
\vspace*{.3in}
\noindent{\bf Supplementary problem 4: }
We have 
$$
f'(x,y) = \left(\begin{array}{cc}
3x^2 - 4xy & -2x^2 + 1 \\
-4x & 3y^2
\end{array}\right)
$$
With a first guess of $(x_0,y_0) = (1,1)$, I find a second guess 
$(x_1,y_1)$ by linearly approximating $f$ about $(1,1)$ and setting the
result equal to $(1.1,-.8)$. 
$$
\left(\begin{array}{c} .1 \\ -.8 \end{array}\right) 
= 
\left(\begin{array}{c} 0 \\ -1 \end{array}\right) 
+
\left(\begin{array}{cc}
-1 & -1 \\
-4 & 3
\end{array}\right)
\left(\begin{array}{c} x_1-1 \\ y_1-1 \end{array}\right) 
$$
which implies that
$$
\left(\begin{array}{c} x_1 \\ y_1 \end{array}\right)
=
\left(\begin{array}{c} 1 \\ 1 \end{array}\right)
+
\left(\begin{array}{cc}
-1 & -1 \\
-4 & 3
\end{array}\right)^{-1}
\left(\begin{array}{c} .1 \\ .2 \end{array}\right)
=
\left(\begin{array}{c} .92857 \\ .97143 \end{array}\right)
$$
To find a better guess it is necessary to repeat this process using the
linear approximation of $f$ near $(x_1,y_1)$ instead of $(x_0,y_0)$.
Rather than type up the details, I have made available the mathematica
worksheet (web viewable version available, too) that I used to do this
computation.  See the course webpage for the link.
