Home»Math Guides»Finding Jacobian of a 3 by 3 Matrix
Jacobian of a 3 Variable Transformation Example (3×3 Matrix Jacobian)
We often need to use the Jacobian when using multivariate transformations.
Example problem: Find the Jacobian of the 3 variable transformation given by the system of equations:
\[\begin{array}{l}x = \frac{u}{v}\\y = \frac{v}{w}\\z = \frac{w}{u}\end{array}\]
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 and fill them 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 3×3 matrix, which can be broken down into a few 2×2 matrix determinants.
See our “How to Find 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 3×3 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 truly is a mess, but at least the last term can be ignored since it’s multiplied by 0.
Just keep simplifying:
\[J = \frac{1}{v}\left[ {\frac{1}{w} \cdot \frac{1}{u} – \left( { – \frac{v}{{{w^2}}} \cdot 0} \right)} \right] + \frac{u}{{{v^2}}}\left[ {0 \cdot \frac{1}{u} – \left( { – \frac{v}{{{w^2}}} \cdot – \frac{w}{{{u^2}}}} \right)} \right]\]
\[J = \frac{1}{{uvw}} – \frac{1}{{uvw}}\]
\[J = 0\]