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Mathematics LibreTexts

4.3: Basic Theory of Homogeneous Linear System

  • Page ID
    17436
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

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    In this section we consider homogeneous linear systems ${\bf y}'=
    A(t){\bf y}$, where $A=A(t)$ is a continuous $n\times n$ matrix
    function on an interval $(a,b)$. The theory of linear homogeneous
    systems has much in common with the theory of linear homogeneous
    scalar equations, which we considered in
    Sections~2.1, 5.1, and 9.1.

    Whenever we refer to solutions of ${\bf y}'=A(t){\bf y}$ we'll mean
    solutions on $(a,b)$. Since ${\bf y}\equiv{\bf 0}$ is obviously a
    solution of ${\bf y}'=A(t){\bf y}$, we call it the {\color{blue}\it trivial\/}
    solution. Any other solution is {\color{blue}\it nontrivial\/}.

    If ${\bf y}_1$, ${\bf y}_2$, \dots, ${\bf y}_n$ are vector functions
    defined on an interval $(a,b)$ and $c_1$, $c_2$, \dots, $c_n$ are
    constants, then
    \begin{equation} \label{eq:10.3.1}
    {\bf y}=c_1{\bf y}_1+c_2{\bf y}_2+\cdots+c_n{\bf y}_n
    \end{equation}
    is a {\color{blue}\it linear combination of\/} ${\bf y}_1$, ${\bf y}_2$, \ldots,${\bf
    y}_n$. It's easy show that if ${\bf
    y}_1$, ${\bf y}_2$, \dots,${\bf y}_n$ are solutions of ${\bf y}'=A(t){\bf
    y}$ on $(a,b)$, then so is any linear combination of
     ${\bf y}_1$, ${\bf y}_2$, \dots, ${\bf y}_n$ (Exercise~\ref{exer:10.3.1}). We say that
    $\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}$ is a {\color{blue}\it fundamental set of
    solutions of ${\bf y}'=A(t){\bf y}$ on\/} $(a,b)$ on if every solution of
    ${\bf y}'=A(t){\bf y}$ on $(a,b)$ can be written as a linear combination of
    ${\bf y}_1$, ${\bf y}_2$, \dots, ${\bf y}_n$, as in \eqref{eq:10.3.1}.
    In this
    case we say that \eqref{eq:10.3.1} is the {\color{blue}\it general solution of ${\bf
    y}'=A(t){\bf y}$ on\/} $(a,b)$.

    It can be shown that if $A$ is continuous on $(a,b)$ then ${\bf
    y}'=A(t){\bf y}$ has infinitely many fundamental sets of solutions on
    $(a,b)$ (Exercises~\ref{exer:10.3.15} and \ref{exer:10.3.16}). The next
    definition will help to characterize fundamental sets of solutions of
    ${\bf y}'=A(t){\bf y}$.


    We say that a set $\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}$ of
    $n$-vector functions is {\color{blue}\it linearly independent\/} on $(a,b)$ if the
    only constants $c_1$, $c_2$, \dots, $c_n$ such that
    \begin{equation} \label{eq:10.3.2}
     c_1{\bf y}_1(t)+c_2{\bf y}_2(t)+\cdots+c_n{\bf y}_n(t)=0,\quad
    a<t<b,
    \end{equation}
    are $c_1=c_2=\cdots=c_n=0$. If \eqref{eq:10.3.2} holds for some set of
    constants $c_1$, $c_2$, \dots, $c_n$ that are not all zero, then $\{{\bf
    y}_1,{\bf y}_2,\dots,{\bf y}_n\}$ is {\color{blue}\it linearly dependent\/} on
    $(a,b)$

    The next theorem is analogous to
    Theorems~\ref{thmtype:5.1.3} and
    \ref{thmtype:9.1.2}.

    \begin{theorem}\color{blue}\label{thmtype:10.3.1}
    Suppose the $n\times n$ matrix $A=A(t)$ is continuous on $(a,b)$.
    Then a set
    $\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}$ of $n$ solutions of ${\bf
    y}'=A(t){\bf y}$ on $(a,b)$ is a fundamental set if and only if it's
    linearly independent on $(a,b)$.
    \end{theorem}

    \begin{example}\label{example:10.3.1} \rm
    Show that the vector functions
    $$
    {\bf y}_1=\left[\begin{array}{c}e^t\\0\\e^{-t}\end{array}\right],\quad
    {\bf
    y}_2=\left[\begin{array}{c}0\\e^{3t}\\1\end{array}\right],
    \mbox{\quad and \quad}{\bf
    y}_3=\left[\begin{array}{c}e^{2t}\\e^{3t}\\0\end{array}\right]
    $$
    are linearly independent on every interval  $(a,b)$.
    \end{example}

    \solution
    Suppose
    $$
    c_1\left[\begin{array}{c}e^t\\0\\e^{-t}\end{array}\right]+
    c_2\left[\begin{array}{c}0\\e^{3t}\\1\end{array}\right]+c_3
    \left[\begin{array}{c}e^{2t}\\e^{3t}\\0\end{array}\right]=
    \left[\begin{array}{c}0\\0\\0\end{array}\right],\quad a<t<b.
    $$
    We must show that $c_1=c_2=c_3=0$. Rewriting this equation in matrix
    form yields
    $$
    \left[\begin{array}{ccc}e^t&0&e^{2t}\\0&e^{3t}&e^{3t}\\e^{-t}&1&0
    \end{array}\right]
    \left[\begin{array}{c}c_1\\c_2\\c_3\end{array}\right]=
    \left[\begin{array}{c}0\\0\\0\end{array}\right],\quad a<t<b.
    $$
    Expanding the determinant of this system in cofactors of the entries
    of the first row yields
    \begin{eqnarray*}
    \left|\begin{array}{ccc}e^t&0&e^{2t}\\0&e^{3t}&e^{3t}\\e^{-t}&1&0
    \end{array}\right|&=&e^t
    \left|\begin{array}{cc}e^{3t}&e^{3t}\\1&0\end{array}\right|-0
    \left|\begin{array}{cc}0&e^{3t}\\e^{-t}&0\end{array}\right|
    +e^{2t}\left|\begin{array}{cc}0&e^{3t}\\e^{-t}&1\end{array}\right| \\
    &=&e^t(-e^{3t})+e^{2t}(-e^{2t})=-2e^{4t}.
    \end{eqnarray*}
    Since this determinant is never zero,
    $c_1=c_2=c_3=0$. \bbox


    We can use the method in
     Example~\ref{example:10.3.1}  to test
    $n$ solutions $\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}$ of any
    $n\times n$ system
    ${\bf y}'=A(t){\bf y}$  for linear independence on an interval $(a,b)$
    on which $A$ is continuous.  To explain this (and for other purposes
    later), it's useful to write a linear combination of
    ${\bf y}_1$, ${\bf y}_2$, \dots, ${\bf y}_n$ in a different way. We first
    write the vector functions in terms of their components as
    $$
    {\bf y}_1=\left[\begin{array}{c} y_{11}\\y_{21}\\ \vdots\\
    y_{n1}\end{array}\right],\quad
    {\bf y}_2=\left[\begin{array}{c} y_{12}\\y_{22}\\ \vdots\\
    y_{n2}\end{array}\right],\dots,\quad
    {\bf y}_n=\left[\begin{array}{c} y_{1n}\\y_{2n}\\ \vdots\\
    y_{nn}\end{array}\right].
    $$
    If
    $$
    {\bf y}=c_1{\bf y}_1+c_2{\bf y}_2+\cdots+c_n{\bf y}_n
    $$
    then
    \enlargethispage{.3in}
    \begin{eqnarray*}
    {\bf y}&=&
    c_1\left[\begin{array}{c} y_{11}\\y_{21}\\ \vdots\\
    y_{n1}\end{array}\right]+
    c_2\left[\begin{array}{c} y_{12}\\y_{22}\\ \vdots\\
    y_{n2}\end{array}\right]+\cdots
    +c_n\left[\begin{array}{c} y_{1n}\\y_{2n}\\ \vdots\\
    y_{nn}\end{array}\right]\\[2\jot]
    &=&\left[\begin{array}{cccc}
    y_{11}&y_{12}&\cdots&y_{1n} \\
    y_{21}&y_{22}&\cdots&y_{2n}\\
    \vdots&\vdots&\ddots&\vdots \\
    y_{n1}&y_{n2}&\cdots&y_{nn} \\
    \end{array}\right]\col cn.
    \end{eqnarray*}
    This shows that
    \begin{equation} \label{eq:10.3.3}
    c_1{\bf y}_1+c_2{\bf y}_2+\cdots+c_n{\bf y}_n=Y{\bf c},
    \end{equation}
    where
    $$
    {\bf c}=\col cn
    $$
    and
    \begin{equation} \label{eq:10.3.4}
    Y=[{\bf y}_1\; {\bf y}_2\; \cdots\; {\bf y}_n]=
    \left[\begin{array}{cccc}
    y_{11}&y_{12}&\cdots&y_{1n} \\
    y_{21}&y_{22}&\cdots&y_{2n}\\
    \vdots&\vdots&\ddots&\vdots \\
    y_{n1}&y_{n2}&\cdots&y_{nn} \\
    \end{array}\right];
    \end{equation}
     that is, the columns of $Y$
    are the vector functions ${\bf y}_1,{\bf y}_2,\dots,{\bf y}_n$.

    For reference below, note that
    \begin{eqnarray*}
    Y'&=&[{\bf y}_1'\; {\bf y}_2'\; \cdots\; {\bf y}_n']\\
    &=&[A{\bf y}_1\; A{\bf y}_2\; \cdots\; A{\bf y}_n]\\
    &=&A[{\bf y}_1\; {\bf y}_2\; \cdots\; {\bf y}_n]=AY;
    \end{eqnarray*}
    that is, $Y$ satisfies the matrix differential equation

    $$
    Y'=AY.
    $$

    The determinant of $Y$,
    \begin{equation} \label{eq:10.3.5}
    W=\left|\begin{array}{cccc}
    y_{11}&y_{12}&\cdots&y_{1n} \\
    y_{21}&y_{22}&\cdots&y_{2n}\\
    \vdots&\vdots&\ddots&\vdots \\
    y_{n1}&y_{n2}&\cdots&y_{nn} \\
    \end{array}\right|
    \end{equation}
    is called the
    \href{http://www-history.mcs.st-and.ac.uk/.../Wronski.html}
    {\color{blue}\it Wronskian\/} of $\{{\bf y}_1,{\bf y}_2,\dots,{\bf
    y}_n\}$. It can be shown (Exercises~\ref{exer:10.3.2} and \ref{exer:10.3.3})
    that this definition is analogous to definitions of the Wronskian of
    scalar functions given in Sections~5.1 and 9.1.
    The next theorem is analogous to
    Theorems~\ref{thmtype:5.1.4} and
    \ref{thmtype:9.1.3}. The proof is sketched in
    Exercise~\ref{exer:10.3.4} for
    $n=2$ and in Exercise~\ref{exer:10.3.5} for general~$n$.

    \begin{theorem}\color{blue}$[$Abel's Formula$]$ \label{thmtype:10.3.2}
    Suppose the $n\times n$ matrix $A=A(t)$ is continuous on $(a,b),$ let
    ${\bf y}_1$, ${\bf y}_2$, \dots, ${\bf y}_n$ be solutions of ${\bf
    y}'=A(t){\bf y}$ on $(a,b),$ and let $t_0$ be in $(a,b)$. Then the
    Wronskian of $\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}$ is given by
    \begin{equation} \label{eq:10.3.6}
    W(t)=W(t_0)\exp\left(
    \int^t_{t_0}\big[a_{11}(s)+a_{22}(s)+\cdots+a_{nn}(s)]\,
    ds\right), \quad  a < t < b.
    \end{equation}
    Therefore$,$  either $W$ has no zeros in  $(a,b)$ or $W\equiv0$
    on  $(a,b).$
    \end{theorem}


    \color{blue}
    \remark{
    The sum of the diagonal entries of a square matrix $A$ is called the
    {\color{blue}\it trace\/} of $A$, denoted by tr$(A)$. Thus, for an $n\times n$
    matrix $A$,
    $$
    \mbox{tr}(A)=a_{11}+a_{22}+\cdots+a_{nn},
    $$
    and  \eqref{eq:10.3.6} can be written as
    $$
    W(t)=W(t_0)\exp\left(
    \int^t_{t_0}\mbox{tr}(A(s))\,
    ds\right), \quad a < t < b.
    $$}
    \color{black}

    The next theorem is analogous to
    Theorems~\ref{thmtype:5.1.6} and
    \ref{thmtype:9.1.4}.

    \begin{theorem}\color{blue}\label{thmtype:10.3.3}
    Suppose the  $n\times n$ matrix $A=A(t)$ is continuous
     on $(a,b)$ and let
    ${\bf y}_1$, ${\bf y}_2$, \dots$,$${\bf y}_n$
    be  solutions of ${\bf y}'=A(t){\bf y}$ on  $(a,b)$.
     Then the following statements are equivalent; that is, they are
    either all true or all false:
    \begin{alist}
    \item % (a)
    The general solution of ${\bf y}'=A(t){\bf y}$ on  $(a,b)$ is
    ${\bf y}=c_1{\bf y}_1+c_2{\bf y}_2+\cdots+c_n{\bf y}_n$,
    where $c_1$, $c_2$, \dots, $c_n$  are arbitrary constants.
    \item % (b)
      $\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}$ is a fundamental
    set of solutions of ${\bf y}'=A(t){\bf y}$  on $(a,b)$.
    \item % (c)
     $\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}$ is linearly
    independent on $(a,b)$.
    \item % (d)
    The Wronskian of  $\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}$ is nonzero
    at some point in $(a,b)$.
    \item % (e)
    The Wronskian of  $\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}$ is nonzero
    at all points in $(a,b)$.
    \end{alist}
    \end{theorem}


    We say that $Y$ in \eqref{eq:10.3.4} is a {\color{blue}\it fundamental matrix\/} for
    ${\bf y}'=A(t){\bf y}$ if any (and therefore all) of the statements
    ({\bf a})-({\bf e}) of Theorem~\ref{thmtype:10.3.2} are true for the
    columns of $Y$. In this case, \eqref{eq:10.3.3} implies that the general
    solution of ${\bf y}'=A(t){\bf y}$ can be written as ${\bf y}=Y{\bf
    c}$, where ${\bf c}$ is an arbitrary constant $n$-vector.

    \begin{example}\label{example:10.3.2} \rm
    The vector functions
    $$
    {\bf y}_1=\twocol {-e^{2t}}{2e^{2t}}\mbox{\quad and \quad}
    {\bf y}_2=\twocol{-e^{-t}}{\phantom{-}e^{-t}}
    $$
    are solutions of the constant coefficient system
    \begin{equation} \label{eq:10.3.7}
    {\bf y}'=\twobytwo{-4}{-3} 65 {\bf y}
    \end{equation}
    on $(-\infty,\infty)$.  (Verify.)
    \begin{alist}
    \item % (a)
    Compute the Wronskian of $\{{\bf y}_1,{\bf y}_2\}$
    directly from the definition \eqref{eq:10.3.5}
    \item % (b)
    Verify Abel's formula \eqref{eq:10.3.6} for the Wronskian of
    $\{{\bf y}_1,{\bf y}_2\}$.
    \item % (c)
    Find the general solution of \eqref{eq:10.3.7}.
    \item % (d)
     Solve the initial value problem
    \begin{equation} \label{eq:10.3.8}
    {\bf y}'=\twobytwo{-4}{-3}65 {\bf y}, \quad {\bf y}(0)=
    \left[\begin{array}{r} 4 \\-5\end{array}\right].
    \end{equation}
    \end{alist}
    \end{example}

    \solutionpart{a}
    From \eqref{eq:10.3.5}
    \begin{equation} \label{eq:10.3.9}
    W(t)=\left|\begin{array}{cc}-e^{2t}&-e^{-t}\\2e^{2t}&\hfill
    e^{-t}\end{array}\right|=
    e^{2t}e^{-t}
    \twobytwo{-1}{-1}21=e^t.
    \end{equation}

    \solutionpart{b}  Here
    $$
    A=\twobytwo{-4}{-3} 65,
    $$
    so
    tr$(A)=-4+5=1$. If $t_0$ is an arbitrary real number then
    \eqref{eq:10.3.6}  implies that
    $$
    W(t)=W(t_0)\exp{\left(\int_{t_0}^t1\,ds\right)}=
    \left|\begin{array}{cc}
    -e^{2t_0}&-e^{-t_0}\\2e^{2t_0}&e^{-t_0}\end{array}\right|e^{(t-t_0)}
    =e^{t_0}e^{t-t_0}=e^t,
    $$
    which is consistent with \eqref{eq:10.3.9}.

    \solutionpart{c} Since $W(t)\ne0$,
    Theorem~\ref{thmtype:10.3.3} implies that $\{{\bf y}_1,{\bf y}_2\}$ is
    a fundamental set of solutions  of \eqref{eq:10.3.7} and
    $$
    Y=\left[\begin{array}{cc}-e^{2t}&-e^{-t}\\2e^{2t}&\hfill
    e^{-t}\end{array}\right]
    $$
    is a fundamental matrix for \eqref{eq:10.3.7}.
     Therefore the general
    solution of \eqref{eq:10.3.7} is
    \begin{equation} \label{eq:10.3.10}
    {\bf y}=c_1{\bf y}_1+c_2{\bf y}_2=
    c_1\twocol {-e^{2t}}{2e^{2t}}+c_2\twocol{-e^{-t}}{e^{-t}}
    =\left[\begin{array}{cc}-e^{2t}&-e^{-t}\\2e^{2t}&\hfill
    e^{-t}\end{array}\right]
    \left[\begin{array}{c}c_1\\c_2\end{array}\right].
    \end{equation}

    \solutionpart{d}
    Setting $t=0$ in \eqref{eq:10.3.10} and imposing the initial condition
    in \eqref{eq:10.3.8} yields
    $$
    c_1\left[\begin{array}{r}-1 \\2\end{array}\right]+c_2
    \left[\begin{array}{r}-1 \\1\end{array}\right]=
    \left[\begin{array}{r} 4 \\-5\end{array}\right].
    $$
    Thus,
    \begin{eqnarray*}
    -c_1-c_2&=&\phantom{-}4 \\
    2c_1+c_2&=&-5.
    \end{eqnarray*}
    The solution of this system is $c_1=-1$, $c_2=-3$.  Substituting these
    values into  \eqref{eq:10.3.10} yields
    $$
    {\bf y}=-\left[\begin{array}{c}-e^{2t} \\ 2e^{2t}\end{array}
    \right]-3
    \left[\begin{array}{c}-e^{-t} \\ e^{-t}\end{array}\right]=
    \left[
    \begin{array}{c} e^{2t}+3e^{-t} \\-2e^{2t}-3e^{-t}
    \end{array}\right]
    $$
    as the solution of  \eqref{eq:10.3.8}.

    \enlargethispage{1in}
    \exercises


    \begin{exerciselist}
    \item\label{exer:10.3.1}
    Prove: If ${\bf y}_1$, ${\bf y}_2$, \dots, ${\bf y}_n$ are solutions of
    ${\bf y}'=A(t){\bf y}$ on  $(a,b)$, then any linear
    combination of ${\bf y}_1$, ${\bf y}_2$, \dots, ${\bf y}_n$ is also a
    solution of ${\bf y}'=A(t){\bf y}$ on $(a,b)$.

    \item\label{exer:10.3.2}
    In Section~5.1 the Wronskian of two
    solutions $y_1$ and $y_2$ of the scalar second order equation
    $$
    P_0(x)y''+P_1(x)y'+P_2(x)y=0
    \eqno{\rm (A)}
    $$
    was defined to be
    $$
    W=\left|\begin{array}{cc} y_1&y_2 \\ y'_1&y'_2\end{array}\right|.
    $$
    \begin{alist}
    \item % (a)
    Rewrite  (A)   as a system of first order equations and show
    that $W$ is
    the Wronskian (as defined in this section) of two solutions of this system.
    \item % (b)
    Apply Eqn.~\eqref{eq:10.3.6} to the system  derived in \part{a}, and
    show that
    $$
    W(x)=W(x_0)\exp\left\{-\int^x_{x_0}{P_1(s)\over P_0(s)}\,
    ds\right\},
    $$
    which is the form of Abel's formula given in
    Theorem~9.1.3.
    \end{alist}


    \item\label{exer:10.3.3}
    In Section~9.1 the Wronskian of $n$
    solutions $y_1$, $y_2$, \dots, $y_n$  of the $n-$th order
    equation
    $$
    P_0(x)y^{(n)}+P_1(x)y^{(n-1)}+\cdots+P_n(x)y=0
    \eqno{\rm (A)}
    $$
    was defined to be
    $$
    W=\left|\begin{array}{cccc}
    y_1&y_2&\cdots&y_n \\[2\jot]
    y'_1&y'_2&\cdots&y_n'\\[2\jot]
    \vdots&\vdots&\ddots&\vdots\\[2\jot]
    y_1^{(n-1)}&y_2^{(n-1)}&\cdots&y_n^{(n-1)}
    \end{array}\right|.
    $$
    \begin{alist}
    \item % (a)
    Rewrite (A) as a system of first order equations and
    show that $W$ is the Wronskian (as defined in this section) of $n$
    solutions of this system.
    \item % (b)
    Apply Eqn.~\eqref{eq:10.3.6} to the system  derived in \part{a}, and
    show that
    $$
    W(x)=W(x_0)\exp\left\{-\int^x_{x_0}{P_1(s)\over P_0(s)}\,
    ds\right\},
    $$
    which is the form of Abel's formula given in
    Theorem~9.1.3.
    \end{alist}


    \item\label{exer:10.3.4}
     Suppose
    $$
    {\bf y}_1=\twocol{y_{11}}{y_{21}}\mbox{\quad and \quad}
    {\bf y}_2=\twocol{y_{12}}{y_{22}}
    $$
    are solutions of the $2\times 2$ system ${\bf y}'=A{\bf y}$ on
    $(a,b)$, and let
    $$
     Y=\twobytwo {y_{11}} {y_{12}} {y_{21}} {y_{22}}\mbox{\quad and \quad}
    W=\left|\begin{array}{cc} y_{11}&y_{12}\\y_{21}&y_{22}\end{array}\right|;
    $$ thus, $W$ is the Wronskian of $\

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    &{y'_{12}}\\ {y_{21}}&
    {y_{22}}\end{array}\right|
    +\left|\begin{array}{cc} {y_{11}}&{y_{12}}\\
     {y'_{21}}&{y'_{22}}\end{array}\right|.
    $$

    \item % (b)
     Use the equation $Y'=A(t)Y$ and
    the definition of matrix multiplication to show that
    $$
    [y'_{11}\quad y'_{12}]=a_{11} [y_{11}\quad y_{12}]+a_{12} [y_{21}
    \quad y_{22}]
    $$
    and
    $$
    [y'_{21}\quad  y'_{22}]=a_{21} [y_{11}\quad y_{12}]+a_{22}
    [y_{21}\quad y_{22}].
    $$
    \item % (c)
     Use  properties of determinants to deduce from \part{a} and \part{a}
     that
    $$
    \left|\begin{array}{cc} {y'_{11}}&{y'_{12}}\\ {y_{21}}&
    {y_{22}}\end{array}\right|=a_{11}W\mbox{\quad and \quad}
    \left|\begin{array}{cc} {y_{11}}&{y_{12}}\\
     {y'_{21}}&{y'_{22}}\end{array}\right|=a_{22}W.
    $$

    \item % (d)
     Conclude from \part{c} that
    $$
    W'=(a_{11}+a_{22})W,
    $$
    and use this to show that if $a<t_0<b$  then
    $$
    W(t)=W(t_0)\exp\left(\int^t_{t_0} \left[a_{11}(s)+a_{22} (s)
    \right]\, ds\right)\quad a<t<b.
    $$
    \end{alist}

    \item\label{exer:10.3.5}
    Suppose  the $n\times n$ matrix $A=A(t)$ is continuous on $(a,b)$. Let
    $$
     Y=
    \left[\begin{array}{cccc} y_{11}&y_{12}&\cdots&y_{1n} \\
    y_{21}&y_{22}&\cdots&y_{2n} \\
    \vdots&\vdots&\ddots&\vdots \\
    y_{n1}&y_{n2}&\cdots&y_{nn}
    \end{array}\right],
    $$
    where the columns of $Y$ are solutions of ${\bf y}'=A(t){\bf y}$. Let
    $$
    r_i=[y_{i1}\, y_{i2}\, \dots\, y_{in}]
    $$
    be the $i$th row of $Y$, and let $W$ be the determinant of $Y$.

    \begin{alist}
    \item % (a)
     Deduce from the definition of  determinant that
    $$
    W'=W_1+W_2+\cdots+W_n,
    $$
    where, for $1 \le m \le n$, the $i$th row of $W_m$ is $r_i$ if $i \ne m$,
    and $r'_m$ if $i=m$.

    \item % (b)
     Use the equation $Y'=A Y$
    and the definition of matrix multiplication to show that
    $$
    r'_m=a_{m1}r_1+a_{m2} r_2+\cdots+a_{mn}r_n.
    $$

    \item % (c)
     Use  properties of determinants to deduce
    from \part{b} that
    $$
    \det (W_m)=a_{mm}W.
    $$

    \item % (d)
     Conclude from \part{a} and \part{c} that
    $$
    W'=(a_{11}+a_{22}+\cdots+a_{nn})W,
    $$
    and use this to  show that  if $a<t_0<b$ then
    $$
    W(t)=W(t_0)\exp\left(
    \int^t_{t_0}\big[a_{11}(s)+a_{22}(s)+\cdots+a_{nn}(s)]\,
    ds\right), \quad a < t < b.
    $$
    \end{alist}

    \item\label{exer:10.3.6}
    Suppose the $n\times n$ matrix $A$ is continuous on $(a,b)$
    and $t_0$ is a point in  $(a,b)$. Let $Y$  be a fundamental matrix for
    ${\bf y}'=A(t){\bf y}$ on  $(a,b)$.
    \begin{alist}
    \item % (a)
    Show that $Y(t_0)$  is invertible.
    \item % (b)
    Show that if ${\bf k}$ is an arbitrary $n$-vector then the solution
    of the initial value problem
    $$
    {\bf y}'=A(t){\bf y},\quad {\bf y}(t_0)={\bf k}
    $$
    is
    $$
    {\bf y}=Y(t)Y^{-1}(t_0){\bf k}.
    $$
    \end{alist}

    \item\label{exer:10.3.7}
    Let
    $$
     A=\twobytwo2442, \quad {\bf y}_1=\left[\begin{array}{c} e^{6t} \\
    e^{6t}
    \end{array}\right], \quad {\bf y}_2=\left[\begin{array}{r}
    e^{-2t} \\
    -e^{-2t}\end{array}\right], \quad {\bf k}=\left[\begin{array}{r}-3
    \\ 9\end{array}\right].
    $$
    \begin{alist}
    \item % (a)
     Verify that $\{{\bf y}_1,{\bf y}_2\}$
    is a fundamental set of solutions for
    ${\bf y}'=A{\bf y}$.

    \item % (b)
    Solve  the initial value problem
    $$
    {\bf y}'=A{\bf y},\quad   {\bf y}(0)={\bf k}.
    \eqno{\rm(A)}
    $$

    \item % (c)
    Use the result of Exercise~\ref{exer:10.3.6}\part{b} to find a formula for
    the solution of (A) for an arbitrary initial vector ${\bf
    k}$.
    \end{alist}


    \item\label{exer:10.3.8}
     Repeat Exercise~\ref{exer:10.3.7}  with
    $$
     A=\twobytwo {-2} {-2} {-5}1, \quad {\bf y}_1=\left[\begin{array}{r}
    e^{-4t} \\ e^{-4t}\end{array}\right], \quad {\bf y}_2=\left[
    \begin{array}{r}-2e^{3t}
    \\ 5e^{3t}\end{array}\right], \quad {\bf k}=\left[\begin{array}{r}
    10 \\-4\end{array}\right].
    $$

    \item\label{exer:10.3.9}
    Repeat Exercise~\ref{exer:10.3.7}  with
    $$
     A=\twobytwo{-4} {-10} 3 7, \quad
    {\bf y}_1=\left[\begin{array}{r}-5e^{2t} \\ 3e^{2t}
    \end{array}\right], \quad
    {\bf y}_2=\left[\begin{array}{r} 2e^t \\-e^t
    \end{array}\right], \quad
    {\bf k}=\left[\begin{array}{r}-19 \\ 11\end{array}
    \right
    ]. $$

    \item\label{exer:10.3.10}
     Repeat Exercise~\ref{exer:10.3.7} with
    $$
     A=\twobytwo 2 1 1 2, \quad {\bf y}_1=\left[\begin{array}{r} e^{3t} \\
    e^{3t}
    \end{array}\right], \quad {\bf y}_2=\left[\begin{array}{r}e^t \\
    -e^t\end{array}\right], \quad {\bf k}=\left[\begin{array}{r} 2 \\ 8
    \end{array}\right].$$

    \item\label{exer:10.3.11}
    Let
    \begin{eqnarray*}
     A&=&\threebythree 3 {-1} {-1} {-2} 3
    24 {-1} {-2}, \\
    {\bf y}_1&=&\left[\begin{array}{c} e^{2t} \\ 0 \\ e^{2t}\end{array}
    \right], \quad
    {\bf y}_2=\left[\begin{array}{c} e^{3t} \\-e^{3t} \\
    e^{3t}\end{array}\right], \quad
    {\bf y}_3=\left[\begin{array}{c} e^{-t} \\-3e^{-t} \\
    7e^{-t}
    \end{array}\right], \quad {\bf k}=\left[\begin{array}{r}
    2 \\-7 \\ 20\end{array}\right].
    \end{eqnarray*}

    \begin{alist}
    \item % (a)
     Verify that $\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$
    is a fundamental set of solutions for
    ${\bf y}'=A{\bf y}$.

    \item % (b)
     Solve the initial value problem
    $$
    {\bf y}'=A{\bf y}, \quad   {\bf y}(0)={\bf k}.
    \eqno{\rm(A)}
    $$

    \item % (c)
    Use the result of Exercise~\ref{exer:10.3.6}\part{b} to find a formula
    for
     the solution of  (A)
    for an arbitrary initial vector ${\bf k}$.
    \end{alist}


    \item\label{exer:10.3.12}
     Repeat Exercise~\ref{exer:10.3.11} with
    \begin{eqnarray*}
     A&=&\threebythree 0 2 2 2 0 2 2 2 0, \\
    {\bf y}_1&=&\left[\begin{array}{c}-e^{-2t} \\ 0 \\ e^{-2t}
    \end{array}\right], \quad
    {\bf y}_2=\left[\begin{array}{c}-e^{-2t} \\ e^{-2t} \\
    0\end{array}\right], \quad
    {\bf y}_3=\left[\begin{array}{c} e^{4t} \\ e^{4t} \\ e^{4t}\end{array}
    \right], \quad
    {\bf k}=\left[\begin{array}{r} 0 \\-9 \\ 12\end{array}
    \right].
    \end{eqnarray*}

    \item\label{exer:10.3.13}
     Repeat Exercise~\ref{exer:10.3.11}  with
    \begin{eqnarray*}
     A&=&\threebythree {-1} 2 3 0 1 6
    0 0 {-2}, \\
    {\bf y}_1&=&\left[\begin{array}{c} e^t \\ e^t \\ 0\end{array}\right],
    \quad
    {\bf y}_2=\left[\begin{array}{c} e^{-t} \\ 0 \\ 0\end{array}\right],
    \quad {\bf y}_3=\left[\begin{array}{c} e^{-2t} \\-2e^{-2t}
    \\ e^{-2t}\end{array}\right], \quad
    {\bf k}=\left[\begin{array}{r} 5 \\ 5 \\-1
    \end{array}\right].
    \end{eqnarray*}

    \item\label{exer:10.3.14}
    Suppose $Y$ and $Z$ are fundamental matrices for the $n\times n$
    system ${\bf y}'=A(t){\bf y}$. Then some of the four matrices
    $YZ^{-1}$, $Y^{-1}Z$, $Z^{-1}Y$, $Z Y^{-1}$ are necessarily
    constant. Identify them and prove that they are constant.

    \item\label{exer:10.3.15}
    Suppose the columns of an $n\times n$ matrix $Y$ are solutions of
    the $n\times n$ system ${\bf y}'=A{\bf y}$ and $C$ is an $n \times n$
    constant matrix.

    \begin{alist}
    \item % (a)
    Show that the matrix $Z=YC$ satisfies the differential equation
    $Z'=AZ$.

    \item % (b)
    Show that $Z$ is a fundamental matrix for ${\bf y}'=A(t){\bf y}$ if
    and only if $C$ is invertible and $Y$ is a fundamental matrix for
    ${\bf y}'=A(t){\bf y}$.
    \end{alist}

    \item\label{exer:10.3.16}
     Suppose the $n\times n$  matrix $A=A(t)$ is
    continuous on $(a,b)$ and $t_0$ is in $(a,b)$.
     For $i=1$, $2$, \dots, $n$, let ${\bf y}_i$ be the solution of the initial
    value
    problem ${\bf y}_i'=A(t){\bf y}_i,\; {\bf y}_i(t_0)={\bf e}_i$, where
    $$
    {\bf e}_1=\left[\begin{array}{c} 1\\0\\ \vdots\\0\end{array}\right],\quad
     {\bf e}_2=\left[\begin{array}{c} 0\\1\\
    \vdots\\0\end{array}\right],\quad\cdots\quad
     {\bf e}_n=\left[\begin{array}{c} 0\\0\\ \vdots\\1\end{array}\right];
    $$
    that is, the $j$th component of ${\bf e}_i$ is $1$ if $j=i$, or $0$ if
    $j\ne i$.

    \begin{alist}
    \item % (a)
     Show that$\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}$ is a fundamental set of
    solutions of ${\bf y}'=A(t){\bf y}$ on  $(a,b)$.

    \item % (b)
     Conclude from \part{a} and Exercise~\ref{exer:10.3.15}  that ${\bf y}'=
    A(t){\bf y}$ has infinitely many fundamental sets of solutions on
    $(a,b)$.
    \end{alist}

    \item\label{exer:10.3.17}
     Show that $Y$ is a
    fundamental matrix for the system ${\bf y}'=A(t){\bf y}$ if and only
    if $Y^{-1}$ is a fundamental matrix for ${\bf y}'=-
    A^T(t){\bf y}$, where $A^T$ denotes the transpose of $A$.
    \hint{See Exercise \ref{exer:10.2.11}.}

    \enlargethispage{1in}
    \item\label{exer:10.3.18}
    Let $Z$ be the fundamental matrix for the constant coefficient system
     ${\bf y}'=A{\bf y}$ such that $Z(0)=I$.
    \begin{alist}
    \item % (a)
    Show that $Z(t)Z(s)=Z(t+s)$ for all $s$ and $t$. \hint{For
    fixed
    $s$ let $\Gamma_1(t)=Z(t)Z(s)$ and $\Gamma_2(t)=Z(t+s)$. Show that
    $\Gamma_1$ and $\Gamma_2$ are both solutions of the matrix initial value
    problem $\Gamma'=A\Gamma,\quad\Gamma(0)=Z(s)$. Then conclude from
    Theorem~\ref{thmtype:10.2.1} that $\Gamma_1=\Gamma_2$.}
    \item % (b)
    Show that $(Z(t))^{-1}=Z(-t)$.
    \item % (c)
    The matrix $Z$ defined above is sometimes denoted by $e^{tA}$. Discuss
     the motivation for this notation.
    \end{alist}


    \end{exerciselist}