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QUESTION {There are certain matrices (in particular the Jacobian,
Hessian and Wronksian) the elements of which consist of functions
and/or their derivatives.}

Let ${\bf u} = (u_1, u_2, \ldots , u_m)$ and ${\bf x} = (x_1, x_2,
\ldots , x_n)$, where each of the coordinate functions $u_r$ is a
function of all the variables $x_s$. The {\em Jacobian matrix} $D$
has $(D)_{rs} = \partial u_r/\partial x_s$.

\medskip

The {\em Jacobian} (or {\em Jacobian determinant}) is the
determinant of this matrix. For example, if $m = n= 2$ then the
Jacobian is denoted by $\partial (u_1, u_2)/\partial (x_1, x_2)$
and in the case when $$
\begin{array}{l}
u_1 = x_1 + x_2, \\ u_2 = x_1x_2^2, \\
\end{array}
$$ then $$ \frac{\partial (u_1, u_2)}{\partial (x_1, x_2)} =
\left|\begin{array}{cc}
 \partial u_1/\partial x_1 & \partial u_1/\partial x_2 \\
 \partial u_2/\partial x_1 & \partial u_2/\partial x_2 \\
\end{array}\right|
= \left|\begin{array}{cc}
 1 & 1 \\
 x_2^2 & 2x_1x_2
\end{array}\right|
= 2x_1x_2 - x_2^2. $$

\medskip

The {\em Hessian matrix} $H$ is defined when $m = 1$ and has
$(H)_{rs} =
\partial ^2u/\partial x_r\partial x_s$; the {\em Hessian} (or {\em
Hessian determinant}) is det$H$. For example, if $u = x^2y^2z^2$
then $$ H = \left[\begin{array}{ccc}
  \partial ^2u/\partial x^2 & \partial ^2u/\partial x\partial y & \partial ^2u/\partial x\partial z \\
  \partial ^2u/\partial y\partial x & \partial ^2u/\partial y^2  & \partial ^2u/\partial y\partial z \\
  \partial ^2u/\partial z\partial x & \partial ^2u/\partial z\partial y & \partial ^2u/\partial z^2
\end{array}\right]
 = \left[\begin{array}{ccc}
 2y^2z^2 & 4xyz^2 & 4xy^2z \\
 4xyz^2 & 2x^2z^2 & 4x^2yz \\
 4xy^2z & 4x^2yz & 2x^2y^2 \\
\end{array}\right].
$$

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[For everyday functions, the mixed partial derivatives are equal,
in which case $H$ is a symmetric matrix.]

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(a) Find the Jacobian matrix and the Jacobian for the following
set of functions: $$
\begin{array}{l}
u = x^2 + y^2 + z^2, \\ v = xy + yz + zx, \\ w = x + y + z. \\
\end{array}
$$ (b) Find the Hessian matrix and Hessian of $u = ax^3 + 3bx^2y +
3cxy^2 + dy^3$.



ANSWER
\begin{description}

\item[(a)]
$$J=\left[\begin{array}{ccc}\frac{\partial u }{\partial x
}&\frac{\partial u}{\partial y}&\frac{\partial u }{\partial z
}\\\frac{\partial v }{\partial x}&\frac{\partial v }{\partial y
}&\frac{\partial v}{\partial z}\\ \frac{\partial w }{\partial x
}&\frac{\partial w }{\partial y}&\frac{\partial w }{\partial z
}\end{array}\right]\
=\left[\begin{array}{ccc}2x&2y&2z\\y+z&x+z&x+y\\1&1&1\end{array}\right]$$

The determinant=0. This can be proved in various ways, e.g.

$\frac{1}{2}$row 1+row 2=$(x+y+z)$row 3.

\item[(b)]
$$H=\left[\begin{array}{cc}\frac{\partial^2u}{\partial x^2
}&\frac{\partial^2u}{\partial x\partial u
}\\\frac{\partial^2u}{\partial y\partial x
}&\frac{\partial^2u}{\partial y^2}\end{array}\right]=
\left[\begin{array}{cc}6ax+6by&6bx+6cy\\6bx+6cy&6cx+6dy\end{array}\right]$$

\begin{eqnarray*}
\textrm{det} H&=&36\{(ax+by)(cx+dy)-(bx+cy)^2\}\\
&=&36\{(ac-b^2)x^2+(ad-bc)xy+(bd-c^2)y^2\}
\end{eqnarray*}


\end{description}




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