matrix

logistic_guy

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Solve.

X=[1353]X\displaystyle \bold{X}' = \begin{bmatrix}1 & 3 \\5 & 3 \end{bmatrix} \bold{X}
 
Solve.X=[1353]X\displaystyle \bold{X}' = \begin{bmatrix}1 & 3 \\5 & 3 \end{bmatrix} \bold{X}
That question is pure rubbish unless you tell us about X & X\displaystyle \bold{X}\ \&\ \bold{X'}
 
That question is pure rubbish unless you tell us about X & X\displaystyle \bold{X}\ \&\ \bold{X'}
Those Russian authors use ambiguous notations even a great mathematician like pka got confused. You, pka, are probably from Japan\displaystyle \text{Japan} and I can understand your situation. It is not your fault if this is the first time you are seeing this!

😩

X=[1353]X\displaystyle \bold{X}' = \begin{bmatrix}1 & 3 \\5 & 3 \end{bmatrix} \bold{X}
This can be written like this:

[x(t)y(t)]=[1353][x(t)y(t)]\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!

😍🤩
 
We start with:

AλI=[1353]λ[1001]=[1353][λ00λ]=[1λ353λ]\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}

💪🗿🗿
 
We start with:

AλI=[1353]λ[1001]=[1353][λ00λ]=[1λ353λ]\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?
 
[1λ353λ]\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.

1λ353λ=(1λ)(3λ)3(5)=0\displaystyle \left| \begin{array}{cc}1 - \lambda & 3 \\5 & 3 - \lambda \end{array} \right| = (1 - \lambda)(3 - \lambda) - 3(5) = 0

This gives:

λ1=2\displaystyle \lambda_1 = -2
λ2=6\displaystyle \lambda_2 = 6
 
Now what? Where are you stuck?
The next step is the most difficult step, yet the most beautiful.

😍

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

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

That is:

X=c1X1+c2X2=c1K1eλ1t+c2K2eλ2t\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 K1\displaystyle \bold{K}_1 and K2\displaystyle \bold{K}_2?

We know that the matrix K1=[k1k2]\displaystyle \bold{K}_1 = \begin{bmatrix}k_1 \\k_2 \end{bmatrix} belongs to the eigenvalue λ1=2\displaystyle \lambda_1 = -2, then we have this system to solve:

(1λ1)k1+3k2=0\displaystyle (1 - \lambda_1)k_1 + 3k_2 = 0
5k1+(3λ1)k2=0\displaystyle 5k_1 + (3 - \lambda_1)k_2 = 0

Or

3k1+3k2=0\displaystyle 3k_1 + 3k_2 = 0
5k1+5k2=0\displaystyle 5k_1 + 5k_2 = 0

If we choose k1=1\displaystyle k_1 = 1, then k2=1\displaystyle k_2 = -1 and we have:

K1=[11]\displaystyle \bold{K_1} = \begin{bmatrix}1 \\-1 \end{bmatrix}

We also know that the matrix K2=[k1k2]\displaystyle \bold{K}_2 = \begin{bmatrix}k_1 \\k_2 \end{bmatrix} belongs to the eigenvalue λ2=6\displaystyle \lambda_2 = 6, then we have this system to solve:

5k1+3k2=0\displaystyle -5k_1 + 3k_2 = 0
5k13k2=0\displaystyle 5k_1 - 3k_2 = 0

We see that k1=35k2\displaystyle k_1 = \frac{3}{5}k_2, so if we choose k2=5\displaystyle k_2 = 5, we get:

K2=[35]\displaystyle \bold{K_2} = \begin{bmatrix}3 \\5 \end{bmatrix}

Finally, we can write the general solution as:

X=c1K1eλ1t+c2K2eλ2t=c1[11]e2t+c2[35]e6t\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)

x(t)=c1e2t+3c2ee6t\displaystyle x(t) = c_1e^{-2t} + 3c_2e^{e^{6t}}
y(t)=c1e2t+5c2ee6t\displaystyle y(t) = -c_1e^{-2t} + 5c_2e^{e^{6t}}

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

x(t)=Ae2t+Bee6t\displaystyle x(t) = Ae^{-2t} + Be^{e^{6t}}
y(t)=Ce2t+Dee6t\displaystyle y(t) = Ce^{-2t} + De^{e^{6t}}

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

💙woman_2.png 💚
 
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