# inverse of a jacobian matrix

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• Nov 27th 2008, 08:09 AM
wantanswers
inverse of a jacobian matrix
What are the steps in finding the inverse of this matrix?
$\displaystyle 3 z^{(k-1)}\hspace{25 mm} siny^{(k-1)}z^{(k-1)}\hspace{15 mm} y^{(k-1)} sin y^{(k-1)}z^{(k-1)}$
$\displaystyle 2x^{(k-1)}\hspace{24 mm} -162(y^{(k-1)}+0.1) \hspace{13 mm}cos z^{(k-1)}$
$\displaystyle -y^{(k-1)}e^{-x^{(k-1)}y^{(k-1)}} \hspace{5 mm}-x^{(k-1)}e^{-x^{(k-1)}y^{(k-1)}} \hspace{13 mm}20$
• Nov 27th 2008, 04:34 PM
mr fantastic
Quote:

Originally Posted by wantanswers
What are the steps in finding the inverse of this matrix?
$\displaystyle 3 z^{(k-1)}\hspace{25 mm} siny^{(k-1)}z^{(k-1)}\hspace{15 mm} y^{(k-1)} sin y^{(k-1)}z^{(k-1)}$
$\displaystyle 2x^{(k-1)}\hspace{24 mm} -162(y^{(k-1)}+0.1) \hspace{13 mm}cos z^{(k-1)}$
$\displaystyle -y^{(k-1)}e^{-x^{(k-1)}y^{(k-1)}} \hspace{5 mm}-x^{(k-1)}e^{-x^{(k-1)}y^{(k-1)}} \hspace{13 mm}20$

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.