Book solution for example 1 is wrong. So I typed corrected solution.
Solve \(\mathbf{\dot{x}}=\begin{pmatrix} 1 & 4\\ 1 & 1 \end{pmatrix} \mathbf{x}+\begin{pmatrix} e^{t}\\ e^{t}\end{pmatrix} \) with \(\mathbf{x}\left ( 0\right ) =\begin{pmatrix} 2\\ 1 \end{pmatrix} \)
Solution\begin{align*} \left ( sI-A\right ) \mathbf{X}\left ( s\right ) & =\mathbf{F}\left ( s\right ) +\mathbf{x}\left ( 0\right ) \\ \left [ \begin{pmatrix} s & 0\\ 0 & s \end{pmatrix} -\begin{pmatrix} 1 & 4\\ 1 & 1 \end{pmatrix} \right ] \begin{pmatrix} x_{1}\left ( s\right ) \\ x_{2}\left ( s\right ) \end{pmatrix} & =\begin{pmatrix} \frac{1}{s-1}\\ \frac{1}{s-1}\end{pmatrix} +\begin{pmatrix} 2\\ 1 \end{pmatrix} \\\begin{pmatrix} s-1 & -4\\ -1 & s-1 \end{pmatrix}\begin{pmatrix} x_{1}\left ( s\right ) \\ x_{2}\left ( s\right ) \end{pmatrix} & =\begin{pmatrix} \frac{1}{s-1}+2\\ \frac{1}{s-1}+1 \end{pmatrix} \end{align*}
Multiplying the second row by \(\left ( s-1\right ) \) and adding the result to the first row to obtain Gaussian elimination. First multiplying second row by \(\left ( s-1\right ) \) gives\begin{align*} \begin{pmatrix} s-1 & -4\\ -\left ( s-1\right ) & \left ( s-1\right ) ^{2}\end{pmatrix}\begin{pmatrix} x_{1}\left ( s\right ) \\ x_{2}\left ( s\right ) \end{pmatrix} & =\begin{pmatrix} \frac{1}{s-1}+2\\ 1+\left ( s-1\right ) \end{pmatrix} \\\begin{pmatrix} s-1 & -4\\ -\left ( s-1\right ) & \left ( s-1\right ) ^{2}\end{pmatrix}\begin{pmatrix} x_{1}\left ( s\right ) \\ x_{2}\left ( s\right ) \end{pmatrix} & =\begin{pmatrix} \frac{1}{s-1}+2\\ s \end{pmatrix} \end{align*}
Now replacing row 2 by row 2 plus row 1 gives\begin{align} \begin{pmatrix} s-1 & -4\\ 0 & \left ( s-1\right ) ^{2}-4 \end{pmatrix}\begin{pmatrix} x_{1}\left ( s\right ) \\ x_{2}\left ( s\right ) \end{pmatrix} & =\begin{pmatrix} \frac{1}{s-1}+2\\ s+\left ( \frac{1}{s-1}+2\right ) \end{pmatrix} \nonumber \\\begin{pmatrix} s-1 & -4\\ 0 & s^{2}-2s-3 \end{pmatrix}\begin{pmatrix} x_{1}\left ( s\right ) \\ x_{2}\left ( s\right ) \end{pmatrix} & =\begin{pmatrix} \frac{1}{s-1}+2\\ \frac{1}{s-1}\left ( s^{2}+s-1\right ) \end{pmatrix} \tag{1} \end{align}
Hence\begin{align} x_{2}\left ( s\right ) & =\frac{1}{s-1}\left ( \frac{s^{2}+s-1}{s^{2}-2s-3}\right ) \nonumber \\ & =\frac{s^{2}+s-1}{\left ( s-1\right ) \left ( s-3\right ) \left ( s+1\right ) } \tag{2} \end{align}
Partial fractions:\begin{align*} \frac{s^{2}+s-1}{\left ( s-1\right ) \left ( s-3\right ) \left ( s+1\right ) } & =\frac{A}{\left ( s-1\right ) }+\frac{B}{\left ( s-3\right ) }+\frac{C}{\left ( s+1\right ) }\\ A & =\left ( \frac{s^{2}+s-1}{\left ( s-3\right ) \left ( s+1\right ) }\right ) _{s=1}=\frac{1+1-1}{\left ( 1-3\right ) \left ( 1+1\right ) }=-\frac{1}{4} \end{align*}
And\[ B=\left ( \frac{s^{2}+s-1}{\left ( s-1\right ) \left ( s+1\right ) }\right ) _{s=3}=\frac{9+3-1}{\left ( 3-1\right ) \left ( 3+1\right ) }=\frac{11}{8}\] And\[ C=\left ( \frac{s^{2}+s-1}{\left ( s-1\right ) \left ( s-3\right ) }\right ) _{s=-1}=\frac{1-1-1}{\left ( -1-1\right ) \left ( -1-3\right ) }=-\frac{1}{8}\] Hence \begin{equation} x_{2}\left ( s\right ) =-\frac{1}{4}\frac{1}{\left ( s-1\right ) }+\frac{11}{8}\frac{1}{\left ( s-3\right ) }-\frac{1}{8}\frac{1}{\left ( s+1\right ) } \tag{3} \end{equation} Therefore\[ x_{2}\left ( t\right ) =-\frac{1}{4}e^{t}+\frac{11}{8}e^{3t}-\frac{1}{8}e^{-t}\] Now we go back to (1) and use the first row to find \(x_{1}\left ( s\right ) \) since we know \(x_{2}\left ( s\right ) \) which is given in (2). This results in\begin{align*} \left ( s-1\right ) x_{1}\left ( s\right ) -4x_{2}\left ( s\right ) & =\frac{1}{s-1}+2\\ \left ( s-1\right ) x_{1}\left ( s\right ) & =\frac{1}{s-1}+2+4x_{2}\\ x_{1}\left ( s\right ) & =\frac{1}{\left ( s-1\right ) ^{2}}+\frac{2}{s-1}+\frac{4}{\left ( s-1\right ) }x_{2}\\ & =\frac{1}{\left ( s-1\right ) ^{2}}+\frac{2}{s-1}+\frac{4}{\left ( s-1\right ) }\left ( \frac{s^{2}+s-1}{\left ( s-1\right ) \left ( s-3\right ) \left ( s+1\right ) }\right ) \\ & =\frac{1}{\left ( s-1\right ) ^{2}}+\frac{2}{s-1}+\frac{4}{\left ( s-1\right ) }\left ( -\frac{1}{4}\frac{1}{\left ( s-1\right ) }+\frac{11}{8}\frac{1}{\left ( s-3\right ) }-\frac{1}{8}\frac{1}{\left ( s+1\right ) }\right ) \\ & =\frac{1}{\left ( s-1\right ) ^{2}}+\frac{2}{s-1}-\frac{1}{\left ( s-1\right ) ^{2}}+\frac{44}{8}\frac{1}{\left ( s-1\right ) \left ( s-3\right ) }-\frac{4}{8}\frac{1}{\left ( s-1\right ) \left ( s+1\right ) }\\ & =\frac{2}{s-1}+\frac{44}{8}\frac{1}{\left ( s-1\right ) \left ( s-3\right ) }-\frac{4}{8}\frac{1}{\left ( s-1\right ) \left ( s+1\right ) }\\ & =\frac{2}{s-1}+\frac{11}{2}\frac{1}{\left ( s-1\right ) \left ( s-3\right ) }-\frac{1}{2}\frac{1}{\left ( s-1\right ) \left ( s+1\right ) }\\ & =\frac{2}{s-1}+\frac{11}{2}\left ( \frac{1}{2}\frac{1}{s-3}-\frac{1}{2}\frac{1}{s-1}\right ) -\frac{1}{2}\left ( \frac{1}{2}\frac{1}{s-1}-\frac{1}{2}\frac{1}{s+1}\right ) \\ & =\frac{2}{s-1}+\frac{11}{4}\frac{1}{s-3}-\frac{11}{4}\frac{1}{s-1}-\frac{1}{4}\frac{1}{s-1}+\frac{1}{4}\frac{1}{s+1}\\ & =\frac{-1}{s-1}+\frac{11}{4}\frac{1}{s-3}+\frac{1}{4}\frac{1}{s+1} \end{align*}
Therefore\[ x_{1}\left ( t\right ) =-e^{t}+\frac{11}{4}e^{3t}+\frac{1}{4}e^{-t}\] We see that book solution is wrong. It gives \(x_{1}\left ( t\right ) =2e^{3t}+\frac{1}{2}e^{t}-\frac{1}{2}e^{-t}\).
Solving the same problem, but using the variation of parameters method:
Since \(A=\begin{pmatrix} 1 & 4\\ 1 & 1 \end{pmatrix} \) then\begin{align*} \det \left ( A-\lambda I\right ) & =0\\ \det \begin{pmatrix} 1-\lambda & 4\\ 1 & 1-\lambda \end{pmatrix} & =0\\ \left ( 1-\lambda \right ) ^{2}-4 & =0 \end{align*}
Hence roots are \(\lambda =-1,\lambda =3\)
\(\lambda =-1\)\begin{align*} \begin{pmatrix} 1-\lambda & 4\\ 1 & 1-\lambda \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\end{pmatrix} & =\begin{pmatrix} 0\\ 0 \end{pmatrix} \\\begin{pmatrix} 2 & 4\\ 1 & 2 \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\end{pmatrix} & =\begin{pmatrix} 0\\ 0 \end{pmatrix} \end{align*}
From first row, \(2v_{1}+4v_{2}=0\) or \(v_{1}=-2v_{2}\). Hence \(\mathbf{v}^{1}=\begin{pmatrix} -2\\ 1 \end{pmatrix} \) and \(\mathbf{x}^{1}\left ( t\right ) =e^{-t}\begin{pmatrix} -2\\ 1 \end{pmatrix} \)
\(\lambda =3\)\begin{align*} \begin{pmatrix} 1-\lambda & 4\\ 1 & 1-\lambda \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\end{pmatrix} & =\begin{pmatrix} 0\\ 0 \end{pmatrix} \\\begin{pmatrix} -2 & 4\\ 1 & -2 \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\end{pmatrix} & =\begin{pmatrix} 0\\ 0 \end{pmatrix} \end{align*}
From first row, \(-2v_{1}+4v_{2}=0\) or \(v_{1}=2v_{2}\). Hence \(\mathbf{v}^{1}=\begin{pmatrix} 2\\ 1 \end{pmatrix} \) and \(\mathbf{x}^{2}\left ( t\right ) =e^{3t}\begin{pmatrix} 2\\ 1 \end{pmatrix} \). Therefore \begin{align*} X\left ( t\right ) & =\begin{pmatrix} -2e^{-t} & 2e^{3t}\\ e^{-t} & e^{3t}\end{pmatrix} \\ X\left ( 0\right ) & =\begin{pmatrix} -2 & 2\\ 1 & 1 \end{pmatrix} \end{align*}
Therefore \(X^{-1}\left ( 0\right ) =\frac{adj\left ( X\left ( 0\right ) \right ) }{\det \left ( X\left ( 0\right ) \right ) }=\frac{\begin{pmatrix} 1 & -1\\ -2 & -2 \end{pmatrix} ^{T}}{-4}=-\frac{1}{4}\begin{pmatrix} 1 & -2\\ -1 & -2 \end{pmatrix} =\begin{pmatrix} -\frac{1}{4} & \frac{1}{2}\\ \frac{1}{4} & \frac{1}{2}\end{pmatrix} \). Hence\begin{align*} e^{At} & =X\left ( t\right ) X^{-1}\left ( 0\right ) \\ & =\begin{pmatrix} -2e^{-t} & 2e^{3t}\\ e^{-t} & e^{3t}\end{pmatrix}\begin{pmatrix} -\frac{1}{4} & \frac{1}{2}\\ \frac{1}{4} & \frac{1}{2}\end{pmatrix} \\ & =\begin{pmatrix} \frac{1}{2}e^{-t}+\frac{1}{2}e^{3t} & -e^{-t}+e^{3t}\\ -\frac{1}{4}e^{-t}+\frac{1}{4}e^{3t} & \frac{1}{2}e^{-t}+\frac{1}{2}e^{3t}\end{pmatrix} \end{align*}
Using (since \(t_{0}=0\))\[ \mathbf{x}\left ( t\right ) =e^{At}\mathbf{x}\left ( 0\right ) +e^{At}\int _{0}^{t}e^{-As}f\left ( s\right ) ds \] But \begin{align*} e^{-As}f\left ( s\right ) & =\begin{pmatrix} \frac{1}{2}e^{s}+\frac{1}{2}e^{-3s} & -e^{s}+e^{-3s}\\ -\frac{1}{4}e^{s}+\frac{1}{4}e^{-3s} & \frac{1}{2}e^{s}+\frac{1}{2}e^{-3s}\end{pmatrix}\begin{pmatrix} e^{s}\\ e^{s}\end{pmatrix} \\ & =\begin{pmatrix} \frac{1}{2}e^{2s}+\frac{1}{2}e^{-2s}-e^{2s}+e^{-2s}\\ -\frac{1}{4}e^{2s}+\frac{1}{4}e^{-2s}+\frac{1}{2}e^{2s}+\frac{1}{2}e^{-2s}\end{pmatrix} \\ & =\begin{pmatrix} -\frac{1}{2}e^{2s}+\frac{3}{2}e^{-2s}\\ \frac{1}{4}e^{2s}+\frac{3}{4}e^{-2s}\end{pmatrix} \end{align*}
Integrating\begin{align*} \int _{0}^{t}e^{-As}f\left ( s\right ) ds & =\begin{pmatrix} \int _{0}^{t}-\frac{1}{2}e^{2s}+\frac{3}{2}e^{-2s}ds\\ \int _{0}^{t}\frac{1}{4}e^{2s}+\frac{3}{4}e^{-2s}ds \end{pmatrix} \\ & =\begin{pmatrix} -\frac{1}{4}\left ( e^{2s}\right ) _{0}^{t}-\frac{3}{4}\left ( e^{-2s}\right ) _{0}^{t}\\ \frac{1}{8}\left ( e^{2s}\right ) _{0}^{t}-\frac{3}{8}\left ( e^{-2s}\right ) _{0}^{t}\end{pmatrix} \\ & =\begin{pmatrix} -\frac{1}{4}\left ( e^{2t}-1\right ) -\frac{3}{4}\left ( e^{-2t}-1\right ) \\ \frac{1}{8}\left ( e^{2t}-1\right ) -\frac{3}{8}\left ( e^{-2t}-1\right ) \end{pmatrix} \\ & =\begin{pmatrix} -\frac{1}{4}e^{2t}+\frac{1}{4}-\frac{3}{4}e^{-2t}+\frac{3}{4}\\ \frac{1}{8}e^{2t}-\frac{1}{8}-\frac{3}{8}e^{-2t}+\frac{3}{8}\end{pmatrix} \\ & =\begin{pmatrix} -\frac{1}{4}e^{2t}-\frac{3}{4}e^{-2t}+1\\ \frac{1}{8}e^{2t}-\frac{3}{8}e^{-2t}+\frac{1}{4}\end{pmatrix} \end{align*}
Hence \begin{align*} e^{At}\int _{0}^{t}e^{-As}f\left ( s\right ) ds & =\begin{pmatrix} \frac{1}{2}e^{-t}+\frac{1}{2}e^{3t} & -e^{-t}+e^{3t}\\ -\frac{1}{8}e^{-t}+\frac{1}{4}e^{3t} & \frac{1}{2}e^{-t}+\frac{1}{2}e^{3t}\end{pmatrix}\begin{pmatrix} -\frac{1}{4}e^{2t}-\frac{3}{4}e^{-2t}+1\\ \frac{1}{8}e^{2t}-\frac{3}{8}e^{-2t}+\frac{1}{4}\end{pmatrix} \\ & =\begin{pmatrix} \frac{1}{4}e^{-t}-e^{t}+\frac{3}{4}e^{3t}\\ -\frac{1}{8}e^{-t}-\frac{1}{4}e^{2t}+\frac{3}{8}e^{4t}\end{pmatrix} \end{align*}
And\begin{align*} e^{At}\mathbf{x}\left ( 0\right ) & =\begin{pmatrix} \frac{1}{2}e^{-t}+\frac{1}{2}e^{3t} & -e^{-t}+e^{3t}\\ -\frac{1}{8}e^{-t}+\frac{1}{4}e^{3t} & \frac{1}{2}e^{-t}+\frac{1}{2}e^{3t}\end{pmatrix}\begin{pmatrix} 2\\ 1 \end{pmatrix} \\ & =\begin{pmatrix} 2e^{3t}\\ e^{3t}\end{pmatrix} \end{align*}
Hence\begin{align*} \mathbf{x}\left ( t\right ) & =\begin{pmatrix} 2e^{3t}\\ e^{3t}\end{pmatrix} +\begin{pmatrix} \frac{1}{4}e^{-t}-e^{t}+\frac{3}{4}e^{3t}\\ -\frac{1}{8}e^{-t}-\frac{1}{4}e^{2t}+\frac{3}{8}e^{4t}\end{pmatrix} \\ & =\begin{pmatrix} \frac{1}{4}e^{-t}-e^{t}+\frac{11}{4}e^{3t}\\ -\frac{1}{8}e^{-t}-\frac{1}{4}e^{2t}+\frac{11}{8}e^{4t}\end{pmatrix} \end{align*}
Therefore\begin{align*} x_{1}\left ( t\right ) & =\frac{1}{4}e^{-t}-e^{t}+\frac{11}{4}e^{3t}\\ x_{2}\left ( t\right ) & =-\frac{1}{8}e^{-t}-\frac{1}{4}e^{2t}+\frac{11}{8}e^{4t} \end{align*}
Which agrees with result using Laplace transform method.