matrix

[logistic_guy]

Solve.

\(\displaystyle \bold{X}' = \begin{bmatrix}1 & 3 \\5 & 3 \end{bmatrix} \bold{X}\)

 

[pka]

Solve.\(\displaystyle \bold{X}' = \begin{bmatrix}1 & 3 \\5 & 3 \end{bmatrix} \bold{X}\)

That question is pure rubbish unless you tell us about \(\displaystyle \bold{X}\ \&\ \bold{X'}\)

 

[logistic_guy]

That question is pure rubbish unless you tell us about \(\displaystyle \bold{X}\ \&\ \bold{X'}\)

Those Russian authors use ambiguous notations even a great mathematician like pka got confused. You, pka, are probably from \(\displaystyle \text{Japan}\) and I can understand your situation. It is not your fault if this is the first time you are seeing this!

\(\displaystyle \bold{X}' = \begin{bmatrix}1 & 3 \\5 & 3 \end{bmatrix} \bold{X}\)

This can be written like this:

\(\displaystyle \begin{bmatrix}x'(t) \\y'(t) \end{bmatrix} = \begin{bmatrix}1 & 3 \\5 & 3 \end{bmatrix} \begin{bmatrix}x(t) \\y(t) \end{bmatrix}\)


My intuition tells me that your field is Banking. And I love Accounting. Yeah I know that they are not the same thing but we both love money, assets, and financial records.



We can play many games together if you are interested. We have a great future as a team!

 

[logistic_guy]

We start with:

\(\displaystyle \bold{A} - \lambda\bold{I} = \begin{bmatrix}1 & 3 \\5 & 3 \end{bmatrix} - \lambda \begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix} = \begin{bmatrix}1 & 3 \\5 & 3 \end{bmatrix} - \begin{bmatrix}\lambda & 0 \\0 & \lambda \end{bmatrix} = \begin{bmatrix}1 - \lambda & 3 \\5 & 3 - \lambda \end{bmatrix}\)

 

[khansaheb]

We start with:

\(\displaystyle \bold{A} - \lambda\bold{I} = \begin{bmatrix}1 & 3 \\5 & 3 \end{bmatrix} - \lambda \begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix} = \begin{bmatrix}1 & 3 \\5 & 3 \end{bmatrix} - \begin{bmatrix}\lambda & 0 \\0 & \lambda \end{bmatrix} = \begin{bmatrix}1 - \lambda & 3 \\5 & 3 - \lambda \end{bmatrix}\)

Now what? Where are you stuck?

 

[logistic_guy]

Now what? Where are you stuck?

 

[logistic_guy]

\(\displaystyle \begin{bmatrix}1 - \lambda & 3 \\5 & 3 - \lambda \end{bmatrix}\)

The next step is to find the determinant of this matrix and set it equal to zero.

\(\displaystyle \left| \begin{array}{cc}1 - \lambda & 3 \\5 & 3 - \lambda \end{array} \right| = (1 - \lambda)(3 - \lambda) - 3(5) = 0\)

This gives:

\(\displaystyle \lambda_1 = -2\)
\(\displaystyle \lambda_2 = 6\)

 

[khansaheb]

Solve.

\(\displaystyle \bold{X}' = \begin{bmatrix}1 & 3 \\5 & 3 \end{bmatrix} \bold{X}\)

Now what? Where are you stuck?

 

[logistic_guy]

Now what? Where are you stuck?

The next step is the most difficult step, yet the most beautiful.



If the eigenvalues \(\displaystyle \lambda\) are distinct real eigenvalues, then the solution to our matrix (or system) is:

\(\displaystyle \bold{X} = c\bold{K}e^{\lambda t}\)

That is:

\(\displaystyle \bold{X} = c_1\bold{X}_1 + c_2\bold{X_2} = c_1\bold{K}_1e^{\lambda_1 t} + c_2\bold{K}_2e^{\lambda_2 t}\)

How to find \(\displaystyle \bold{K}_1\) and \(\displaystyle \bold{K}_2\)?

We know that the matrix \(\displaystyle \bold{K}_1 = \begin{bmatrix}k_1 \\k_2 \end{bmatrix}\) belongs to the eigenvalue \(\displaystyle \lambda_1 = -2\), then we have this system to solve:

\(\displaystyle (1 - \lambda_1)k_1 + 3k_2 = 0\)
\(\displaystyle 5k_1 + (3 - \lambda_1)k_2 = 0\)

Or

\(\displaystyle 3k_1 + 3k_2 = 0\)
\(\displaystyle 5k_1 + 5k_2 = 0\)

If we choose \(\displaystyle k_1 = 1\), then \(\displaystyle k_2 = -1\) and we have:

\(\displaystyle \bold{K_1} = \begin{bmatrix}1 \\-1 \end{bmatrix}\)

We also know that the matrix \(\displaystyle \bold{K}_2 = \begin{bmatrix}k_1 \\k_2 \end{bmatrix}\) belongs to the eigenvalue \(\displaystyle \lambda_2 = 6\), then we have this system to solve:

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

We see that \(\displaystyle k_1 = \frac{3}{5}k_2\), so if we choose \(\displaystyle k_2 = 5\), we get:

\(\displaystyle \bold{K_2} = \begin{bmatrix}3 \\5 \end{bmatrix}\)

Finally, we can write the general solution as:

\(\displaystyle \bold{X} = c_1\bold{K}_1e^{\lambda_1 t} + c_2\bold{K}_2e^{\lambda_2 t} = c_1\begin{bmatrix}1 \\-1 \end{bmatrix}e^{-2t} + c_2\begin{bmatrix}3 \\5 \end{bmatrix}e^{6t}\) (Matrix form)

Or (Normal form)

\(\displaystyle x(t) = c_1e^{-2t} + 3c_2e^{e^{6t}}\)
\(\displaystyle y(t) = -c_1e^{-2t} + 5c_2e^{e^{6t}}\)

We can also write the system solution in a more clean way like:

\(\displaystyle x(t) = Ae^{-2t} + Be^{e^{6t}}\)
\(\displaystyle y(t) = Ce^{-2t} + De^{e^{6t}}\)

But we have to keep track of what those \(\displaystyle 4\) constants are. I myself as a professional skydiver, prefer the Matrix form!