i'm need to find the general solution for matrix

[logistic_guy]

here is the question

Find the general solution of $\displaystyle \ \bold{X}' = \begin{bmatrix}-4 & 0 & 6 & 0 \\ 0 & -5 & 0 & -4 \\ -1 & 0 & 1 & 0 \\ 0 & 3 & 0 & 2 \end{bmatrix}\bold{X}$.

Hint: $\displaystyle \bold{X} = e^{\bold{A}t}\bold{C}$.


my attemb
i don't never solve 4 rows matrix

 

[pka]

Find the general solution of $\displaystyle \ \bold{X}' = \begin{bmatrix}-4 & 0 & 6 & 0 \\ 0 & -5 & 0 & -4 \\ -1 & 0 & 1 & 0 \\ 0 & 3 & 0 & 2 \end{bmatrix}\bold{X}$. Hint: $\displaystyle \bold{X} = e^{\bold{A}t}\bold{C}$.

Can you explain what general solution means?
The matrix $\begin{bmatrix}-4 & 0 & 6 & 0 \\ 0 & -5 & 0 & -4 \\ -1 & 0 & 1 & 0 \\ 0 & 3 & 0 & 2 \end{bmatrix}$ is a $4\times 4$ thus $\bf{X}$ a $4\times k$
Now I have no idea what $e^{\bold{A}t}$ or $\bf{C}$ mean.
Please help us clarify the terms.

 

[logistic_guy]

Can you explain what general solution means?
The matrix $\begin{bmatrix}-4 & 0 & 6 & 0 \\ 0 & -5 & 0 & -4 \\ -1 & 0 & 1 & 0 \\ 0 & 3 & 0 & 2 \end{bmatrix}$ is a $4\times 4$ thus $\bf{X}$ a $4\times k$
Now I have no idea what $e^{\bold{A}t}$ or $\bf{C}$ mean.
Please help us clarify the terms.

thank

$\displaystyle \bold{A} = \begin{bmatrix}-4 & 0 & 6 & 0 \\ 0 & -5 & 0 & -4 \\ -1 & 0 & 1 & 0 \\ 0 & 3 & 0 & 2 \end{bmatrix}$

$\displaystyle \bold{C} = \begin{bmatrix}c_1 \\ c_2 \\ c_3 \\ c_4 \end{bmatrix}$

this $\displaystyle e^{\bold{A}t}$ i don't understand what this

the general solution

$\displaystyle x(t) = c_1x_1(t) + c_2x_2(t) + c_3x_3(t) + c_4x_4(t)$
$\displaystyle y(t) = c_1y_1(t) + c_2y_2(t) + c_3y_3(t) + c_4y_4(t)$
$\displaystyle z(t) = c_1z_1(t) + c_2z_2(t) + c_3z_3(t) + c_4z_4(t)$
$\displaystyle h(t) = c_1h_1(t) + c_2h_2(t) + c_3h_3(t) + c_4h_4(t)$

 

[mario99]

The main purpose of this thread is to show the OP how to get $e^{\bold{A}t}$ which he has written it as $e^{\bold{P}t}$. The OP has not replied to this thread since yesterday and the only thing that I can think of is that he is having difficulties of accessing this math website as it happened to me before. (Or he got what he wanted from somewhere else.)

For the sake of completeness, I will show the method in this post. To find $e^{\bold{A}t}$, you need two matrices, $\bold{A}$ our main matrix and $\bold{I}$ the identity matrix. You also need a little skill in finding inverse matrices as well as finding inverse Laplace Transform. You don't need to be super in inverse Laplace Transform as everything you need is available in the inverse Laplace Transform table.

$\displaystyle \bold{A} = \begin{bmatrix}-6 & 2 \\-2 & -10 \end{bmatrix} \$ and $\displaystyle \ \bold{I} = \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} \$

The first step is to solve this matrix:

$s\bold{I} - \bold{A}$

where $s$ is the parameter of the Laplace Transform.

$s\bold{I} - \bold{A} = \displaystyle s\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix}-6 & 2 \\-2 & -10 \end{bmatrix} = \displaystyle \begin{bmatrix}s & 0 \\ 0 & s \end{bmatrix} + \begin{bmatrix}6 & -2 \\2 & 10 \end{bmatrix} = \begin{bmatrix} s+ 6 & -2 \\2 & s + 10 \end{bmatrix}$

The next step is to find the inverse of the matrix above:

$\displaystyle (s\bold{I} - \bold{A})^{-1} = \begin{bmatrix}\frac{s + 10}{s^2 + 16s + 64} & \frac{2}{s^2 + 16s + 64} \\[7pt] \frac{-2}{s^2 + 16s + 64} & \frac{s + 6}{s^2 + 16s + 64} \end{bmatrix} = \begin{bmatrix}\frac{s + 10}{(s + 8)^2} & \frac{2}{(s + 8)^2} \\[7pt] \frac{-2}{(s + 8)^2} & \frac{s + 6}{(s + 8)^2} \end{bmatrix}$

Now you need to take inverse Laplace Transform of each element of the matrix above. With a little or no manipulation of each element, you can get the inverse Laplace Transform from the table.

And finally, we get what the OP was asking for:

$e^{\bold{P}t} = e^{\bold{A}t} = \begin{bmatrix}(1 + 2t)e^{-8t} & 2te^{-8t} \\[7pt] -2te^{-8t} & (1 - 2t)e^{-8t} \end{bmatrix}$

To solve the exponential matrix $e^{\bold{A}t}$, you may want to look at my post #22.

 

[khansaheb]

To solve the exponential matrix $e^{\bold{A}t}$, you may want to look at my post #22.

Where is post #22? This thread has only 5 posts up until now!

 

[mario99]

Where is post #22? This thread has only 5 posts up until now!

Post #4 already contains post #22. But if you wanna see the original post #22 click on mario99 said: in post #4.

 

[logistic_guy]

To solve the exponential matrix $e^{\bold{A}t}$, you may want to look at my post #22.

i'll copy

$\displaystyle s\bold{I} - \bold{A} = s\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} - \begin{bmatrix}-4 & 0 & 6 & 0 \\ 0 & -5 & 0 & -4 \\ -1 & 0 & 1 & 0 \\ 0 & 3 & 0 & 2 \end{bmatrix} = \begin{bmatrix}s & 0 & 0 & 0 \\ 0 & s & 0 & 0 \\ 0 & 0 & s & 0 \\ 0 & 0 & 0 & s \end{bmatrix} + \begin{bmatrix}4 & 0 & -6 & 0 \\ 0 & 5 & 0 & 4 \\ 1 & 0 & -1 & 0 \\ 0 & -3 & 0 & -2 \end{bmatrix}$

$\displaystyle = \begin{bmatrix}s + 4 & 0 & -6 & 0 \\ 0 & s + 5 & 0 & 4 \\ 1 & 0 & s -1 & 0 \\ 0 & -3 & 0 & s -2 \end{bmatrix}$

how to take this inverse?

 

[khansaheb]

i'll copy

$\displaystyle s\bold{I} - \bold{A} = s\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} - \begin{bmatrix}-4 & 0 & 6 & 0 \\ 0 & -5 & 0 & -4 \\ -1 & 0 & 1 & 0 \\ 0 & 3 & 0 & 2 \end{bmatrix} = \begin{bmatrix}s & 0 & 0 & 0 \\ 0 & s & 0 & 0 \\ 0 & 0 & s & 0 \\ 0 & 0 & 0 & s \end{bmatrix} + \begin{bmatrix}4 & 0 & -6 & 0 \\ 0 & 5 & 0 & 4 \\ 1 & 0 & -1 & 0 \\ 0 & -3 & 0 & -2 \end{bmatrix}$

$\displaystyle = \begin{bmatrix}s + 4 & 0 & -6 & 0 \\ 0 & s + 5 & 0 & 4 \\ 1 & 0 & s -1 & 0 \\ 0 & -3 & 0 & s -2 \end{bmatrix}$

how to take this inverse?

The same way you will calculate the inverse of any-other non-singular matrix!!

 

[mario99]

how to take this inverse?

Matrix calculator

Matrix addition, multiplication, inversion, determinant and rank calculation, transposing, bringing to diagonal, row echelon form, exponentiation, LU Decomposition, QR-decomposition, Singular Value Decomposition (SVD), solving of systems of linear equations with solution steps

matrixcalc.org

 

[Metronome]

I don't think I ever learned this method for ODE systems; I did other things related to the matrix exponent like finding generalized eigenvectors and so forth. Is there an intuition behind using $(sI - A)^{-1}$? It looks a lot like $A - \lambda I$. Is this some sort of $s$ domain eigenspace computation?

 

[mario99]

I don't think I ever learned this method for ODE systems; I did other things related to the matrix exponent like finding generalized eigenvectors and so forth. Is there an intuition behind using $(sI - A)^{-1}$? It looks a lot like $A - \lambda I$. Is this some sort of $s$ domain eigenspace computation?

If you have studied Laplace transform before, it all would make sense to you.



It all began when a bald man said that: if $x = ce^{at}$ is a solution to $x' = ax$, then $\bold{X} = e^{\bold{A}t}\bold{C}$ must be a solution to $\bold{X}' = \bold{A}\bold{X}$.

He wanted to write $e^{\bold{A}t}$ in a different form, so he tried to solve the following initial value problem by Laplace transform:

$\bold{X}' = \bold{A}\bold{X}, \ \ \ \bold{X}(0) = \bold{I}$

Take the Laplace transform of both sides.

$s\bold{x(s)} - \bold{X}(0) = \bold{A}\bold{x}(s)$

Rearrange.

$(s - \bold{A})\bold{x}(s) = \bold{I}$

This arrangement is wrong because $s$ is a scalar and $\bold{A}$ is a matrix. So we have to multiply $s$ by $\bold{I}$ to properly apply $s$ to each component of the matrix $\bold{x}(s)$.

Therefore, the correct arrangement is:

$(s\bold{I} - \bold{A})\bold{x}(s) = \bold{I}$

Multiply both sides by $(s\bold{I} - \bold{A})^{-1}$

$\bold{x}(s) = \bold{I}(s\bold{I} - \bold{A})^{-1} = (s\bold{I} - \bold{A})^{-1}$

We already know that:

$\bold{X}(t) = e^{\bold{A}t}$

Then

$e^{\bold{A}t} = \mathcal{L}^{-1}\{\bold{x}(s)\} = \mathcal{L}^{-1}\{(s\bold{I} - \bold{A})^{-1}\}$

 

[fresh_42]

Matrix calculator

Matrix addition, multiplication, inversion, determinant and rank calculation, transposing, bringing to diagonal, row echelon form, exponentiation, LU Decomposition, QR-decomposition, Singular Value Decomposition (SVD), solving of systems of linear equations with solution steps

matrixcalc.org

FWIW: Here is a matrix calculator that swallows even variables as matrix entries:

Matrix Calculator

Free matrix calculator - solve matrix operations and functions step-by-step

www.symbolab.com

Example.