inverse of a jacobian matrix

**wantanswers** · November 27th 2008, 08:09 AM

What are the steps in finding the inverse of this matrix?
$ 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)} $
$ 2x^{(k-1)}\hspace{24 mm} -162(y^{(k-1)}+0.1) \hspace{13 mm}cos z^{(k-1)} $
$ -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 $

**mr fantastic** · November 27th 2008, 04:34 PM

Originally Posted by wantanswers

What are the steps in finding the inverse of this matrix?
$ 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)} $
$ 2x^{(k-1)}\hspace{24 mm} -162(y^{(k-1)}+0.1) \hspace{13 mm}cos z^{(k-1)} $
$ -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.

Thread: inverse of a jacobian matrix

LinkBack

Thread Tools

Display

inverse of a jacobian matrix

Similar Math Help Forum Discussions

Jacobian Matrix

jacobian matrix

Eigenvectors of a Jacobian matrix

jacobian matrix. derivatives

Jacobian matrix exercise

Search Tags