Question

Find the general solution of x'=(3, 2, -2, -2)x. (This is an matrix, 3 and 2 on the left, -2 and -2 on the right.)

Answer 1

\[
x' = \begin{bmatrix}
3&-2\\2&-2
\end{bmatrix}x
\]Your first step is to find the eigen vectors and values.

Answer 2

How?

Answer 3

In any linear algebra class, should have have to know how to find the eigen vectors and values by the time you're doing matrix differential equations.

Answer 4

To get eigen values of a matrix \(A\), solve for the scalar \(\lambda\) where: \[
\det(A-\lambda I) = 0
\]In this case \(I\) is the identity matrix.
Do you know how to do matrix operations?

Answer 5

In this case \[
\begin{split}
A-\lambda I &= \begin{bmatrix}
3&-2\\2&-2
\end{bmatrix} - \lambda \begin{bmatrix}
1 &0\\0&1 \\
\end{bmatrix} \\
&= \begin{bmatrix}
3&-2\\2&-2
\end{bmatrix} -
\begin{bmatrix}
\lambda &0\\0&\lambda
\end{bmatrix} \\
&= \begin{bmatrix}
3-\lambda &-2\\2&-2-\lambda
\end{bmatrix}
\end{split}


\]

Answer 6

Thus: \[\begin{split}
\det(A-\lambda I) &= (3-\lambda )(-2-\lambda) - (-2)(2) \\
&= (-6 - \lambda +\lambda^2)- (-4) \\
&= \lambda^2-\lambda -2

\end{split}
\]So to find the Eigen values, solve for \(\lambda\) where: \[
\lambda^2-\lambda -2 = 0
\]

Answer 7

@Idealist, you can find the eigenvalues from here. Do you know how to find eigenvectors?

Answer 8

No.

Answer 9

Let's work on this tomorrow. It's too late.

Answer 10

Okay.