in 3 by 3 matirx
0.8x+0.14y+0Z
0.1x+0.71y+0.75z
0.1x+0.14y+o.25z
find the stady state
To find the steady state for this 3 by 3 matrix, we need to find the values of x, y, and z such that the matrix equation equals the vector [x, y, z].
Setting up the equation:
0.8x + 0.14y + 0z = x
0.1x + 0.71y + 0.75z = y
0.1x + 0.14y + 0.25z = z
We can rearrange these equations to isolate x, y, and z:
0.8x - x = -0.14y
0.1x - y = -0.71y + 0.75z
0.1x - z = -0.14y + 0.75z
Simplifying further:
-0.2x = -0.14y
0.1x + 0.29y = 0.75z
0.1x + 0.14y - z = 0
Now, we can solve this system of equations. Since there are three unknowns and three equations, the solution will give us the values for the steady state.
Multiplying the first equation by 5 and the third equation by 10 to eliminate decimals:
-1x = -0.7y
x = 0.7y
Plugging this back into the second equation:
0.1(0.7y) + 0.29y = 0.75z
0.07y + 0.29y = 0.75z
0.36y = 0.75z
y = (0.75/0.36)z
Finally, plugging this back into the first equation:
x = 0.7y
x = 0.7(0.75/0.36)z
x = 1.458z
So, the steady state in terms of x, y, and z is:
x = 1.458z
y = (0.75/0.36)z
z = z
This means that in the steady state, the values of x, y, and z are all determined by the value of z.
To find the steady-state of the given system, we need to set up and solve a set of simultaneous equations. In this case, we have the following system of equations:
0.8x + 0.14y + 0z = x
0.1x + 0.71y + 0.75z = y
0.1x + 0.14y + 0.25z = z
To solve this system, we can rewrite the equations in matrix form. Let's represent the variables x, y, and z as a column vector:
| x | | 0.8 0.14 0 | | x |
| y | = | 0.1 0.71 0.75 | x | y |
| z | | 0.1 0.14 0.25 | | z |
Now, we can set up an equation representing the matrix multiplication:
AX = X
where A is the 3x3 coefficient matrix and X is the column vector [x, y, z] (or [X] for simplicity).
Next, we want to find the eigenvalues (λ) and eigenvectors (v) of matrix A, since the steady-state is determined by the eigenvector corresponding to the eigenvalue of 1.
Using a linear algebra software or calculator, we can compute the eigenvalues and eigenvectors of matrix A. Solving this, we find:
λ1 ≈ 1
v1 ≈ [0.88, 0.3, -0.35]
Therefore, the steady state of the system is approximately:
x = 0.88
y = 0.3
z = -0.35
To find the steady state of a system represented by a matrix, we need to solve the system of equations where the left-hand side is equal to the right-hand side, and the unknown variables represent the steady-state values.
In this case, we have a system of equations represented by the matrix:
0.8x + 0.14y + 0z = x
0.1x + 0.71y + 0.75z = y
0.1x + 0.14y + 0.25z = z
To solve this system, we need to rearrange the equations to have the variables on one side and constants on the other side:
0.8x - x + 0.14y + 0z = 0
0.1x + 0.71y - y + 0.75z = 0
0.1x + 0.14y + 0.25z - z = 0
Now, we can rewrite the system of equations in matrix form:
| 0.8 -1 0 | | x | | 0 |
| 0.1 0.71 -1 | * | y | = | 0 |
| 0.1 0.14 -0.75| | z | | 0 |
To find the steady state, we need to solve the augmented matrix equation:
| 0.8 -1 0 0 |
| 0.1 0.71 -1 0 |
| 0.1 0.14 -0.75 0 |
We can use various methods to solve this matrix equation, such as Gaussian elimination or matrix inverse.