matrix - 4

[logistic_guy]

Solve.

\(\displaystyle \bold{X}' = \begin{bmatrix}0 & 6 & 0 \\1 & 0 & 1 \\ 1 & 1 & 0\end{bmatrix}\bold{X}\)

 

[logistic_guy]

\(\displaystyle \begin{vmatrix}-\lambda & 6 & 0 \\1 & -\lambda & 1 \\1 & 1 & -\lambda \\\end{vmatrix} = -\lambda \bigg[(-\lambda)(-\lambda) - (1)(1)\bigg] - 6\bigg[(1)(-\lambda) - (1)(1)\bigg] + 0\bigg[(1)(1) - (-\lambda)(1)\bigg] = 0\)


\(\displaystyle -\lambda \bigg[\lambda^2 - 1\bigg] - 6\bigg[-\lambda - 1\bigg] = 0\)


\(\displaystyle -\lambda^3 + \lambda + 6\lambda + 6 = 0\)


\(\displaystyle \lambda^3 - 7\lambda - 6 = 0\)

At this point, I need to sharpen my skills in \(\displaystyle \textcolor{indigo}{\text{algebra}}\). One way to solve this equation is by trial and error. We will try this, if it didn't work, we will try something else.

When \(\displaystyle \lambda = 1\), I don't get zero.

When \(\displaystyle \lambda = -1\), I get zero

Then,

\(\displaystyle (\lambda + 1)\) is a factor. And when I use long division I get:

\(\displaystyle (\lambda + 1)(\lambda + 2)(\lambda - 3) = 0\)

This gives:

\(\displaystyle \lambda_1 = -1\)
\(\displaystyle \lambda_2 = -2\)
\(\displaystyle \lambda_3 = 3\)

 

[logistic_guy]

For \(\displaystyle \lambda_1 = -1, \)we have:

\(\displaystyle k_1 + 6k_2 = 0\)
\(\displaystyle k_1 + k_2 + k_3 = 0\)

This gives:

\(\displaystyle \bold{X}_1 = c_1\begin{bmatrix}k_1 \\k_2 \\ k_3\end{bmatrix}e^{\lambda_1 t} = c_1\begin{bmatrix}6 \\-1 \\ -5\end{bmatrix}e^{-t}\)

 

[logistic_guy]

For \(\displaystyle \lambda_2 = -2, \)we have:

\(\displaystyle 2k_1 + 6k_2 = 0\)
\(\displaystyle k_1 + 2k_2 + k_3 = 0\)
\(\displaystyle k_1 + k_2 + 2k_3 = 0\)

This gives:

\(\displaystyle \bold{X}_2 = c_2\begin{bmatrix}k_1 \\k_2 \\ k_3\end{bmatrix}e^{\lambda_2 t} = c_2\begin{bmatrix}3 \\-1 \\ -1\end{bmatrix}e^{-2t}\)

 

[logistic_guy]

For \(\displaystyle \lambda_3 = 3, \)we have:

\(\displaystyle -3k_1 + 6k_2 = 0\)
\(\displaystyle k_1 - 3k_2 + k_3 = 0\)
\(\displaystyle k_1 + k_2 - 3k_3 = 0\)

This gives:

\(\displaystyle \bold{X}_3 = c_3\begin{bmatrix}k_1 \\k_2 \\ k_3\end{bmatrix}e^{\lambda_3 t} = c_3\begin{bmatrix}2 \\1 \\ 1\end{bmatrix}e^{3t}\)

 

[logistic_guy]

Solve.

\(\displaystyle \bold{X}' = \begin{bmatrix}0 & 6 & 0 \\1 & 0 & 1 \\ 1 & 1 & 0\end{bmatrix}\bold{X}\)

The solution to this system is:

\(\displaystyle \bold{X} = \bold{X}_1 + \bold{X}_2 + \bold{X}_3 = c_1\begin{bmatrix}6 \\-1 \\ -5\end{bmatrix}e^{-t} + c_2\begin{bmatrix}3 \\-1 \\ -1\end{bmatrix}e^{-2t} + c_3\begin{bmatrix}2 \\1 \\ 1\end{bmatrix}e^{3t}\)