matrix - 6

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

[imath]\bold{X}' = \displaystyle \begin{pmatrix} \phantom{-}2 & \phantom{-}5 & \phantom{-}1 \\[5pt] -5 & -6 & \phantom{-}4 \\[5pt] \phantom{-}0 & \phantom{-}0 & \phantom{-}2 \end{pmatrix}\bold{X}[/imath]

 

[khansaheb]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

[imath]\bold{X}' = \displaystyle \begin{pmatrix} \phantom{-}2 & \phantom{-}5 & \phantom{-}1 \\[5pt] -5 & -6 & \phantom{-}4 \\[5pt] \phantom{-}0 & \phantom{-}0 & \phantom{-}2 \end{pmatrix}\bold{X}[/imath]

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem

 

[logistic_guy]

We start by finding the determinant of the matrix \(\displaystyle \bold{A} - \lambda\bold{I}\) and set it equal to zero. In other words, we will find the eigenvalues.

\(\displaystyle \left| \begin{array}{ccc} \phantom{-}2 - \lambda & \phantom{-}5 & \phantom{-}1\\ -5 & -6 - \lambda & \phantom{-}4\\ \phantom{-}0 & \phantom{-}0 & \phantom{-}2 - \lambda\end{array} \right| = 0\)


\(\displaystyle (2 - \lambda)\bigg[(-6 - \lambda)(2 - \lambda) - (4)(0)\bigg] - 5\bigg[(-5)(2 - \lambda) - (4)(0)\bigg] + \bigg[(-5)(0) - (-6 - \lambda)(0)\bigg] = 0\)


\(\displaystyle (2 - \lambda)\bigg[(-6 - \lambda)(2 - \lambda)\bigg] - 5\bigg[(-5)(2 - \lambda)\bigg] = 0\)


\(\displaystyle (2 - \lambda)(-6 - \lambda)(2 - \lambda) + 25(2 - \lambda) = 0\)


\(\displaystyle (2 - \lambda)(\lambda^2 + \lambda + 13) = 0\)

This gives:

\(\displaystyle \lambda_1 = 2\)
\(\displaystyle \lambda_2 = -2 + 3\text{i}\)
\(\displaystyle \lambda_3 = -2 - 3\text{i}\)

 

[logistic_guy]

Now we are going to find \(\displaystyle \bold{K}_1\).

We have this system to solve:

\(\displaystyle 5k_2 + k_3 = 0\)
\(\displaystyle -5k_1 - 8k_2 + 4k_3 = 0\)

We can choose anything for \(\displaystyle k_1, k_2\), and \(\displaystyle k_3\) that satisfies these two equations. But we will try to choose values for \(\displaystyle k\)s that are not fractions. In this way the solution will be more beautiful \(\displaystyle \longrightarrow\) more organized.

If we rearrange the second equation, we get:

\(\displaystyle k_1 = -\frac{8}{5}k_2 + \frac{4}{5}k_3\)

If we look at the first equation, I can choose \(\displaystyle k_2 = 1\) which gives \(\displaystyle k_3 = -5\) but this will make \(\displaystyle k_1\) as a fraction.

The best choice is to choose \(\displaystyle k_2 = -5\). This makes \(\displaystyle k_3 = 25\).

Then,

\(\displaystyle k_1 = -\frac{8}{5}(-5) + \frac{4}{5}(25) = 28\)

This gives:

[imath]\displaystyle \bold{K}_1 = \begin{pmatrix}\phantom{-}28 \\[5pt] -5 \\[5pt] \phantom{-}25 \end{pmatrix}[/imath]

 

[logistic_guy]

Let us try to find \(\displaystyle \bold{K}_2\).

We have this system to solve:

\(\displaystyle (4 - 3i)k_1 + 5k_2 + k_3 = 0\)
\(\displaystyle -5k_1 - (4 + 3i)k_2 + 4k_3 = 0\)
\(\displaystyle (4 - 3i)k_3 = 0\)

This gives:

\(\displaystyle k_3 = 0\)

If we solve the second equation for \(\displaystyle k_1\), we get:

\(\displaystyle k_1 = -\frac{4 + 3i}{5}k_2\)

To get rid of fraction, we can choose:

\(\displaystyle k_2 = -5\)

Then,

\(\displaystyle k_1 = 4 + 3i\)


[imath]\displaystyle \bold{K}_2 = \begin{pmatrix}4 + 3i \\[5pt] -5 \\[5pt] \phantom{-}0 \end{pmatrix}[/imath]

 

[logistic_guy]

We continue.

[imath]\displaystyle \bold{K}_2 = \begin{pmatrix}\phantom{-}4\\[5pt] -5 \\[5pt] \phantom{-}0 \end{pmatrix} + i\begin{pmatrix}3\\[5pt] 0 \\[5pt] 0 \end{pmatrix}[/imath]

 

[logistic_guy]

[imath]\displaystyle \bold{K}_2 = \begin{pmatrix}\phantom{-}4\\[5pt] -5 \\[5pt] \phantom{-}0 \end{pmatrix} + i\begin{pmatrix}3\\[5pt] 0 \\[5pt] 0 \end{pmatrix}[/imath]

We will use the following Theorem \(\displaystyle (\textcolor{blue}{\bold{Theorem}} \textcolor{red}{\bold{69}})\):

\(\displaystyle \bold{X}_1 = (\bold{B}_1\cos \beta t - \bold{B}_2\sin \beta t)e^{\alpha t}\)
\(\displaystyle \bold{X}_2 = (\bold{B}_2\cos \beta t + \bold{B}_1\sin \beta t)e^{\alpha t}\)

where \(\displaystyle \bold{B}_1\) and \(\displaystyle \bold{B}_2\) are the real and imaginary parts of \(\displaystyle \bold{K}_1\) respectively. And \(\displaystyle \alpha\) and \(\displaystyle \beta\) are the real and imaginary parts of \(\displaystyle \lambda_1\) respectively, where \(\displaystyle \lambda_1 = \alpha + i\beta\) is the eigenvalue of \(\displaystyle \bold{K}_1\).

For our purpose we are finding \(\displaystyle \bold{X}_2\) and \(\displaystyle \bold{X}_3\) and we are working with \(\displaystyle \lambda_2\) and \(\displaystyle \bold{K}_2\).

Then,

[imath]\displaystyle \bold{X}_2 = \begin{pmatrix}4\cos 3t - 3\sin 3t\\[5pt] -5\cos 3t \\[5pt] \phantom{-}0 \end{pmatrix}e^{-2t}[/imath]


[imath]\displaystyle \bold{X}_3 = \begin{pmatrix}3\cos 3t + 4\sin 3t\\[5pt] -5\sin 3t \\[5pt] \phantom{-}0 \end{pmatrix}e^{-2t}[/imath]

 

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{FINALLY}}\)

The general solution is:

\(\displaystyle \bold{X} = c_1\bold{X}_1 + c_2\bold{X}_2 + c_3\bold{X}_3\)

Or

[imath]\displaystyle \bold{X} = \textcolor{red}{c_1}\begin{pmatrix}\phantom{-}28 \\[5pt] -5 \\[5pt] \phantom{-}25 \end{pmatrix}e^{2t} + \textcolor{blue}{c_2}\begin{pmatrix}4\cos 3t - 3\sin 3t\\[5pt] -5\cos 3t \\[5pt] \phantom{-}0\end{pmatrix}e^{-2t} + \textcolor{green}{c_3}\begin{pmatrix}3\cos 3t + 4\sin 3t\\[5pt] -5\sin 3t \\[5pt] \phantom{-}0 \end{pmatrix}e^{-2t}[/imath]