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\begin{center}
{Mathematical Physics --- PHZ 3113}\\\vspace {2mm} 
Classwork 13 (April 10, 2013) \\ 
Solutions Linear Equations
\end{center}
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After finding solutions: Check that they are correct!

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\noindent 1. Find all solutions of the equations
\begin{eqnarray} \label{Eq1a}
   x + 2y &=& 3\,,\\ \label{Eq1b} 3x +  y &=& 2\,.
\end{eqnarray}
\bigskip
Solution:
\begin{eqnarray} \nonumber
  \left| \begin{matrix} 1 & 2 \cr
                        3 & 1 \end{matrix} \right|\ =\ -5\,,~~
  \left| \begin{matrix} 3 & 2 \cr
                        2 & 1 \end{matrix} \right|\ =\ -1\,,~~
  \left| \begin{matrix} 1 & 3 \cr
                        3 & 2 \end{matrix} \right|\ =\ -7
\end{eqnarray}
and by Cramer's rule
\begin{eqnarray} \nonumber
   x = \frac{1}{5}\,,~~~ y = \frac{7}{5}\,.
\end{eqnarray}
Check: Insert $x$ and $y$ into (\ref{Eq1a}) and (\ref{Eq1b}).
Alternative way to the solution: Gauss elimination.
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\bigskip
\noindent 2. Find all solutions of the equations
\begin{eqnarray} \label{Eq2a}
   x + 2y &=& 3\,,\\ \label{Eq2b} 2x + 4y &=& 6\,.
\end{eqnarray}
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Solution:
\begin{eqnarray} \nonumber
  \left| \begin{matrix} 1 & 2 \cr
                        2 & 4 \end{matrix} \right|\ =\ 0\,,~~
  \left| \begin{matrix} 3 & 2 \cr
                        6 & 4 \end{matrix} \right|\ =\ 0\,,~~
  \left| \begin{matrix} 1 & 3 \cr
                        2 & 6 \end{matrix} \right|\ =\ 0
\end{eqnarray}
and the two equations are linearly dependent. Dividing (\ref{Eq2b})
by 2 gives (\ref{Eq2a}) again. Therefore the solution is
\begin{eqnarray} \nonumber
  y &=& \frac{3}{2} - \frac{x}{2}
\end{eqnarray}
for general $x$.
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\bigskip
\noindent 3. Find all solutions of the equations
\begin{eqnarray} \label{Eq3a}
   x + 2y &=& 3\,,\\ \label{Eq3b} 2x + 4y &=& 4\,.
\end{eqnarray}
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Solution:
\begin{eqnarray} \nonumber
  \left| \begin{matrix} 1 & 2 \cr
                        2 & 4 \end{matrix} \right|\ =\ 0\,,~~
  \left| \begin{matrix} 3 & 2 \cr
                        4 & 4 \end{matrix} \right|\ =\ 4\,,~~
  \left| \begin{matrix} 1 & 3 \cr
                        2 & 4 \end{matrix} \right|\ =\ -2
\end{eqnarray}
and there are no solutions. The equations contradict one another.
Subtracting (\ref{Eq3a}) from (\ref{Eq3b}) gives
\begin{eqnarray} 
   x + 2y &=& 1\,,
\end{eqnarray}
which disagrees with (\ref{Eq3a}).
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\bigskip
\noindent 4. Find all solutions of the equations
\begin{eqnarray} 
   x + 2y &=& 0\,,\\ 3x +  y &=& 0\,.
\end{eqnarray}
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Solution: As the determinant
\begin{eqnarray} \nonumber
  \left| \begin{matrix} 1 & 2 \cr
                        3 & 1 \end{matrix} \right|\ =\ -5
\end{eqnarray}
is non-zero there is only the trivial solution $x=y=0$.
\bigskip

\bigskip
\noindent 5. Find all solutions of the equations
\begin{eqnarray} \label{Eq5a}
   x + 2y &=& 0\,,\\ \label{Eq5b} 2x + 4y &=& 0\,.
\end{eqnarray}
\bigskip
Solution: From the deterninant
\begin{eqnarray} \nonumber
  \left| \begin{matrix} 1 & 2 \cr
                        2 & 4 \end{matrix} \right|\ =\ 0\,,~~
\end{eqnarray}
we read off 
\begin{eqnarray} \nonumber
  \frac{x}{y} &=& \frac{4}{-2}\ =\ -2\
\end{eqnarray}
which, of course, follows also directly from either (\ref{Eq5a})
or (\ref{Eq5b}).
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\noindent 6. Find all solutions of the equations
\begin{eqnarray} \label{Eq6a}
   x + 2y +  z &=& 0\,,\\ \label{Eq6b} 
  2x + 4y +  z &=& 0\,,\\ \label{Eq6c}
   x +  y + 2z &=& 0\,.
\end{eqnarray}
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Solution: As the determinant
\begin{eqnarray} \nonumber
  \left| \begin{matrix} 1 & 2 & 1 \cr
                        2 & 4 & 1 \cr
                        1 & 1 & 2 \end{matrix} \right|\ =\ 
  8+2+2-1-8-4\ =\ -1
\end{eqnarray}
is non-zero there is only the trivial solution $x=y=z=0$.
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\noindent 7. Find all solutions of the equations
\begin{eqnarray} \label{Eq7a}
   x + 2y +  z &=& 0\,,\\ \label{Eq7b} 
  2x + 4y + 2z &=& 0\,,\\ \label{Eq7c}
   x +  y + 2z &=& 0\,.
\end{eqnarray}
\bigskip
Solution: From the determinant
\begin{eqnarray} \nonumber
  \left| \begin{matrix} 1 & 2 & 1 \cr
                        2 & 4 & 2 \cr
                        1 & 1 & 2 \end{matrix} \right|\ =\ 
  8+4+2-2-8-4\ =\ 0
\end{eqnarray}
we read off 
\begin{eqnarray} \nonumber
  \frac{x}{y} &=& - \frac{\left| \begin{matrix}  4 & 2 \cr
                         1 & 2 \end{matrix} \right|}{
   \left| \begin{matrix} 2 & 2 \cr
                         1 & 2 \end{matrix} \right|}\ =
   - \frac{6}{2}\ =\ -3\,,\\  \label{xy}  &~& \\ \nonumber
  \frac{x}{z} &=& + \frac{\left| \begin{matrix}  4 & 2 \cr
                         1 & 2 \end{matrix} \right|}{
   \left| \begin{matrix} 2 & 4 \cr
                         1 & 1 \end{matrix} \right|}\ =
   - \frac{6}{-2}\ =\ -3\,,\\ \label{xz} 
\end{eqnarray}
\begin{eqnarray} \nonumber
  \frac{y}{z} &=& - \frac{\left| \begin{matrix}  2 & 2 \cr
                         1 & 2 \end{matrix} \right|}{
   \left| \begin{matrix} 2 & 4 \cr
                         1 & 1 \end{matrix} \right|}\ =
   - \frac{2}{-2}\ =\ +1\,.\\ \label{yz}
\end{eqnarray}
Multiplying (\ref{xy}) with (\ref{yz}) we find consistency with 
(\ref{xz}). Choosing the normalization $y=1$ a solution is
\begin{eqnarray} \nonumber
  x\ =\ -3\,,~~~ y\ =\ 1\,,~~~ z\ =\ 1\,.
\end{eqnarray}
Inserting these numbers into equations (\ref{Eq7a}), (\ref{Eq7b}) and 
(\ref{Eq7c}), we check that the solution is correct.
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