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\begin{document}
\headers{Math343}{Quiz 3}{Sep 12, 2025}

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\noindent
\hfill Name: \underline{\hspace{3.5in}} 

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\noindent
\textbf{Directions:} Show all work.  No credit for answers without work.

\begin{enumerate}

\item\ppoints{3}{1} True/False.  Justify your answers.
\begin{enumerate}
	\item There is a vector $\vect{u}\in\RR^n$ such that for each vector $\vect{v}\in\RR^n$, we have $\vect{u}\in\Span\{\vect{v}\}$.  
\vfill
	
	\item If $\vect{u}$ and $\vect{v}$ are both linear combinations of the vectors $\vect{a}_1,\ldots,\vect{a}_p$, then so is $\vect{u} + \vect{v}$.
\vfill
	
	\item For all vectors $\vect{u}$ and $\vect{v}$, we have that $\Span\{\vect{u}\} = \Span\{\vect{v}\}$ implies $\vect{u} = \vect{v}$.
\vfill
\end{enumerate}

\item\points{1} Determine if $\vect{b}$ is in the span of $\vect{a_1},\vect{a_2},\vect{a_3}$ where $\vect{a_1} = \kgmmatrix{1}{-1 \\ -3 \\ 2}$, $\vect{a_2} = \kgmmatrix{1}{1 \\ 4 \\ 5}$, $\vect{a_3} = \kgmmatrix{1}{1 \\ 6 \\ 19}$, and $\vect{b} = \kgmmatrix{1}{1 \\ 0 \\ 0}$.
\vspace{6cm}

\newpage

\item\ppoints{2}{3} Given $A$ and $\vect{b}$ below, solve for $\vect{x}$ in the matrix equation $A\vect{x} = \vect{b}$.
\begin{enumerate}
\item $A = \kgmmatrix{3}{
5 & 3 & -1\\
0 & 1 & -2\\
-2 & 1 & 1
}$, $\vect{b} = \kgmmatrix{1}{31 \\ -8 \\ -15}$.
\vfill

\item $A = \kgmmatrix{3}{ 
2 & -1 & 2 \\ 
1 & 0 & 2 \\
3 & -1 & -1 \\
4 & 1 & 0\\
}$, $\vect{b} = \kgmmatrix{1}{23 \\ 13 \\ 11 \\ 5}$.
\vfill
\end{enumerate}


	

\end{enumerate}
\end{document}
