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Math Help - inverse of a jacobian matrix

  1. #1
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    inverse of a jacobian matrix

    What are the steps in finding the inverse of this matrix?
    <br />
3 z^{(k-1)}\hspace{25 mm} siny^{(k-1)}z^{(k-1)}\hspace{15 mm} y^{(k-1)} sin y^{(k-1)}z^{(k-1)}<br />
    <br />
2x^{(k-1)}\hspace{24 mm} -162(y^{(k-1)}+0.1) \hspace{13 mm}cos z^{(k-1)}<br />
    <br />
-y^{(k-1)}e^{-x^{(k-1)}y^{(k-1)}} \hspace{5 mm}-x^{(k-1)}e^{-x^{(k-1)}y^{(k-1)}} \hspace{13 mm}20<br />
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  2. #2
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    Quote Originally Posted by wantanswers View Post
    What are the steps in finding the inverse of this matrix?
    <br />
3 z^{(k-1)}\hspace{25 mm} siny^{(k-1)}z^{(k-1)}\hspace{15 mm} y^{(k-1)} sin y^{(k-1)}z^{(k-1)}<br />
    <br />
2x^{(k-1)}\hspace{24 mm} -162(y^{(k-1)}+0.1) \hspace{13 mm}cos z^{(k-1)}<br />
    <br />
-y^{(k-1)}e^{-x^{(k-1)}y^{(k-1)}} \hspace{5 mm}-x^{(k-1)}e^{-x^{(k-1)}y^{(k-1)}} \hspace{13 mm}20<br />
    It's a 3x3 matrix and so the inverse is found in the way you've been taught find the inverse of any 3x3 matrix. Simple but tedious in this case.
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