06-240/Term Test: Difference between revisions
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2. 1) <math>\frac{1}{2+3i}+\frac{1}{2-3i}=\frac{(2-3i)+(2+3i)}{(2+3i)(2-3i)}=\frac{4}{2^2+3^3}=\frac{4}{13}</math> |
2. 1) <math>\frac{1}{2+3i}+\frac{1}{2-3i}=\frac{(2-3i)+(2+3i)}{(2+3i)(2-3i)}=\frac{4}{2^2+3^3}=\frac{4}{13}</math> |
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<math>\frac{1}{2+3i}-\frac{1}{2-3i}=\frac{(2-3i)-(2+3i)}{(2+3i)(2-3i)}=\frac{-6i}{2^2+3^3}=\frac{-6}{13}i</math> |
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Revision as of 19:46, 25 October 2006
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The Test
Front Page
Do not turn this page until instructed.
Math 240 Algebra I - Term Test
University of Toronto, October 24, 2006
Solve the 5 problems on the other side of this page.
Each of the problems is worth 20 points.
You have an hour and 45 minutes.
Notes.
- No outside material other than stationary and a basic calculator is allowed.
- We will have an extra hour of class time in our regular class room on Thursday, replacing the first tutorial hour.
- The final exam date was posted by the faculty - it will take place on Wednesday December 13 from 2PM until 5PM at room 3 of the Clara Benson Building, 320 Huron Street (south west of Harbord cross Huron, home of the Faculty of Physical Education and Health).
Questions Page
Solve the following 5 problems. Each of the problems is worth 20 points. You have an hour and 45 minutes.
Problem 1. Let [math]\displaystyle{ F }[/math] be a field with zero element [math]\displaystyle{ 0_F }[/math], let [math]\displaystyle{ V }[/math] be a vector space with zero element [math]\displaystyle{ 0_V }[/math] and let [math]\displaystyle{ v\in V }[/math] be some vector. Using only the axioms of fields and vector spaces, prove that [math]\displaystyle{ 0_F\cdot v=0_V }[/math].
Problem 2.
- In the field {\mathbb C} of complex numbers, compute
[math]\displaystyle{ \frac{1}{2+3i}+\frac{1}{2-3i} }[/math] and [math]\displaystyle{ \frac{1}{2+3i}-\frac{1}{2-3i} }[/math]. - Working in the field [math]\displaystyle{ {\mathbb Z}/7 }[/math] of integers modulo 7, make a table showing the values of [math]\displaystyle{ a^{-1} }[/math] for every [math]\displaystyle{ a\neq 0 }[/math].
Problem 3. Let [math]\displaystyle{ V }[/math] be a vector space and let [math]\displaystyle{ W_1 }[/math] and [math]\displaystyle{ W_2 }[/math] be subspaces of [math]\displaystyle{ V }[/math]. Prove that [math]\displaystyle{ W_1\cup W_2 }[/math] is a subspace of [math]\displaystyle{ V }[/math] iff [math]\displaystyle{ W_1\subset W_2 }[/math] or [math]\displaystyle{ W_2\subset W_1 }[/math].
Problem 4. In the vector space [math]\displaystyle{ M_{2\times 2}({\mathbb Q}) }[/math], decide if the matrix [math]\displaystyle{ \begin{pmatrix}1&2\\-3&4\end{pmatrix} }[/math] is a linear combination of the elements of [math]\displaystyle{ S=\left\{\begin{pmatrix}1&0\\-1&0\end{pmatrix},\ \begin{pmatrix}0&1\\0&1\end{pmatrix},\ \begin{pmatrix}1&1\\0&0\end{pmatrix}\right\} }[/math].
Problem 5. Let [math]\displaystyle{ V }[/math] be a finite dimensional vector space and let [math]\displaystyle{ W_1 }[/math] and [math]\displaystyle{ W_2 }[/math] be subspaces of [math]\displaystyle{ V }[/math] for which [math]\displaystyle{ W_1\cap W_2=\{0\} }[/math]. Denote the linear span of [math]\displaystyle{ W_1\cup W_2 }[/math] by [math]\displaystyle{ W_1+W_2 }[/math]. Prove that [math]\displaystyle{ \dim(W_1+W_2)=\dim W_1 + \dim W_2 }[/math].
Solution Set
Students are most welcome to post a solution set here.
1. [math]\displaystyle{ 0_F\cdot v=(0_F+0_F)\cdot v }[/math] (by F3)
[math]\displaystyle{ (0_F+0_F)\cdot v=0_F\cdot v+0_F\cdot v }[/math] (by VS8)
By VS4, [math]\displaystyle{ \exists\ (0_F\cdot v)' s.t. (0_F\cdot v)+(0_F\cdot v)'=0_V }[/math]
Add [math]\displaystyle{ (0_F\cdot v)' }[/math] to both sides of [math]\displaystyle{ 0_F\cdot v=0_F\cdot v+0_F\cdot v }[/math]
[math]\displaystyle{ (0_F\cdot v)'+(0_F\cdot v)=[(0_F\cdot v)'+0_F\cdot v]+0_F\cdot v }[/math]
[math]\displaystyle{ 0_V=0_V+0_F\cdot v }[/math] (by construction)
[math]\displaystyle{ 0_V=0_F\cdot v }[/math] (by VS3)
2. 1) [math]\displaystyle{ \frac{1}{2+3i}+\frac{1}{2-3i}=\frac{(2-3i)+(2+3i)}{(2+3i)(2-3i)}=\frac{4}{2^2+3^3}=\frac{4}{13} }[/math]
[math]\displaystyle{ \frac{1}{2+3i}-\frac{1}{2-3i}=\frac{(2-3i)-(2+3i)}{(2+3i)(2-3i)}=\frac{-6i}{2^2+3^3}=\frac{-6}{13}i }[/math]