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\begin{document}
\examtitle{MATH 2270-003}{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}{%
Is $\begin{pmatrix}1\\ 2\end{pmatrix}$ a unit vector? Explain.
}{}{%
No. A unit vector has length 1, but
\[\left|\left|\begin{pmatrix}1\\
      2\end{pmatrix}\right|\right|=\sqrt{\begin{pmatrix}1\\
    2\end{pmatrix}\cdot \begin{pmatrix}1\\
    2\end{pmatrix}}=\sqrt{5}\neq 1\]
}

\vfill





\problem{5}{
The matrix $A$ is given below. What is the elimination matrix that
eliminates the entries below the second pivot.
\[
A=\begin{pmatrix}
   1 & 1 & 1 & 1\\
  0 & 0 & 2 & 0\\
   0 & 0 & 4 & 0\\
   0 & 0 & 1 & 1
\end{pmatrix}\]
}{}{
The second pivot is the 2 in row 2 column 3. Use this to eliminate the
entries in the 3,3 and 4,3 positions using the elimination matrix
\[\begin{pmatrix}
   1 & 0 & 0 & 0\\
  0 & 1 & 0 & 0\\
   0 &  -2 &1& 0\\
   0 &  -\frac{1}{2}&0 & 1
\end{pmatrix}\]

}


\vfill
\anotherpage

\problem{5}{
$E$ and $F$ are given below.
What is the inverse of $EF$?
\[E=\begin{pmatrix}

   1 & 0 & 0\\
    -5 & 1 & 0\\
    1 & 0 & 1
\end{pmatrix}\]
\[
F=\begin{pmatrix}
   1 & 0 & 0\\
    0 & 1 & 0\\
   0 & -2 & 1
\end{pmatrix}\]
}{}{
$ (EF)^{-1}=F^{-1}E^{-1}$. Now $E$ and $F$ are elimination matrices, so we
know how to invert them, just switch the signs on the off-diagonal entries:
\[
E^{-1}=\begin{pmatrix}
     1 & 0 & 0\\
    +5 & 1 & 0\\
    -1 & 0 & 1
\end{pmatrix}\]
\[F^{-1}=\begin{pmatrix}
   1 & 0 & 0\\
    0 & 1 & 0\\
   0 & +2 & 1
\end{pmatrix}\]
}

\vfill
\anotherpage

\problem{5}{
Describe in words the effect of multiplying $EA$ where $E$ is the matrix
in the previous problem and $A$ is a $3\times 5$ matrix.
}{}{
$E$ is an elimination matrix. It replaces Row 2 with Row 2 -- 5 Row
  1 and replaces Row 3 with Row 3 + Row 1.
}

\anotherpage
\problem{15}{
Let $A=\begin{pmatrix} 1 & 2\\ 3 & 4\end{pmatrix}$. Let $B=\begin{pmatrix} 6 &
  3\\ 2 & 1\end{pmatrix} $. Let $C= \begin{pmatrix} 1 & 7\\ 1 &
  2\end{pmatrix}$.
Calculate the following:
\[AB\]
\[BA\]
\[C^2\]
\[A^T\begin{pmatrix}1\\2\end{pmatrix}\]
\[A(2B-C)\]
}{}{
\[AB=\begin{pmatrix} 10&5\\24&13 \end{pmatrix}\]
\[BA=\begin{pmatrix} 15&24\\5&8\end{pmatrix}\]
\[C^2=\begin{pmatrix}8&21\\3&11\end{pmatrix}\]
\[A^T\begin{pmatrix}1\\2\end{pmatrix}=\begin{pmatrix} 7\\10\end{pmatrix}\]
\[A(2B-C)=\begin{pmatrix} 17&-1\\45&-3\end{pmatrix}\]
}

\anotherpage

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

\[
\begin{pmatrix}
  1\\ 0
\end{pmatrix},
\begin{pmatrix}
  2\\ 2
\end{pmatrix}
\text{ in } \mathbb{R}^2\]

\[
\begin{pmatrix}
  1\\ -1 \\ 1
\end{pmatrix},
\begin{pmatrix}
  -1\\ 1 \\ 1
\end{pmatrix}
\text{ in } \mathbb{R}^3\]
}{}{
\[
\begin{pmatrix}
  1\\ 2
\end{pmatrix},
\begin{pmatrix}
  2\\ 2
\end{pmatrix}
\text{ in } \mathbb{R}^2\]
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.

\[
\begin{pmatrix}
  1\\ -1 \\ 1
\end{pmatrix},
\begin{pmatrix}
  -1\\ 1 \\ 1
\end{pmatrix}
\text{ in } \mathbb{R}^3\]
These two vectors do not form a basis. The dimension of $\mathbb{R}^3$
is three, so any basis must contain exactly three vectors.
}

\anotherpage
\problem{25}{
Find all solutions to the equation $A\bar{x}=\bar{b}$ for
\[A=
\begin{pmatrix}
  1 & 2 & 3 &4\\
1 & 2 & 4 &6\\
2&4&8&12
\end{pmatrix}
\]
\[\bar{b}=
\begin{pmatrix}
  4\\5\\10
\end{pmatrix}
\]
}{}{
The augmented matrix for this system of equations is:
\[\begin{pmatrix}
  1 & 2 & 3 &4&4\\
1 & 2 & 4 &6&5\\
2&4&8&12&10
\end{pmatrix}
\]
Put this is reduced row echelon from by performing Gauss-Jordan Elimination.

Eliminate entries below first pivot:
\[\begin{pmatrix}1&0&0\\-1&1&0\\-2&0&1\end{pmatrix}\begin{pmatrix}
  1 & 2 & 3 &4&4\\
1 & 2 & 4 &6&5\\
2&4&8&12&10
\end{pmatrix}=\begin{pmatrix}
  1 & 2 & 3 &4&4\\
0 & 0 & 1 &2&1\\
0&0&2&4&2
\end{pmatrix}\]

Eliminate entries below second pivot:
\[
\begin{pmatrix}1&0&0\\0&1&0\\0&-2&1\end{pmatrix}
\begin{pmatrix}
  1 & 2 & 3 &4&4\\
0 & 0 & 1 &2&1\\
0&0&2&4&2
\end{pmatrix}=\begin{pmatrix}
  1 & 2 & 3 &4&4\\
0 & 0 & 1 &2&1\\
0&0&0&0&0
\end{pmatrix}\]

Eliminate entries above second pivot:
\[
\begin{pmatrix}1&-3&0\\0&1&0\\0&0&1\end{pmatrix}
\begin{pmatrix}
  1 & 2 & 3 &4&4\\
0 & 0 & 1 &2&1\\
0&0&0&0&0
\end{pmatrix}=\begin{pmatrix}
  1 & 2 & 0 &-2&1\\
0 & 0 & 1 &2&1\\
0&0&0&0&0
\end{pmatrix}
\]
Solutions to the original equation are the same as solutions to the
new equation :
\[\begin{pmatrix}
  1 & 2 & 0 &-2\\
0 & 0 & 1 &2\\
0&0&0&0
\end{pmatrix}\bar{x}=\begin{pmatrix}1\\1\\0\end{pmatrix}\]

Columns 1 and 3 are pivot columns; columns 2 and 4 are free
columns, so $x_2$ and $x_4$ are the free variables.

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

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.

$\bar{s}_2$ is the special solution obtained by setting $x_2=1$ and
$x_4=0$ and solving
\[\begin{pmatrix}
  1 & 2 & 0 &-2\\
0 & 0 & 1 &2\\
0&0&0&0
\end{pmatrix}\begin{pmatrix}x_1\\1\\x_3\\0\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\]
 So $\bar{s}_3=\begin{pmatrix}-2\\1\\0\\0\end{pmatrix}$

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

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

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:
\[\bar{x}=\begin{pmatrix}1\\0\\1\\0\end{pmatrix}+x_3 \begin{pmatrix}-2\\1\\0\\0\end{pmatrix}+x_4 \begin{pmatrix}2\\0\\-2\\1\end{pmatrix}\]
where $x_3$ and $x_4$ are arbitrary real numbers.
}

\anotherpage
\problem{15}{
Find all solutions to the equation $A\bar{x}=\bar{b}$ for
\[A=
\begin{pmatrix}
  1 & 2 & 3 &4\\
2 & 4 & 6 &8
\end{pmatrix}
\]
\[\bar{b}=
\begin{pmatrix}
  1\\3
\end{pmatrix}
\]
}{}{
All columns of $A$ are multiples of column 1, so the column space is
the span of column 1. However, $\bar{b}$ is not a scalar multiple of
column 1, so $\bar{b}$ is not in the column space of $A$. Thus, there
are no solutions.
}

\vfill

\problem{15}{%
Consider a system of 3 linear equations in five variables. How many
solutions can there be? Explain the possibilities.
}{}{%
The coefficient matrix would be $3\times 5$, so it could have at most
3 pivots, hence at least two free variables.

It is possible there are no solutions, but if there are solutions then
there are infinitely many. They would be given by one particular
solution plus any vector in the null space, which is at least two dimensional.
}

\phantom{page}
\vfill \begin{center} No new questions beyond this point.\end{center}
\anotherpage


\phantom{page}
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% Problems End Here % ------------------------------------------------------- %

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