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 [imath]\begin{bmatrix}-4 & 0 & 6 & 0 \\ 0 & -5 & 0 & -4 \\ -1 & 0 & 1 & 0 \\ 0 & 3 & 0 & 2 \end{bmatrix}[/imath] is a [imath]4\times 4[/imath] thus [imath]\bf{X}[/imath] a [imath]4\times k[/imath]
Now I have no idea what [imath]e^{\bold{A}t}[/imath] or [imath]\bf{C}[/imath] mean.
Please help us clarify the terms.

 

[logistic_guy]

Can you explain what general solution means?
The matrix [imath]\begin{bmatrix}-4 & 0 & 6 & 0 \\ 0 & -5 & 0 & -4 \\ -1 & 0 & 1 & 0 \\ 0 & 3 & 0 & 2 \end{bmatrix}[/imath] is a [imath]4\times 4[/imath] thus [imath]\bf{X}[/imath] a [imath]4\times k[/imath]
Now I have no idea what [imath]e^{\bold{A}t}[/imath] or [imath]\bf{C}[/imath] 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 [imath]e^{\bold{A}t}[/imath] which he has written it as [imath]e^{\bold{P}t}[/imath]. 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 [imath]e^{\bold{A}t}[/imath], you need two matrices, [imath]\bold{A}[/imath] our main matrix and [imath]\bold{I}[/imath] 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.

[imath]\displaystyle \bold{A} = \begin{bmatrix}-6 & 2 \\-2 & -10 \end{bmatrix} \ [/imath] and [imath]\displaystyle \ \bold{I} = \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} \ [/imath]

The first step is to solve this matrix:

[imath]s\bold{I} - \bold{A}[/imath]

where [imath]s[/imath] is the parameter of the Laplace Transform.

[imath]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}[/imath]

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

[imath]\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}[/imath]

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:

[imath]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}[/imath]

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