Thread: inverse of a jacobian matrix

1. inverse of a jacobian matrix

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
$

2. 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.

inverse of Jacobian matrix

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