### Jacobian of a 3 Variable Transformation Example (3x3 Jacobian)

Find the Jacobian of the 3 variable transformation given by $$x = \frac{u}{v}$$, $$y = \frac{v}{w}$$ and $$z = \frac{w}{u}$$.

The Jacobian is given by:

$J = \left| {\begin{array}{*{20}{c}}{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}&{\frac{{\partial x}}{{\partial w}}}\\{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}&{\frac{{\partial y}}{{\partial w}}}\\{\frac{{\partial z}}{{\partial u}}}&{\frac{{\partial z}}{{\partial v}}}&{\frac{{\partial z}}{{\partial w}}}\end{array}} \right|$

So work out all the partial derivatives all fill it in!

$J = \left| {\begin{array}{*{20}{c}}{\frac{1}{v}}&{\frac{{ - u}}{{{v^2}}}}&0\\0&{\frac{1}{w}}&{ - \frac{v}{{{w^2}}}}\\{ - \frac{w}{{{u^2}}}}&0&{\frac{1}{u}}\end{array}} \right|$

Now you need to find the determinant of a 3x3 matrix, which can be broken down into a few 2x2 matrix determinants. See my "Determinants" guide if you don't know how to do this, but in general, recall that:

$\left| {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}} \right| = {a_1}\left| {\begin{array}{*{20}{c}}{{b_1}}&{{b_2}}\\{{c_2}}&{{c_3}}\end{array}} \right| - {a_2}\left| {\begin{array}{*{20}{c}}{{b_1}}&{{b_3}}\\{{c_1}}&{{c_3}}\end{array}} \right| + {a_3}\left| {\begin{array}{*{20}{c}}{{b_1}}&{{b_2}}\\{{c_1}}&{{c_2}}\end{array}} \right|$

Remember that you can "travel" along any column or row of the 3x3 matrix to find the determinant. It's best to take advantage of where the zeros are, but in general we can just use the formula above since the first row has a 0.

$\left| {\begin{array}{*{20}{c}}{\frac{1}{v}}&{\frac{{ - u}}{{{v^2}}}}&0\\0&{\frac{1}{w}}&{ - \frac{v}{{{w^2}}}}\\{ - \frac{w}{{{u^2}}}}&0&{\frac{1}{u}}\end{array}} \right| = \frac{1}{v}\left| {\begin{array}{*{20}{c}}{\frac{1}{w}}&{ - \frac{v}{{{w^2}}}}\\0&{\frac{1}{u}}\end{array}} \right| - \left( {\frac{{ - u}}{{{v^2}}}} \right)\left| {\begin{array}{*{20}{c}}0&{ - \frac{v}{{{w^2}}}}\\{ - \frac{w}{{{v^2}}}}&{\frac{1}{u}}\end{array}} \right| + (0)\left| {\begin{array}{*{20}{c}}0&{\frac{1}{w}}\\{ - \frac{w}{{{u^2}}}}&{\frac{1}{u}}\end{array}} \right|$

It looks like a mess and it is a mess, but at least the last term can be ignored since it's multiplied by 0. Just keep simplifying:

$J = \frac{1}{v}\left[ {\frac{1}{w} \cdot \frac{1}{u} - \left( { - \frac{v}{{{w^2}}} \cdot 0} \right)} \right] + \frac{u}{{{v^2}}}\left[ {0 \cdot \frac{1}{u} - \left( { - \frac{v}{{{w^2}}} \cdot - \frac{w}{{{u^2}}}} \right)} \right]$

$J = \frac{1}{{uvw}} - \frac{1}{{uvw}}$

$J = 0$

Please see our other examples on finding the Jacobian for a multivariate transformation:

Why Multiply By Extra r In Polar Coordinates?