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Question

# Construct the general solution of x'=Ax involving complex eigenfunctions and then obtain the general real solution. Describe the shapes of typical trajectories.$A=\left[ \begin{array}{rrr}{30} & {64} & {23} \\ {-11} & {-23} & {-9} \\ {6} & {15} & {4}\end{array}\right]$

Solution

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Let

$A=\left[\begin{array}{rrr}{30} & {64} & {23} \\ {-11} & {-23} & {-9} \\ {6} & {15} & {4}\end{array}\right]$

Then

$(A-xI_3)=\begin{bmatrix} \phantom{-}x-30 & {-64} & {-23} \\ {\phantom{-}11} & {\phantom{-}x+23} & {\phantom{-}9} \\ {-6} & {-15} & {\phantom{-}x-4} \end{bmatrix}$

Let characteristic polynomial of $A$ be $p(x)$. Then

\begin{align*} p(x) &=\text{ det}(xI_3-A)\\ &=x^3-11x^2+39x-29 \end{align*}

Thus the characteristic polynomial is $x^3-11x^2+39x-29$.

Now $x^3-11x^2+39x-29=(x-1)[x-(5-2i)][x-(5+2i)]$. This implies that the roots of $p(x)$ are $1,5-2i,5+2i$. Hence the eigenvalues of $A$ are $1,5-2i,5+2i$.

Now let

$X_1=\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$

be an eigenvector corresponding to the eigenvalue $5+2i$. Therefore $AX_1=(5+2i)X_1$ which means $[(5+2i)I_3-A]X_1=0$. From this equation we get that

$\begin{bmatrix} -25+2i & {-64} & {-23} \\ {\phantom{-}11} & {\phantom{-}28+2i} & {\phantom{-}9} \\ {-6} & {-15} & {-1-2i} \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

By Gaussian Elimination, we have $x_1=\frac{23-34i}{3}x_3$ and $x_2=\frac{-9+14i}{3}x_3$. Therefore

$\begin{bmatrix} (23-34i)r\\ (-9+14i)r\\ 3r \end{bmatrix}$

is an eigenvector corresponding to the eigenvalue $5+2i$ for any non-zero real number $r$. By putting $r=1$, we get that

$\mathbf{v_1}=\begin{bmatrix} 23-34i\\ -9+14i\\ 3 \end{bmatrix}$

is an eigenvector corresponding to the eigenvalue $5+2i$.

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