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\begin{document}
\examtitle{MATH 2270-003}{Sample Exam 1}{Fall 2010}

\studentinfo
\instructions{You may not use a calculator. Your solutions must include enough justification that another person could understand
 and be convinced by your argument.\\
There are extra blank pages at the end of the booklet. If you need more room to work a problem please note the page number where your work continues.\\}
\finishfirstpage
% Problems Start Here % ----------------------------------------------------- %
\problem{5}{%
What is the magnitude of the vector $\begin{pmatrix}1\\ 2\end{pmatrix}$?
}{}{%
\[\left|\left|\begin{pmatrix}1\\ 2\end{pmatrix}\right|\right|=\sqrt{\begin{pmatrix}1\\ 2\end{pmatrix}\cdot \begin{pmatrix}1\\ 2\end{pmatrix}}=\sqrt{5}\]
}

\problem{5}{%
What is the unit vector that has the same direction as $\begin{pmatrix}1\\ 2\end{pmatrix}$?
}{}{%
\[\frac{\begin{pmatrix}1\\ 2\end{pmatrix}}{\left|\left|\begin{pmatrix}1\\ 2\end{pmatrix}\right|\right|}=\begin{pmatrix}\frac{1}{\sqrt{5}}\\ \frac{2}{\sqrt{5}}\end{pmatrix}\]
}

\problem{5}{%
What is the coefficient matrix corresponding to the following system
of linear equations?
 
 \begin{align*}
    \phantom{2}x_1    + \phantom{2}x_2   + \phantom{2}x_3  + \phantom{2}x_4 & =b_1\\
 4x_1    + 2x_2   + 3x_3  \phantom{+ 4x_4} & =b_2\\
 \phantom{2x_1 }  \phantom{+ 2x_2 +}    4x_3  + 5x_4 & =b_3
  \end{align*}
}{}{%
\[
\begin{pmatrix} 1 & 1& 1& 1\\ 4 & 2&3&0\\0&0&4&5\end{pmatrix}\]
}

\problem{5}{
Let $A$ be a $3\times 4$ matrix. What is the elimination matrix that
replaces Row 2 of $A$ with $Row 2- Row 1$ and replaces $Row 3$ of $A$
by $Row 3-2Row 1$?
}{}{
\[\begin{pmatrix} 1 & 0 & 0\\ -1 & 1 & 0 \\ -2 & 0 & 1\end{pmatrix}\]
}

\problem{5}{
What is the inverse of the following matrix?
\[
E=\begin{pmatrix}
  1 & 0 & 0 & 0 & 0\\
  0 & 1 & 0 & 0 & 0\\
 0 & 0 & 1 & 0 & 0\\
  0 & 0 & -5 & 1 & 0\\
  0 & 0 & 1 & 0 & 1
\end{pmatrix}\]
}{}{$E$ is an elimination matrix. It replaces Row 4 with Row 4- 5 Row
  3 and replaces Row 5 with Row 3 + Row 5. The inverse is the
  elimination matrix that reverses these steps, replacing Row 4 with
  Row 4 + 5Row 3 and replacing Row 5 with Row 5 - Row 3. 
\[
E^{-1}=\begin{pmatrix}
  1 & 0 & 0 & 0 & 0\\
  0 & 1 & 0 & 0 & 0\\
 0 & 0 & 1 & 0 & 0\\
  0 & 0 & 5 & 1 & 0\\
  0 & 0 & -1 & 0 & 1
\end{pmatrix}\]
}

\problem{5}{
Descibe in word the effect of multiplying $EA$ where $E$ is the matrix
in the previous problem and $A$ is a $5\times 5$ matrix.
}{}{
$E$ is an elimination matrix. It replaces Row 4 with Row 4- 5 Row
  3 and replaces Row 5 with Row 3 + Row 5.
}

\anotherpage
\problem{15}{
Let $A=\begin{pmatrix} 1 & 2\\ 3 & 4\end{pmatrix}$. Let $B=\begin{pmatrix} 1 &
  1\\ 1 & 1\end{pmatrix} $. Let $C= \begin{pmatrix} 1 & 1\\ 1 &
  0\end{pmatrix}$.
Calculate the following:
\[A(BC)\]
\[(AB)C\]
\[C^2\]
\[2A+B-C\]
\[A(B-C)\]
}{}{
\[A(BC)=\begin{pmatrix} 6 & 3 \\ 14 &7 \end{pmatrix}\]
\[(AB)C=\begin{pmatrix} 6& 3 \\14 &7\end{pmatrix}\]
\[C^2=\begin{pmatrix} 2&1 \\ 1&1\end{pmatrix}\]
\[2A+B-C=\begin{pmatrix} 2& 4\\6 &9\end{pmatrix}\]
\[A(B-C)=\begin{pmatrix} 0&2 \\0 &4\end{pmatrix}\]
}

\problem{20}{
Prove or disprove that the following sets of vectors form a basis for
the indicated vector space:

\[
\begin{pmatrix}
  1\\ 2
\end{pmatrix},
\begin{pmatrix}
  2\\ 2
\end{pmatrix}
\text{ in } \mathbb{R}^2\]

\[
\begin{pmatrix}
  1\\ 0 \\ 0
\end{pmatrix},
\begin{pmatrix}
  0\\ 2 \\ 0
\end{pmatrix},
\begin{pmatrix}
  1\\ -1 \\ 1
\end{pmatrix},
\begin{pmatrix}
  -1\\ 1 \\ 1
\end{pmatrix}
\text{ in } \mathbb{R}^3\]
}{}{
\[
\begin{pmatrix}
  1\\ 2
\end{pmatrix},
\begin{pmatrix}
  2\\ 2
\end{pmatrix}
\text{ in } \mathbb{R}^2\]
These two vectors form a basis for $\mathbb{R}^2$.
These two vectors are not scalar multiples of each other, so they are
linearly independent. $\mathbb{R}^2$ is two dimensional, so an two
linearly independent vectors form a basis.

\[
\begin{pmatrix}
  1\\ 0 \\ 0
\end{pmatrix},
\begin{pmatrix}
  0\\ 2 \\ 0
\end{pmatrix},
\begin{pmatrix}
  1\\ -1 \\ 1
\end{pmatrix},
\begin{pmatrix}
  -1\\ 1 \\ 1
\end{pmatrix}
\text{ in } \mathbb{R}^3\]
These four vectors do not form a basis. A basis consists of linearly
independent vectors. $\mathbb{R}^3$ is three dimensional, so no set of
more than three vectors can ever be linearly independent in $\mathbb{R}^3$.
}

\anotherpage
\problem{25}{
Find all solutions to the equation $A\bar{x}=\bar{b}$ for 
\[A=
\begin{pmatrix}
  1 & 2 & 3 &4\\
5 & 6 & 7 &8
\end{pmatrix}
\]
\[\bar{b}=
\begin{pmatrix}
  0\\4
\end{pmatrix}
\]
}{}{
The augmented matrix for this system of equations is:
\[\begin{pmatrix}
  1 & 2 & 3 &4& 0\\
5 & 6 & 7 &8& 4
\end{pmatrix}\]
Put this is reduced row echelon from by performing Gauss-Jordan Elimination:
\[\begin{pmatrix}1 & 0\\-5&1\end{pmatrix}\begin{pmatrix}
  1 & 2 & 3 &4& 0\\
5 & 6 & 7 &8& 4
\end{pmatrix}=\begin{pmatrix}
  1 & 2 & 3 &4& 0\\
0 & -4 & -8 &-12& 4
\end{pmatrix}
\]

\[\begin{pmatrix}1 & \frac{1}{2}\\0&1\end{pmatrix}
\begin{pmatrix}
  1 & 2 & 3 &4& 0\\
0 & -4 & -8 &-12& 4
\end{pmatrix}
=
\begin{pmatrix}
  1 & 0 & -1 &-2& 2\\
0 & -4 & -8 &-12& 4
\end{pmatrix}
\]
\[\begin{pmatrix}1 & 0\\0&-\frac{1}{4}\end{pmatrix}
\begin{pmatrix}
  1 & 0 & -1 &-2& 2\\
0 & -4 & -8 &-12& 4
\end{pmatrix}
=
\begin{pmatrix}
  1 & 0 & -1 &-2& \phantom{-}2\\
0 & 1& \phantom{-}2 &\phantom{-}3& -1
\end{pmatrix}
\]

Solutions to the original equation are the same as solutions to the
new equation :
\[\begin{pmatrix}
  1 & 0 & -1 &-2\\
0 & 1& \phantom{-}2 &\phantom{-}3
\end{pmatrix}\bar{x}=\begin{pmatrix}2\\-1\end{pmatrix}\]

Columns 1 and 2 are pivot columns; columns 3 and 4 are free
columns, so $x_3$ and $x_4$ are the free variables.

A particular solution is obtained by setting $x_3=x_4=0$ and solving
for $x_1$ and $x_2$.
\[\begin{pmatrix}
  1 & 0 & -1 &-2\\
0 & 1& \phantom{-}2 &\phantom{-}3
\end{pmatrix}\begin{pmatrix}x_1\\x_2\\0\\0\end{pmatrix}=\begin{pmatrix}2\\-1\end{pmatrix}\]
The particular solution is
\[\bar{x}_p=\begin{pmatrix}2\\-1\\0\\0\end{pmatrix}\]

The other solutions differ from $\bar{x}_p$ by vectors in the
nullspace of $A$. A basis for the nullspace is given by the ``special
solutions'' obtained by setting one free variable equal to 1, the
others equal to 0, and solving for the non-free variables.

$\bar{s}_3$ is the special solution obtained by setting $x_3=1$ and
$x_4=0$ and solving
\[\begin{pmatrix}
  1 & 0 & -1 &-2\\
0 & 1& \phantom{-}2 &\phantom{-}3
\end{pmatrix}\begin{pmatrix}x_1\\x_2\\1\\0\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\]
 So $\bar{s}_3=\begin{pmatrix}1\\-2\\1\\0\end{pmatrix}$

Similarly, $s_4$ is obtained by setting $x_3=0$ and $x_4=1$ and solving
\[\begin{pmatrix}
  1 & 0 & -1 &-2\\
0 & 1& \phantom{-}2 &\phantom{-}3
\end{pmatrix}\begin{pmatrix}x_1\\x_2\\0\\1\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\]

So $\bar{s}_4=\begin{pmatrix}2\\-3\\0\\1\end{pmatrix}$

Finally, all solutions are of the form
$\bar{x}_p+x_3\bar{s}_3+x_4\bar{s}_4$, so all solutions to the
equation are all vectors:
\[\bar{x}=\begin{pmatrix}2\\-1\\0\\0\end{pmatrix}+x_3 \begin{pmatrix}1\\-2\\1\\0\end{pmatrix}+x_4 \begin{pmatrix}2\\-3\\0\\1\end{pmatrix}\]
where $x_3$ and $x_4$ are arbitrary real numbers.
}

\problem{20}{
Prove that a system of 4 linear equations in 5 unknowns has either 0
or infinitely many solutions.
}{}{
There are more unknowns than equations. Either the equations are
inconsistent and there are no solutions, or there is a free variable.
We may choose this free variable to be any real number and find a
corresponding solution, so this is infinitely many solutions.
}


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% Problems End Here % ------------------------------------------------------- %

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