anonymous
  • anonymous
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.)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[ x' = \begin{bmatrix} 3&-2\\2&-2 \end{bmatrix}x \]Your first step is to find the eigen vectors and values.
anonymous
  • anonymous
How?
anonymous
  • anonymous
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.

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anonymous
  • anonymous
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?
anonymous
  • anonymous
In this case \[ \begin{split} A-\lambda I &= \begin{bmatrix} 3&-2\\2&-2 \end{bmatrix} - \lambda \begin{bmatrix} 1 &0\\0&1 \\ \end{bmatrix} \\ &= \begin{bmatrix} 3&-2\\2&-2 \end{bmatrix} - \begin{bmatrix} \lambda &0\\0&\lambda \end{bmatrix} \\ &= \begin{bmatrix} 3-\lambda &-2\\2&-2-\lambda \end{bmatrix} \end{split} \]
anonymous
  • anonymous
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: \[ \lambda^2-\lambda -2 = 0 \]
anonymous
  • anonymous
@Idealist, you can find the eigenvalues from here. Do you know how to find eigenvectors?
anonymous
  • anonymous
No.
anonymous
  • anonymous
Let's work on this tomorrow. It's too late.
anonymous
  • anonymous
Okay.

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