4.3E: Exercises

Exercise \(\PageIndex{1}\)

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 \((a,b)\).

Exercise \(\PageIndex{2}\)

In Section 2.1 the Wronskian of two solutions \(y_1\) and \(y_2\) of the scalar second order equation

\begin{equation} \label{eq:4.3E.1}
P_0(x)y'' + P_1(x)y' + P_2(x)y = 0
\end{equation}

was defined to be

\begin{eqnarray*}
W = \left| \begin{array} \\ y_1 & y_2 \\ y'_1 & y'_2 \end{array} \right|.
\end{eqnarray*}

(a) Rewrite \eqref{eq:4.3E.1} 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.

(b) Apply Equation \((4.3.6)\) to the system derived in part (a), and show that

\begin{eqnarray*}
W(x) = W(x_0)\exp\left\{-\int^x_{x_0}{P_1(s)\over P_0(s)}\,ds\right\},
\end{eqnarray*}

which is the form of Abel's formula given in Theorem \((3.1.3\))

Answer

\part{a}
The system equivalent of (A) is (B) \({\bf
y}'=-\sisplaystyle{1\over P_0(x)}\twobytwo01{P_2(x)}{P_1(x)}{\bf y}\), where
\({\bf y}=\twocol{y}{y'}\). Let \({\bf y}_1=\twocol{y_1}{y_1'}\) and
 \({\bf y}_1=\twocol{y_2}{y_2'}\). Then the Wronskian of \(\{{\bf
y}_1,{\bf y}_2\}\) as defined in this section is
\(\twobytwo{y_1}{y_2}{y_1'}{y_2'}=W\).

\part{b}
The trace of the matrix in (B) is \(-P_1(x)/P_0(x)\), so Eqn.~4.3.6 implies
that
\(W(x)=W(x_0)\sisplaystyle\exp\left\{-\int^x_{x_0}{P_1(s)\over P_0(s)}\,
ds\right\}\).
 

Exercise \(\PageIndex{3}\)

In Section 3.1, the Wronskian of \(n\) solutions \(y_1\), \(y_2\), \(\dots\), \(y_n\) of the \(n\)th order equation

\begin{equation} \label{eq:4.3E.2}
P_0(x)y^{(n)} + P_1(x)y^{(n-1)} + \cdots + P_n(x)y = 0
\end{equation}

was defined to be

\begin{eqnarray*}
W = \left| \begin{array} \\ y_1 & y_2 & \cdots & y_n \\ y'_1 & y'_2 & \cdots & y'_n \\ \vdots & \vdots & \ddots & \vdots \\ y_1^{(n-1)} & y_2^{(n-1)} & \cdots & y_n^{(n-1)} \end{array} \right|.
\end{eqnarray*}

(a) Rewrite \eqref{eq:4.3E.2} 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.

(b) Apply Equation \((4.3.6)\) to the system derived in part (a), and show that

\begin{eqnarray*}
W(x) = W(x_0)\exp\left\{-\int^x_{x_0}{P_1(s)\over P_0(s)}\,ds\right\},
\end{eqnarray*}

which is the form of Abel's formula given in Theorem \((3.1.3\)).

Exercise \(\PageIndex{4}\)

Suppose

\begin{eqnarray*}
{\bf y}_1 = \left[ \begin{array} \\ y_{11} \\ y_{21} \end{array} \right] \quad \mbox{and} \quad {\bf y}_2 = \left[ \begin{array} \\ y_{12} \\ y_{22} \end{array} \right]
\end{eqnarray*}

are solutions of the \(2\times 2\) system \({\bf y}'=A{\bf y}\) on \((a,b)\), and let

\begin{eqnarray*}
Y = \left| \begin{array} \\ y_{11} & y_{12} \\ y_{21} & y_{22} \end{array} \right] \quad \mbox{and} \quad W = \left| \begin{array} y_{11} & y_{12} \\ y_{21} & y_{22} \end{array} \right|;
\end{eqnarray*}

thus, \(W\) is the Wronskian of \(\{{\bf y}_1,{\bf y}_2\}\).

(a) Deduce from the definition of determinant that

\begin{eqnarray*}
W' = \left| \begin{array} \\ y'_{11} & y'_{12} \\ y_{21} & y_{22} \end{array} \right| + \left| \begin{array} \\ y_{11} & y_{12} \\ y'_{21} & y'_{22} \end{array} \right|.
\end{eqnarray*}

(b) Use the equation \(Y'=A(t)Y\) and the definition of matrix multiplication to show that

\begin{eqnarray*}
[y'_{11} \quad y'_{12}] = a_{11} [y_{11} \quad y_{12}] + a_{12} [y_{21} \quad y_{22}]
\end{eqnarray*}

and

\begin{eqnarray*}
[y'_{21} \quad y'_{22}] = a_{21} [y_{11} \quad y_{12}] + a_{22} [y_{21} \quad y_{22}].
\end{eqnarray*}

(c) Use properties of determinants to deduce from part (a) and part (b) that

\begin{eqnarray*}
\left| \begin{array} \\ y'_{11} & y'_{12} \\ y_{21} & y_{22} \end{array} \right| = a_{11} W \quad \mbox{and} \quad \left| \begin{array} \\ y_{11} & y_{12} \\ y'_{21} & y'_{22} \end{array} \right| = a_{22}W.
\end{eqnarray*}

(d) Conclude from part (c) that

\begin{eqnarray*}
W' = (a_{11} + a_{22}) W,
\end{eqnarray*}

and use this to show that if \(a<t_0<b\) then

\begin{eqnarray*}
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{eqnarray*}

Answer

\part{a} Since
\(F=\sum\pm f_{1i_1}f_{2i_2},\dots,f_{ni_n}\),
\begin{eqnarray*}
F'&=&\sum\pm f'_{1i_1}f_{2i_2},\dots,f_{ni_n}
+\sum\pm f_{1i_1}f'_{2i_2},\dots,f_{ni_n}+\cdots+
\sum\pm f_{1i_1}f_{2i_2},\dots,f'_{ni_n}\\
&=&F_1+F_2+\cdots+F_n.
\end{eqnarray*}

\part{c}
\(
\left|\begin{array}{cc}
y_{11}'&y_{12}'\\y_{21}&y_{22}
\end{array}\right|=
\left|\begin{array}{cc}
a_{11}y_{11}+a_{12}y_{21}&a_{11}y_{12}+a_{12}y_{22}\\
y_{21}&y_{22}
\end{array}\right|=
a_{11}\left|\begin{array}{cc}
y_{11}&y_{12}\\y_{21}&y_{22}
\end{array}\right|
+a_{12}\left|\begin{array}{cc}
y_{21}&y_{22}\\y_{21}&y_{22}
\end{array}\right|=a_{11}W+a_{12}0=a_{11}W\).
Similarly,
\(
\left[\begin{array}{cc}
y_{11}&y_{12}\\y_{21}'&y_{22}'
\end{array}\right]=a_{22}W\).

Exercise \(\PageIndex{5}\)

Suppose the \(n\times n\) matrix \(A=A(t)\) is continuous on \((a,b)\). Let

\begin{eqnarray*}
Y = \left[\begin{array} \\ 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{eqnarray*}

where the columns of \(Y\) are solutions of \({\bf y}'=A(t){\bf y}\). Let

\begin{eqnarray*}
r_i = [y_{i1} \, y_{i2} \, \dots \, y_{in}]
\end{eqnarray*}

be the \(i\)th row of \(Y\), and let \(W\) be the determinant of \(Y\).

(a) Deduce from the definition of determinant that

\begin{eqnarray*}
W' = W_1 + W_2 + \cdots + W_n,
\end{eqnarray*}

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 \(i=m\).

(b) Use the equation \(Y'=A Y\) and the definition of matrix multiplication to show that

\begin{eqnarray*}
r'_m = a_{m1} r_1 + a_{m2} r_2 + \cdots + a_{mn} r_n.
\end{eqnarray*}

(c) Use properties of determinants to deduce from part (b) that

\begin{eqnarray*}
\det (W_m) = a_{mm}W.
\end{eqnarray*}

(d) Conclude from part (a) and part (c) that

\begin{eqnarray*}
W' = (a_{11} + a_{22} + \cdots + a_{nn}W,
\end{eqnarray*}

and use this to show that if \(a<t_0<b\) then

\begin{eqnarray*}
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{eqnarray*}

Exercise \(\PageIndex{6}\)

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 \((a,b)\).

(a) Show that \(Y(t_0)\) is invertible.

(b) Show that if \({\bf k}\) is an arbitrary \(n\)-vector then the solution of the initial value problem

\begin{eqnarray*}
{\bf y}' = A(t){\bf y}, \quad {\bf y}(t_0) = {\bf k}
\end{eqnarray*}

is

\begin{eqnarray*}
{\bf y} = Y(t)Y^{-1}(t_0){\bf k}.
\end{eqnarray*}

Answer

\part{a} From  the equivalence of Theorem~4.3.3\part{b}and \part{e},
\(Y(t_0)\) is invertible.

\part{b} From  the equivalence of Theorem~10.3.3\part{a}and \part{b},
the solution of the initial value problem is \({\bf y}=Y(t){\bf
c}\), where \({\bf c}\) is a constant vector. To satisfy \({\bf
y}(t_0)={\bf k}\), we must have
\(Y(t_0){\bf c}={\bf k}\), so \({\bf c}=Y^{-1}(t_0){\bf k}\)
and \({\bf y}=Y^{-1}(t_0)Y(t){\bf k}\).

Exercise \(\PageIndex{7}\)

Let

\begin{eqnarray*}
A = \left[ \begin{array} \\ 2 & 4 \\ 4 & 2 \end{array} \right] \quad {\bf y}_1 = \left[ \begin{array} \\ e^{6t} \\ e^{6t} \end{array} \right], \quad {\bf y}_2 = \left[ \begin{array} \\ e^{-2t} \\ -e^{-2t} \end{array} \right], \quad {\bf k} = \left[ \begin{array} \\ -3 \\ 9 \end{array} \right].
\end{eqnarray*}

(a) Verify that \(\{{\bf y}_1,{\bf y}_2\}\) is a fundamental set of solutions for \({\bf y}'=A{\bf y}\).

(b) Solve the initial value problem

\begin{equation} \label{eq:4.3E.3}
{\bf y}' = A{\bf y}, \quad {\bf y}(0) = {\bf k}.
\end{equation}

(c) Use the result of Exercise \((4.3E.6)\) part (b) to find a formula for the solution of \eqref{eq:4.3E.3} for an arbitrary initial vector \({\bf k}\).

Exercise \(\PageIndex{8}\)

Repeat Exercise \((4.3E7)\) with

\begin{eqnarray*}
A = \left[ \begin{array} \\ -2 & -2 \\ -5 & 1 \end{array} \right], \quad {\bf y}_1 = \left[ \begin{array} \\ e^{-4t} \\ e^{-4t} \end{array} \right], \quad {\bf y}_2 = \left[ \begin{array} \\ -2e^{3t} \\ 5e^{3t} \end{array} \right], \quad {\bf k} = \left[ \begin{array} \\ 10 \\ -4 \end{array} \right].
\end{eqnarray*}

Answer

Add texts here. Do not delete this text first.

Exercise \(\PageIndex{9}\)

Repeat Exercise \((4.3E.7)\) with

\begin{eqnarray*}
A = \left[ \begin{array} \\ -4 & -10 \\ 3 & 7 \end{array} \right], \quad {\bf y}_1 = \left[ \begin{array} \\ -5e^{2t} \\ 3e^{2t} \end{array} \right], \quad {\bf y}_2 = \left[ \begin{array} \\ 2e^t \\ -e^t \end{array} \right], \quad {\bf k} = \left[ \begin{array} \\ -19 \\ 11 \end{array} \right].
\end{eqnarray*}

Exercise \(\PageIndex{10}\)

Repeat Exercise \((4.3E.7)\) with

\begin{eqnarray*}
A = \left[ \begin{array} \\ 2 & 1 \\ 1 & 2 \end{array} \right], \quad {\bf y}_1 = \left[ \begin{array} \\ e^{3t} \\ e^{3t} \end{array} \right], \quad {\bf y}_2 = \left[ \begin{array} \\ e^t \\ -e^t \end{array} \right], \quad {\bf k} = \left[ \begin{array} \\ 2 \\ 8 \end{array} \right].
\end{eqnarray*}

Answer

\part{b}
\({\bf y}_1=c_1\left[\begin{array}{r} e^{3t} \\e^{3t}
\end{array}\right]+c_2\left[\begin{array}{r}e^t \\
-e^t\end{array}\right]\),
where
\(\begin{array}{rcl}
c_1+c_2&=&2\\c_1-c_2&=&8
\end{array},
\)
 so \(c_1=5\), \(c_2=-3\), and
\({\bf y}=\sisplaystyle\left[\begin{array}{c}5e^{3t}-3e^t\\5e^{3t}+3e^t
\end{array}\right]\).

\part{c} \(Y(t)=\left[\begin{array}{cr}e^{3t}&e^t\\
3e^{3t}&-e^t\end{array}\right]\);
\(Y(0)=\left[\begin{array}{rr}1&1\\1&-1\end{array}\right]\);

\(Y^{-1}(0)=
\sisplaystyle{1\over2}\left[\begin{array}{cr}1&1\\1&-1\end{array}\right]\);
\({\bf y}=Y(t)Y^{-1}(0){\bf k}=
\sisplaystyle{1\over2}\left[\begin{array}{cc}e^{3t}+e^t&e^{3t}-e^t
\\e^{3t}-e^t&e^{3t}+e^t\end{array}\right]{\bf k}\).

Exercise \(\PageIndex{11}\)

Let

\begin{eqnarray*}
A &=& \left[ \begin{array} \\ 3 & -1 & -1 \\ -2 & 3 & 2 \\ 4 & -1 & -2 \end{array} \right], \\ {\bf y}_1 &=& \left[ \begin{array} \\ e^{2t} \\ 0 \\ e^{2t} \end{array} \right], \quad {\bf y}_2 = \left[ \begin{array} \\ e^{3t} \\ -e^{3t} \\ e^{3t} \end{array} \right], \quad {\bf y}_3 = \left[ \begin{array} \\ e^{-t} \\ -3e^{-t} \\ 7e^{-t} \end{array} \right], \quad {\bf k} = \left[ \begin{array} \\ 2 \\ -7 \\ 20 \end{array} \right].
\end{eqnarray*}

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

(b) Solve the initial value problem

\begin{equation} \label{eq:4.3E.4}
{\bf y}' = A{\bf y}, \quad {\bf y}(0) = {\bf k}.
\end{equation}

(c) Use the result of Exercise \((4.3E.6)\) part (b) to find a formula for the solution of \eqref{eq:4.3E.4} for an arbitrary initial vector \({\bf k}\).

Exercise \(\PageIndex{12}\)

Repeat Exercise \((4.3E.11)\) with

\begin{eqnarray*}
A &=& \left[ \begin{array} \\ 0 & 2 & 2 \\ 2 & 0 & 2 \\ 2 & 2 & 0 \end{array} \right], \\ {\bf y}_1 &=& \left[ \begin{array} \\ -e^{-2t} \\ 0 \\ e^{-2t} \end{array} \right], \quad {\bf y}_2 = \left[ \begin{array} \\ -e^{-2t} \\ e^{-2t} \\ 0 \end{array} \right], \quad {\bf y}_3 = \left[ \begin{array} \\ e^{4t} \\ e^{4t} \\ e^{4t} \end{array} \right], \quad {\bf k} = \left[ \begin{array} \\ 0 \\ -9 \\ 12 \end{array} \right].
\end{eqnarray*}

Answer

\part{b}
\({\bf y}=c_1\left[\begin{array}{c}-e^{-2t} \\ 0 \\ e^{-2t}
\end{array}\right]+c_2
\left[\begin{array}{c}-e^{-2t} \\ e^{-2t} \\
0\end{array}\right]+
c_3\left[\begin{array}{c} e^{4t} \\ e^{4t} \\ e^{4t}\end{array}
\right]\),
where
\(\begin{array}{rcr}
-c_1-c_2+c_3&=&0\\c_2+c_3&=&-9\\c_1+c_3&=&12
\end{array},
\)
 so \(c_1=11\), \(c_2=-10\), \(c_3=1\), and
\({\bf y
}=\sisplaystyle{1\over3}\left[\begin{array}{c}-e^{-2t}+e^{4t}\\-10e^{-2t}+e^{4t}\\
11e^{-2t}+e^{4t}\end{array}\right]\).

\part{c} \(Y(t)=
\left[\begin{array}{ccc}-e^{-2t}&-e^{-2t}&e^{4t}\\
0&e^{-2t}&e^{4t}\\e^{-2t}&0&e^{4t}\end{array}\right]\);
\(Y(0)=
\left[\begin{array}{rrr}-1&-1&1\\0&1&1\\1&0&1\end{array}\right]\);
\(Y^{-1}(0)=
\sisplaystyle{1\over3}\left[\begin{array}{rrr}-1&-1&2\\-1&2&-1
\\1&1&1\end{array}\right]\);
\({\bf y}=Y(t)Y^{-1}(0){\bf k}=
\sisplaystyle{1\over3}\left[\begin{array}{ccc}2e^{-2t}+e^{4t}&-e^{-2t}+e^{4t}
&-e^{-2t}+e^{4t}
\\-e^{-2t}+e^{4t}&2e^{-2t}+e^{4t}&-e^{-2t}+e^{4t}\\
-e^{-2t}+e^{4t}&-e^{-2t}+e^{4t}&2e^{-2t}+e^{4t}
\end{array}\right]{\bf k}\).
 

Exercise \(\PageIndex{13}\)

Repeat Exercise \((4.3E.11)\) with

\begin{eqnarray*}
A &=& \left[ \begin{array} \\ -1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -2 \end{array} \right], \\ {\bf y}_1 &=& \left[ \begin{array} \\ e^t \\ e^t \\ 0 \end{array} \right], \quad {\bf y}_2 = \left[ \begin{array} \\ e^{-t} \\ 0 \\ 0 \end{array} \right], \quad {\bf y}_3 = \left[ \begin{array} \\ e^{-2t} \\ -2e^{-2t} \\ e^{-2t} \end{array} \right], \quad {\bf k} = \left[ \begin{array} 5 \\ 5 \\ -1 \end{array} \right].
\end{eqnarray*}

Exercise \(\PageIndex{14}\)

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.

Answer

If \(Y\)  and \(Z\) are both fundamental matrices for \({\bf y}'=A(t){\bf y
}\), then \(Z=CY\), where \(c\) is a constant invertible matrix. Therefore,
\(ZY^{-1}=C\) and \(YZ^{-1}=C^{-1}\).

Exercise \(\PageIndex{15}\)

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.

(a) Show that the matrix \(Z=YC\) satisfies the differential equation \(Z'=AZ\).

(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 \({\bf y}'=A(t){\bf y}\).

Exercise \(\PageIndex{16}\)

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

\begin{eqnarray*}
{\bf e}_1 = \left[ \begin{array} \\ 1 \\ 0 \\ \vdots \\ 0 \end{array} \right], \quad {\bf e}_2 = \left[ \begin{array} \\ 0 \\ 1 \\ \vdots \\ 0 \end{array} \right], \quad \cdots \quad {\bf e}_n = \left[ \begin{array} \\ 0 \\ 0 \\ \vdots \\ 1 \end{array} \right];
\end{eqnarray*}

that is, the \(j\)th component of \({\bf e}_i\) is \(1\) if \(j=i\), or \(0\) if \(j\ne i\).

(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 \((a,b)\).

(b) Conclude from part (a) and Exercise \((4.3E.15)\) that \({\bf y}'=A(t){\bf y}\) has infinitely many fundamental sets of solutions on \((a,b)\).

Answer

\part{a}
The Wronskian  of \(\{{\bf y}_1,{\bf y}_2,\dots,{\bf y}_n\}\) equals
one when \(t=t_0\). Apply Theorem~10.3.3.

\part{b}
Let \(Y\) be the matrix with columns \(\{{\bf y}_1,{\bf y}_2,\dots,{\bf
y}_n\}\). From \part{a}, \(Y\) is a fundamental matrix for \({\bf
y}'=A(t){\bf y}\) on \)(a,b)\). From Exercise~10.3.15\part{b},
so is \(Z=YC\) if \(c\) is any invertible constant matrix.

Exercise \(\PageIndex{17}\)

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 \((4.2E.11)\).

Exercise \(\PageIndex{18}\)

Let \(Z\) be the fundamental matrix for the constant coefficient system \({\bf y}'=A{\bf y}\) such that \(Z(0)=I\).

(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 \((4.2.1)\) that \(\Gamma_1=\Gamma_2\).

(b) Show that \((Z(t))^{-1}=Z(-t)\).

(c) The matrix \(Z\) defined above is sometimes denoted by \(e^{tA}\). Discuss the motivation for this notation.

Answer

\part{a}
\(\Gamma_1'(t)=Z'(t)Z(s)=AZ(t)Z(s)=A\Gamma(t)\) and \(\Gamma_1(0)=Z(s)\),
since \(Z(0)=I\).
\(\Gamma_2'(t)=Z'(t+s)=AZ(t+s)=A\Gamma_2(t)\) (since \(a\) is constant)
and \(\Gamma_2(0)=Z(s)\). Applying Theorem~10.2.1  to the columns of
\(\Gamma_1\) and \(\Gamma_2\)  shows that \(\Gamma_1=\Gamma_2\).

\part{b} With \(s=-t\), \part{a} implies that \(Z(t)Z(-t)=Z(0)=I\);
therefore \((Z(t))^{-1}=Z(-t)\).

\part{c} \(e^{0\cdot A}=I\) is analogous to \(e^{o\cdot a}=e^{0}=1\) when
\(a\) is a scalar, while \(e^{(t+s)A}=e^{tA}e^{sA}\)  is analagous
to \(e^{(t+s)a}=e^{ta}e^{sa}\) when \(a\) is a scalar.