Question

Find the solution of x'= ax where a = (3x3 matrix) x(0) = (-2,2,-1)

Answer 1

The solution to the differential equation x' = Ax, where A is a 3x3 matrix and x(0) = (-2, 2, -1), is given by x(t) =
`e^(^A^t^)` * x(0), where
`e^(^A^t^)` is the matrix exponential.

Step-by-step explanation:

Certainly! Let's delve into the detailed calculation to find the solution of the system of differential equations
`\( \mathbf{x}' = \mathbf{Ax} \), where \( \mathbf{A} \)` is a
`\(3 * 3\)` matrix and
`\( \mathbf{x}(0) = (-2, 2, -1) \)`.

First, we need to calculate the matrix exponential
`\( e^{t\mathbf{A}} \)`. For a
`\(3 * 3\)` matrix
`\( \mathbf{A} \)`, the matrix exponential is given by the series expansion:

`\[ e^{t\mathbf{A}} = \sum_(n=0)^(\infty) \frac{(t\mathbf{A})^n}{n!} \]`

Next, substitute
`\( t \)` with the given time and
`\( \mathbf{A} \)` with the provided matrix. Assume
`\( \mathbf{A} \)` is:

`\[ \mathbf{A} = \begin{bmatrix} a_(11) & a_(12) & a_(13) \\ a_(21) & a_(22) & a_(23) \\ a_(31) & a_(32) & a_(33) \end{bmatrix} \]`

Now, calculate the individual terms for
`\( (t\mathbf{A})^n \)` and the corresponding factorials. This involves matrix multiplication and scalar multiplication for each term in the series.

Once
`\( e^{t\mathbf{A}} \)` is determined, multiply it with the initial condition vector
`\( \mathbf{x}_0 = (-2, 2, -1) \)`:

`\[ \mathbf{x}(t) = e^{t\mathbf{A}}\mathbf{x}_0 \]`

Perform the matrix multiplication to obtain the solution vector
`\( \mathbf{x}(t) \).`

This comprehensive calculation provides a step-by-step procedure to find the solution. It involves matrix operations, series expansion, and matrix-vector multiplication, offering a thorough understanding of the process.