Astha Jain

2022-07-02

user_27qwe

To find a fundamental matrix of the given system and apply it to find a solution satisfying the initial conditions, we can follow the steps below:
Step 1: Writing the system in matrix form
The given system can be written in matrix form as:
${x}^{\prime }=\left[\begin{array}{ccc}5& 0& -6\\ 1& -1& -1\\ 2& -2& -2\end{array}\right]x$
where $x$ is the column vector $\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right]$.
Step 2: Finding the eigenvalues and eigenvectors
To find the fundamental matrix, we need to find the eigenvalues and eigenvectors of the coefficient matrix.
The characteristic equation is given by:
$\text{det}\left(A-\lambda I\right)=0$
where $A$ is the coefficient matrix, $\lambda$ is the eigenvalue, and $I$ is the identity matrix.
Substituting the given coefficient matrix $A$, we have:
$\text{det}\left(\left[\begin{array}{ccc}5& 0& -6\\ 1& -1& -1\\ 2& -2& -2\end{array}\right]-\lambda \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\right)=0$
Simplifying the determinant equation, we get:
$\text{det}\left[\begin{array}{ccc}5-\lambda & 0& -6\\ 1& -1-\lambda & -1\\ 2& -2& -2-\lambda \end{array}\right]=0$
Expanding the determinant along the first row, we have:
$\left(5-\lambda \right)\left(|\begin{array}{cc}-1-\lambda & -1\\ -2& -2-\lambda \end{array}|\right)-0+\left(-6\right)\left(|\begin{array}{cc}1& -1-\lambda \\ 2& -2\end{array}|\right)=0$
Simplifying further, we obtain:
$\left(5-\lambda \right)\left({\lambda }^{2}+3\lambda +2\right)-6\left(-2-\lambda \right)=0$
Expanding and rearranging the equation, we get:
${\lambda }^{3}+6{\lambda }^{2}+9\lambda +4=0$
Factoring the equation, we find that $\lambda =-1$ is a repeated root. So the characteristic equation becomes:
$\left(\lambda +1{\right)}^{2}\left(\lambda +4\right)=0$
Hence, the eigenvalues are ${\lambda }_{1}=-1$ (with algebraic multiplicity 2) and ${\lambda }_{2}=-4$.
Next, we need to find the eigenvectors corresponding to these eigenvalues.
For ${\lambda }_{1}=-1$, we solve the equation $\left(A-{\lambda }_{1}I\right){v}_{1}=0$:
$\left[\begin{array}{ccc}6& 0& -6\\ 1& 0& -1\\ 2& 0& -1\end{array}\right]\left[\begin{array}{c}{v}_{11}\\ {v}_{21}\\ {v}_{31}\end{array}\right]=0$
Row reducing the augmented matrix, we get:
$\left[\begin{array}{ccc}1& 0& -1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}{v}_{11}\\ {v}_{21}\\ {v}_{31}\end{array}\right]=0$
From the reduced row-echelon form, we can see that ${v}_{21}$ and ${v}_{31}$ are free variables. Let ${v}_{21}=t$ and ${v}_{31}=s$ (where $t$ and $s$ are arbitrary constants).
Therefore, the eigenvector corresponding to ${\lambda }_{1}=-1$ is:
${v}_{1}=\left[\begin{array}{c}{v}_{11}\\ {v}_{21}\\ {v}_{31}\end{array}\right]=\left[\begin{array}{c}s\\ t\\ s\end{array}\right]$
For ${\lambda }_{2}=-4$, we solve the equation $\left(A-{\lambda }_{2}I\right){v}_{2}=0$:
$\left[\begin{array}{ccc}9& 0& -6\\ 1& 3& -1\\ 2& 2& 2\end{array}\right]\left[\begin{array}{c}{v}_{12}\\ {v}_{22}\\ {v}_{32}\end{array}\right]=0$
Row reducing the augmented matrix, we get:
$\left[\begin{array}{ccc}1& 0& -\frac{2}{3}\\ 0& 1& -\frac{1}{3}\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}{v}_{12}\\ {v}_{22}\\ {v}_{32}\end{array}\right]=0$
From the reduced row-echelon form, we can see that ${v}_{12}$, ${v}_{22}$, and ${v}_{32}$ are all free variables. Let ${v}_{12}=p$, ${v}_{22}=q$, and ${v}_{32}=r$ (where $p$, $q$, and $r$ are arbitrary constants).
Therefore, the eigenvector corresponding to ${\lambda }_{2}=-4$ is:
${v}_{2}=\left[\begin{array}{c}{v}_{12}\\ {v}_{22}\\ {v}_{32}\end{array}\right]=\left[\begin{array}{c}p\\ q\\ r\end{array}\right]$
Step 3: Constructing the fundamental matrix
The fundamental matrix $\Phi \left(t\right)$ is constructed using the eigenvectors ${v}_{1}$ and ${v}_{2}$ corresponding to the eigenvalues ${\lambda }_{1}$ and ${\lambda }_{2}$, respectively.
We have:
$\Phi \left(t\right)=\left[\begin{array}{cc}{v}_{1}& {v}_{2}\end{array}\right]=\left[\begin{array}{cc}s& p\\ t& q\\ s& r\end{array}\right]$
Step 4: Applying the initial condition
To find a solution satisfying the initial condition $x\left(0\right)=\left[\begin{array}{c}1\\ 0\\ 2\end{array}\right]$, we use the formula:
$x\left(t\right)=\Phi \left(t\right)\Phi \left(0{\right)}^{-1}{x}_{0}$
We are given $x\left(0\right)=\left[\begin{array}{c}1\\ 0\\ 2\end{array}\right]$, so we substitute the values into the formula:
$x\left(t\right)=\left[\begin{array}{cc}s& p\\ t& q\\ s& r\end{array}\right]{\left[}^{\begin{array}{cc}{v}_{1}& {v}_{2}\end{array}}\left[\begin{array}{c}1\\ 0\\ 2\end{array}\right]$
To find the inverse of the matrix $\left[\begin{array}{cc}{v}_{1}& {v}_{2}\end{array}\right]$, we can use the formula for the inverse of a $2×2$ matrix:
${\left[}^{\begin{array}{cc}a& b\\ c& d\end{array}}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$
Applying this formula, we find:
${\left[}^{\begin{array}{cc}{v}_{1}& {v}_{2}\end{array}}=\frac{1}{sq-pt}\left[\begin{array}{cc}q& -p\\ -r& s\end{array}\right]$
Substituting this back into the formula for $x\left(t\right)$, we get:
$x\left(t\right)=\left[\begin{array}{cc}s& p\\ t& q\\ s& r\end{array}\right]\frac{1}{sq-pt}\left[\begin{array}{cc}q& -p\\ -r& s\end{array}\right]\left[\begin{array}{c}1\\ 0\\ 2\end{array}\right]$
Simplifying the expression, we have:
$x\left(t\right)=\frac{1}{sq-pt}\left[\begin{array}{cc}sq+p\left(-r\right)& s\left(-p\right)+p\left(s\right)\\ tq+q\left(-r\right)& t\left(-p\right)+q\left(s\right)\\ sq+p\left(2r\right)& s\left(-p\right)+p\left(2s\right)\end{array}\right]\left[\begin{array}{c}1\\ 0\\ 2\end{array}\right]$
Performing the matrix multiplication, we get:
$x\left(t\right)=\frac{1}{sq-pt}\left[\begin{array}{c}3p-2r\\ t\left(q-p\right)\\ 2p+4r\end{array}\right]$
Therefore, the solution satisfying the initial condition $x\left(0\right)=\left[\begin{array}{c}1\\ 0\\ 2\end{array}\right]$ is given by:
$x\left(t\right)=\frac{1}{sq-pt}\left[\begin{array}{c}3p-2r\\ t\left(q-p\right)\\ 2p+4r\end{array}\right]$

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