS 1105, PARTIAL 3
SOLUTIONS TO SOME PRACTICE PROBLEMS
JOHN GOODRICK
4.1, exercise 31: The direction vector between the vertices (1, 0, 1) and
(2, 1, 6) is (2 1, 1 0, 6 1) = (3, 1, 5), and similarly the direction vector
between the vertices (3,