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\title{Introduction to linear systems}
\date{April 16, 2021}

\begin{document}
\pagestyle{empty}

\ 

\vspace{-.1in}

\maketitle

\vspace{-.2in}

\prob{[Completion]
Rewrite the system in matrix form and verify that the given vector function is a solution.
\begin{enumerate}[a)]
\item \begin{align*}
y_1' &= -4y_1-10y_2 \\
y_2' &= 3y_1+7y_2
\end{align*}
\[ \mathbf{y} = \begin{bmatrix} -5 \\ 3 \end{bmatrix} e^{2t} + \begin{bmatrix} 2 \\ -1 \end{bmatrix} e^t \]

\vspace{.25in}

\item \begin{align*}
y_1' &= -y_1+ 2y_2+3y_3 \\
y_2' &=  y_2+6y_3 \\
y_3' &=   -2y_3
\end{align*}
\[ \mathbf{y} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} e^{t} + \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} e^{-t} + \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} e^{-2t}\]

\vspace{.25in}

\end{enumerate}
}

\dfn{Matrix multiplication: $\displaystyle
\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x & y \\ z & w \end{bmatrix} = \begin{bmatrix} ax+bz & ay+bw \\ cx+dz & cy+dw \end{bmatrix} 
$
}

\prob{[Completion]
Verify that $Y' = AY$ for
$\displaystyle Y = \begin{bmatrix} e^{6t} & e^{-2t} \\  e^{6t} & -e^{-2t} \end{bmatrix}$ and $\displaystyle A = \begin{bmatrix} 2 & 4 \\ 4 & 2 \end{bmatrix}$.

}

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