MAT 1302 Midterm: sm09-1
University of Ottawa
Department of Mathematics and Statistics
MAT 1302B: Mathematical Methods II
Instructor: Alistair Savage
First Midterm Exam – February 6, 2009
Surname First Name
Student # DGD (1–4)
Instructions:
(1) You have 80 minutes to complete this exam.
(2) The number of points available for each question is indicated in square brackets.
(3) You must show your work and justify your answers to receive full marks. Partial
marks may be awarded for making sufficient progress towards a solution.
(4) All work to be considered for grading should be written in the space provided. The
reverse side of pages is for scrap work. If you find that you need extra space in order
to answer a particular question, you should continue on the reverse side of the page
and indicate this clearly. Otherwise, the work written on the reverse side of pages
will not be considered for marks.
(5) Write your student number at the top of each page in the space provided.
(6) No notes, books, calculators or scrap paper are allowed.
(7) You should write in pen, not pencil.
(8) The final page of the exam may be used for scrap work.
(9) Good luck!
Please do not write in the table below.
Question 1 2 3 4 5 6 Total
Maximum 4 6 5 5 5 5 30
Score
Student # MAT 1302B First Midterm Exam, February 6, 2009
1. [4] Find all solutions to the vector equation
x1
1
0
−1
+x2
−3
2
5
+x3
3
−2
−1
=
2
4
−2
.
Solution: We reduce the corresponding augmented matrix to echelon form.
1−3 3 2
0 2 −2 4
−1 5 −1−2
R3+R1→R3
−−−−−−−→
1−3 3 2
0 2 −2 4
0 2 2 0
R3−R2→R3
−−−−−−−→
1−3 3 2
0 2 −2 4
0 0 4 −4
1
2R2→R2
1
4R3→R3
−−−−−→
1−3 3 2
0 1 −1 2
0 0 1 −1
R1−3R3→R1
R2+R3→R2
−−−−−−−→
1−3 0 5
0 1 0 1
001−1
R1+3R2→R1
−−−−−−−→
1 0 0 8
0 1 0 1
0 0 1 −1
Therefore the unique solution to the vector equation is
x1= 8, x2= 1, x3=−1.
Page 2 of 8
Student # MAT 1302B First Midterm Exam, February 6, 2009
2. [6] Is the following linear system consistent? If so, find the general solution.
x1+ 12 = 3x2+ 3x3−x4
x1+ 2x2+ 5x3+ 2 = x2+ 3x4+ 6
−x1+x2−x3+x4−4 = 0
−3x2+x4=−12 −x1+ 3x3
Solution: We first write the system in standard form:
x1−3x2−3x3+x4=−12
x1+x2+ 5x3−3x4= 4
−x1+x2−x3+x4= 4
x1−3x2−3x3+x4=−12
We then write the augmented matrix and row reduce:
1−3−3 1 −12
115−3 4
−1 1 −1 1 4
1−3−3 1 −12
R2−R1→R2
R3+R1→R3
R4−R1→R4
−−−−−−−→
1−3−3 1 −12
0 4 8 −4 16
0−2−4 2 −8
0 0 0 0 0
R3−1
2R2→R3
−−−−−−−−→
1−3−3 1 −12
0 4 8 −416
0 0 0 0 0
0 0 0 0 0
1
4R2→R2
−−−−−→
1−3−3 1 −12
0 1 2 −14
0 0 0 0 0
0 0 0 0 0
R1+3R2→R1
−−−−−−−→
1 0 3 −20
0 1 2 −1 4
0 0 0 0 0
0 0 0 0 0
Since the rightmost column is not a pivot column, the system is consistent. We return to
equation notation.
x1+ 3x3−2x4= 0
x2+ 2x3−x4= 4
The basic variables are x1and x2and the free variables are x3and x4. Solving for the basic
variables in terms of the free variables, we obtain the general solution to the system:
x1=−3x3+ 2x4
x2=−2x3+x4+ 4
x3, x4free
Page 3 of 8
20
MAT 1302 Full Course Notes
Verified Note
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Document Summary
Dgd (1 4) (1) you have 80 minutes to complete this exam. (2) the number of points available for each question is indicated in square brackets. (3) you must show your work and justify your answers to receive full marks. Partial marks may be awarded for making su cient progress towards a solution. (4) all work to be considered for grading should be written in the space provided. The reverse side of pages is for scrap work. If you nd that you need extra space in order to answer a particular question, you should continue on the reverse side of the page and indicate this clearly. Please do not write in the table below. Mat 1302b first midterm exam, february 6, 2009: [4] find all solutions to the vector equation. Solution: we reduce the corresponding augmented matrix to echelon form. Therefore the unique solution to the vector equation is x1 = 8, x2 = 1, x3 = 1.
