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+    \hypertarget{matrixarrays}{%
+\section{matrixArrays}\label{matrixarrays}}
+
+In computer programming, an array is a structure for storing and
+retrieving data. We often talk about an array as if it were a grid in
+space, with each cell storing one element of the data. For instance, if
+each element of the data were a number, we might visualize a
+``one-dimensional'' array like a list:
+
+\begin{longtable}[]{@{}llll@{}}
+\toprule
+1 & 5 & 2 & 0 \\
+\midrule
+\endhead
+\bottomrule
+\end{longtable}
+
+A two-dimensional array would be like a table:
+
+\begin{longtable}[]{@{}llll@{}}
+\toprule
+1 & 5 & 2 & 0 \\
+\midrule
+\endhead
+8 & 3 & 6 & 1 \\
+1 & 7 & 2 & 9 \\
+\bottomrule
+\end{longtable}
+
+A three-dimensional array would be like a set of tables, perhaps stacked
+as though they were printed on separate pages. If we visualize the
+position of each element as a position in space. Then we can represent
+the value of the element as a property. In other words, if we were to
+analyze the stress concentration of an aluminum block, the property
+would be stress.
+
+\begin{itemize}
+\tightlist
+\item
+  From
+  \href{https://numpy.org/doc/2.2/user/absolute_beginners.html}{Numpy
+  documentation}
+
+\end{itemize}
+
+If the load on this block changes over time, then we may want to add a
+4th dimension i.e.~additional sets of 3-D arrays for each time
+increment. As you can see - the more dimensions we add, the more
+complicated of a problem we have to solve. It is possible to increase
+the number of dimensions to the n-th order. This course we will not be
+going beyond dimensional analysis.
+
+\begin{center}\rule{0.5\linewidth}{0.5pt}\end{center}
+
+\hypertarget{numpy---the-pythons-array-library}{%
+\section{Numpy - the python's array
+library}\label{numpy---the-pythons-array-library}}
+
+In this tutorial we will be introducing arrays and we will be using the
+numpy library. Arrays, lists, vectors, matrices, sets - You might've
+heard of them before, they all store data. In programming, an array is a
+variable that can hold more than one value at a time. We will be using
+the Numpy python library to create arrays. Since we already have
+installed Numpy previously, we can start using the package.
+
+Before importing our first package, let's as ourselves \emph{what is a
+package?} A package can be thought of as pre-written python code that we
+can re-use. This means the for every script that we write in python we
+need to tell it to use a certain package. We call this importing a
+package.
+
+\hypertarget{importing-numpy}{%
+\subsection{Importing Numpy}\label{importing-numpy}}
+
+When using packages in python, we need to let it know what package we
+will be using. This is called importing. To import numpy we need to
+declare it a the start of a script as follows:
+
+\begin{Shaded}
+\begin{Highlighting}[]
+\ImportTok{import}\NormalTok{ numpy }\ImportTok{as}\NormalTok{ np}
+\end{Highlighting}
+\end{Shaded}
+
+\begin{itemize}
+\tightlist
+\item
+  \texttt{import} - calls for a library to use, in our case it is Numpy.
+\item
+  \texttt{as} - gives the library an alias in your script. It's common
+  convention in Python programming to make the code shorter and more
+  readable. We will be using \emph{np} as it's a standard using in many
+  projects.
+\end{itemize}
+
+\begin{center}\rule{0.5\linewidth}{0.5pt}\end{center}
+
+\hypertarget{creating-arrays}{%
+\section{Creating arrays}\label{creating-arrays}}
+
+Now that we have imported the library we can create a one dimensional
+array or \emph{vector} with three elements.
+
+\begin{Shaded}
+\begin{Highlighting}[]
+\NormalTok{x }\OperatorTok{=}\NormalTok{ np.array([}\DecValTok{1}\NormalTok{,}\DecValTok{2}\NormalTok{,}\DecValTok{3}\NormalTok{])}
+\end{Highlighting}
+\end{Shaded}
+
+To create a \emph{matrix} we can nest the arrays to create a two
+dimensional array. This is done as follows.
+
+\begin{Shaded}
+\begin{Highlighting}[]
+\NormalTok{matrix }\OperatorTok{=}\NormalTok{ np.array([[}\DecValTok{1}\NormalTok{,}\DecValTok{2}\NormalTok{,}\DecValTok{3}\NormalTok{],}
+\NormalTok{                   [}\DecValTok{4}\NormalTok{,}\DecValTok{5}\NormalTok{,}\DecValTok{6}\NormalTok{],}
+\NormalTok{                   [}\DecValTok{7}\NormalTok{,}\DecValTok{8}\NormalTok{,}\DecValTok{9}\NormalTok{]])}
+\end{Highlighting}
+\end{Shaded}
+
+\emph{Note: for every array we nest, we get a new dimension in our data
+structure.}
+
+    \hypertarget{display-arrays}{%
+\section{Display arrays}\label{display-arrays}}
+
+Using command print("") Accessing particular elements of an array
+\ldots..
+
+    \hypertarget{practice-problem}{%
+\section{Practice Problem}\label{practice-problem}}
+
+Problem statement
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{1}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{k+kn}{import} \PY{n+nn}{numpy} \PY{k}{as} \PY{n+nn}{np}
+
+\PY{n}{x} \PY{o}{=} \PY{n}{np}\PY{o}{.}\PY{n}{array}\PY{p}{(}\PY{p}{[}\PY{l+m+mi}{7}\PY{p}{,} \PY{l+m+mi}{10} \PY{p}{,}\PY{l+m+mi}{12}\PY{p}{]}\PY{p}{)}
+
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{x}\PY{p}{)}
+
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{x}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{]}\PY{p}{)}
+\end{Verbatim}
+\end{tcolorbox}
+
+    \begin{Verbatim}[commandchars=\\\{\}]
+[ 7 10 12]
+10
+    \end{Verbatim}
+
+    \hypertarget{numpy-array-creation-functions}{%
+\subsection{Numpy array creation
+functions}\label{numpy-array-creation-functions}}
+
+Numpy comes with some built-in function that we can use to create arrays
+quickly. Here are a couple of functions that are commonly used in
+python.
+
+\hypertarget{np.arange}{%
+\subsubsection{np.arange}\label{np.arange}}
+
+The \texttt{np.arange()} function returns an array with evenly spaced
+values within a specified range. It is similar to the built-in
+\texttt{range()} function in Python but returns a Numpy array instead of
+a list. The parameters for this function are the start value
+(inclusive), the stop value (exclusive), and the step size. If the step
+size is not provided, it defaults to 1.
+
+\begin{Shaded}
+\begin{Highlighting}[]
+\OperatorTok{\textgreater{}\textgreater{}\textgreater{}}\NormalTok{ np.arange(}\DecValTok{4}\NormalTok{)}
+\NormalTok{array([}\FloatTok{0.}\NormalTok{ , }\FloatTok{1.}\NormalTok{, }\FloatTok{2.}\NormalTok{, }\FloatTok{3.}\NormalTok{ ])}
+\end{Highlighting}
+\end{Shaded}
+
+In this example, \texttt{np.arange(4)} generates an array starting from
+0 and ending before 4, with a step size of 1.
+
+\hypertarget{np.linspace}{%
+\subsubsection{np.linspace}\label{np.linspace}}
+
+The \texttt{np.linspace()} function returns an array of evenly spaced
+values over a specified range. Unlike \texttt{np.arange()}, which uses a
+step size to define the spacing between elements, \texttt{np.linspace()}
+uses the number of values you want to generate and calculates the
+spacing automatically. It accepts three parameters: the start value, the
+stop value, and the number of samples.
+
+\begin{Shaded}
+\begin{Highlighting}[]
+\OperatorTok{\textgreater{}\textgreater{}\textgreater{}}\NormalTok{ np.linspace(}\FloatTok{1.}\NormalTok{, }\FloatTok{4.}\NormalTok{, }\DecValTok{6}\NormalTok{)}
+\NormalTok{array([}\FloatTok{1.}\NormalTok{ ,  }\FloatTok{1.6}\NormalTok{,  }\FloatTok{2.2}\NormalTok{,  }\FloatTok{2.8}\NormalTok{,  }\FloatTok{3.4}\NormalTok{,  }\FloatTok{4.}\NormalTok{ ])}
+\end{Highlighting}
+\end{Shaded}
+
+In this example, \texttt{np.linspace(1.,\ 4.,\ 6)} generates 6 evenly
+spaced values between 1. and 4., including both endpoints.
+
+Try this and see what happens:
+
+\begin{Shaded}
+\begin{Highlighting}[]
+\NormalTok{x }\OperatorTok{=}\NormalTok{ np.linspace(}\DecValTok{0}\NormalTok{,}\DecValTok{100}\NormalTok{,}\DecValTok{101}\NormalTok{)}
+\NormalTok{y }\OperatorTok{=}\NormalTok{ np.sin(x)}
+\end{Highlighting}
+\end{Shaded}
+
+\hypertarget{other-useful-functions}{%
+\subsubsection{Other useful functions}\label{other-useful-functions}}
+
+\begin{itemize}
+\tightlist
+\item
+  \texttt{np.zeros()}
+\item
+  \texttt{np.ones()}
+\item
+  \texttt{np.eye()}
+\end{itemize}
+
+    \hypertarget{practice-problem}{%
+\subsection{Practice problem}\label{practice-problem}}
+
+Problem statement below
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{2}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{n}{y}\PY{o}{=}\PY{n}{np}\PY{o}{.}\PY{n}{linspace}\PY{p}{(}\PY{l+m+mi}{10}\PY{p}{,}\PY{l+m+mi}{20}\PY{p}{,}\PY{l+m+mi}{5}\PY{p}{)}
+\PY{n+nb}{print}\PY{p}{(}\PY{n}{y}\PY{p}{)}
+\end{Verbatim}
+\end{tcolorbox}
+
+    \begin{Verbatim}[commandchars=\\\{\}]
+[10.  12.5 15.  17.5 20. ]
+    \end{Verbatim}
+
+    \hypertarget{working-with-arrays}{%
+\subsection{Working with Arrays}\label{working-with-arrays}}
+
+Now that we have been introduced to some ways to create arrays using the
+Numpy functions let's start using them.
+
+\hypertarget{indexing}{%
+\subsubsection{Indexing}\label{indexing}}
+
+Indexing in Python allows you to access specific elements within an
+array based on their position. This means you can directly retrieve and
+manipulate individual items as needed.
+
+Python uses \textbf{zero-based indexing}, meaning the first element is
+at position \textbf{0} rather than \textbf{1}. This approach is common
+in many programming languages. For example, in a list with five
+elements, the first element is at index \texttt{0}, followed by elements
+at indices \texttt{1}, \texttt{2}, \texttt{3}, and \texttt{4}.
+
+Here's an example of data from a rocket test stand where thrust was
+recorded as a function of time.
+
+\begin{Shaded}
+\begin{Highlighting}[]
+\NormalTok{thrust\_lbf }\OperatorTok{=}\NormalTok{ np.array(}\FloatTok{0.603355}\NormalTok{, }\FloatTok{2.019083}\NormalTok{, }\FloatTok{2.808092}\NormalTok{, }\FloatTok{4.054973}\NormalTok{, }\FloatTok{1.136618}\NormalTok{, }\FloatTok{0.943668}\NormalTok{)}
+
+\OperatorTok{\textgreater{}\textgreater{}\textgreater{}}\NormalTok{ thrust\_lbs[}\DecValTok{3}\NormalTok{]}
+\end{Highlighting}
+\end{Shaded}
+
+Due to the nature of zero-based indexing. If we want to call the value
+\texttt{4.054973} that will be the 3rd index.
+
+\hypertarget{operations-on-arrays}{%
+\subsubsection{Operations on arrays}\label{operations-on-arrays}}
+
+\begin{itemize}
+\tightlist
+\item
+  Arithmetic operations (\texttt{+}, \texttt{-}, \texttt{*}, \texttt{/},
+  \texttt{**})
+\item
+  \texttt{np.add()}, \texttt{np.subtract()}, \texttt{np.multiply()},
+  \texttt{np.divide()}
+\item
+  \texttt{np.dot()} for dot product
+\item
+  \texttt{np.matmul()} for matrix multiplication
+\item
+  \texttt{np.linalg.inv()}, \texttt{np.linalg.det()} for linear algebra
+\end{itemize}
+
+\hypertarget{statistics}{%
+\paragraph{Statistics}\label{statistics}}
+
+\begin{itemize}
+\tightlist
+\item
+  \texttt{np.mean()}, \texttt{np.median()}, \texttt{np.std()},
+  \texttt{np.var()}
+\item
+  \texttt{np.min()}, \texttt{np.max()}, \texttt{np.argmin()},
+  \texttt{np.argmax()}
+\item
+  Summation along axes: \texttt{np.sum(arr,\ axis=0)}
+\end{itemize}
+
+\hypertarget{combining-arrays}{%
+\paragraph{Combining arrays}\label{combining-arrays}}
+
+\begin{itemize}
+\tightlist
+\item
+  Concatenation: \texttt{np.concatenate((arr1,\ arr2),\ axis=0)}
+\item
+  Stacking: \texttt{np.vstack()}, \texttt{np.hstack()}
+\item
+  Splitting: \texttt{np.split()}
+\end{itemize}
+
+    \hypertarget{exercise}{%
+\section{Exercise}\label{exercise}}
+
+Let's solve a statics problem given the following problem
+
+A simply supported bridge of length L=20L = 20L=20 m is subjected to
+three point loads:
+
+\begin{itemize}
+\tightlist
+\item
+  \(P1=1010 kN\) at \(x=5m\)
+\item
+  \(P2=15 kN\) at \(x=10m\)
+\item
+  \(P3=20 kN\) at \(x=15m\)
+\end{itemize}
+
+The bridge is supported by two reaction forces at points AAA (left
+support) and BBB (right support). We assume the bridge is in static
+equilibrium, meaning the sum of forces and sum of moments about any
+point must be zero.
+
+\hypertarget{equilibrium-equations}{%
+\paragraph{Equilibrium Equations:}\label{equilibrium-equations}}
+
+\begin{enumerate}
+\def\labelenumi{\arabic{enumi}.}
+\item
+  \textbf{Sum of Forces in the Vertical Direction}:
+
+  \(R_A + R_B - P_1 - P_2 - P_3 = 0\)
+\item
+  \textbf{Sum of Moments About Point A}:
+
+  \(5 P_1 + 10 P_2 + 15 P_3 - 20 R_B = 0\)
+\item
+  \textbf{Sum of Moments About Point B}:
+
+  \(20 R_A - 15 P_3 - 10 P_2 - 5 P_1 = 0\)
+\end{enumerate}
+
+\hypertarget{system-of-equations}{%
+\paragraph{System of Equations:}\label{system-of-equations}}
+
+\[
+\begin{cases}
+R_A + R_B - 10 - 15 - 20 = 0 \\
+5 \cdot 10 + 10 \cdot 15 + 15 \cdot 20 - 20 R_B = 0 \\
+20 R_A - 5 \cdot 10 - 10 \cdot 15 - 15 \cdot 20 = 0
+\end{cases}
+\]
+
+    \hypertarget{solution}{%
+\subsubsection{Solution}\label{solution}}
+
+    \begin{tcolorbox}[breakable, size=fbox, boxrule=1pt, pad at break*=1mm,colback=cellbackground, colframe=cellborder]
+\prompt{In}{incolor}{3}{\boxspacing}
+\begin{Verbatim}[commandchars=\\\{\}]
+\PY{k+kn}{import} \PY{n+nn}{numpy} \PY{k}{as} \PY{n+nn}{np}
+
+\PY{c+c1}{\PYZsh{} Define the coefficient matrix A}
+\PY{n}{A} \PY{o}{=} \PY{n}{np}\PY{o}{.}\PY{n}{array}\PY{p}{(}\PY{p}{[}
+    \PY{p}{[}\PY{l+m+mi}{1}\PY{p}{,} \PY{l+m+mi}{1}\PY{p}{]}\PY{p}{,}
+    \PY{p}{[}\PY{l+m+mi}{0}\PY{p}{,} \PY{o}{\PYZhy{}}\PY{l+m+mi}{20}\PY{p}{]}\PY{p}{,}
+    \PY{p}{[}\PY{l+m+mi}{20}\PY{p}{,} \PY{l+m+mi}{0}\PY{p}{]}
+\PY{p}{]}\PY{p}{)}
+
+\PY{c+c1}{\PYZsh{} Define the right\PYZhy{}hand side vector b}
+\PY{n}{b} \PY{o}{=} \PY{n}{np}\PY{o}{.}\PY{n}{array}\PY{p}{(}\PY{p}{[}
+    \PY{l+m+mi}{45}\PY{p}{,}
+    \PY{l+m+mi}{5}\PY{o}{*}\PY{l+m+mi}{10} \PY{o}{+} \PY{l+m+mi}{10}\PY{o}{*}\PY{l+m+mi}{15} \PY{o}{+} \PY{l+m+mi}{15}\PY{o}{*}\PY{l+m+mi}{20}\PY{p}{,}
+    \PY{l+m+mi}{5}\PY{o}{*}\PY{l+m+mi}{10} \PY{o}{+} \PY{l+m+mi}{10}\PY{o}{*}\PY{l+m+mi}{15} \PY{o}{+} \PY{l+m+mi}{15}\PY{o}{*}\PY{l+m+mi}{20}
+\PY{p}{]}\PY{p}{)}
+
+\PY{c+c1}{\PYZsh{} Solve the system of equations Ax = b}
+\PY{c+c1}{\PYZsh{} Using least squares to handle potential overdetermination}
+\PY{n}{x} \PY{o}{=} \PY{n}{np}\PY{o}{.}\PY{n}{linalg}\PY{o}{.}\PY{n}{lstsq}\PY{p}{(}\PY{n}{A}\PY{p}{,} \PY{n}{b}\PY{p}{,} \PY{n}{rcond}\PY{o}{=}\PY{k+kc}{None}\PY{p}{)}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{]}
+
+\PY{c+c1}{\PYZsh{} Display the results}
+\PY{n+nb}{print}\PY{p}{(}\PY{l+s+sa}{f}\PY{l+s+s2}{\PYZdq{}}\PY{l+s+s2}{Reaction force at A (R\PYZus{}A): }\PY{l+s+si}{\PYZob{}}\PY{n}{x}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{]}\PY{l+s+si}{:}\PY{l+s+s2}{.2f}\PY{l+s+si}{\PYZcb{}}\PY{l+s+s2}{ kN}\PY{l+s+s2}{\PYZdq{}}\PY{p}{)}
+\PY{n+nb}{print}\PY{p}{(}\PY{l+s+sa}{f}\PY{l+s+s2}{\PYZdq{}}\PY{l+s+s2}{Reaction force at B (R\PYZus{}B): }\PY{l+s+si}{\PYZob{}}\PY{n}{x}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{]}\PY{l+s+si}{:}\PY{l+s+s2}{.2f}\PY{l+s+si}{\PYZcb{}}\PY{l+s+s2}{ kN}\PY{l+s+s2}{\PYZdq{}}\PY{p}{)}
+\end{Verbatim}
+\end{tcolorbox}
+
+    \begin{Verbatim}[commandchars=\\\{\}]
+Reaction force at A (R\_A): 25.11 kN
+Reaction force at B (R\_B): -24.89 kN
+    \end{Verbatim}
+
+
+    % Add a bibliography block to the postdoc