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\section*{Strang: Chapter 5}
\begin{description}

\item[{\bf Section 5.1}.] Exercises 1, 3, 8, 14, 21, 27, 33

\item[{\bf Section 5.2}.] Exercises 2, 5, 7, 11, 12, 16, 20, 23, 32\\
{\bf 5.2-2}: Use exercise 1 in part (a).
{\bf 5.2-12}: Find the cofactor matrix for $A$. Then compare the inverse of $A$ with $AC^T$.

\item[{\bf Problem week10-1}.] The first three Legendre polynomials are $1$, $x$, and $x^2 - 1$. Choose $c$ so that the fourth
polynomial $x^3 - cx$ is orthogonal to the first three. All integrals go from $-1$ to $1$. See Exercise 8.5-4.

\item[{\bf Problem week10-2}.] Graph the square wave. Then graph by hand the sum of two sine terms in its series,
or graph by machine the sum of $2$, $3$, and $10$ terms. The famous Gibbs phenomenon
is the oscillation that overshoots the jump (this doesn't die down with more terms). See Exercise 8.5-7.

\item[{\bf Section 5.3}.] Exercises 2, 4, 6, 9, 17, 28, 33, 36


\end{description}


\subsection*{Some Answers}

{\bf 5.1}. Exercises 1, 8, 14, 21, 27  have textbook answers.

{\bf 5.1-3}. (a) False: $\det(I + I)$ is not $1+1$ (b) True: The product rule extends to $ABC$ (use
it twice) (c) False: $\det(4A)$ is $4^n \det(A)$ (d) False: $A=\twobytwo{0&0}{0&1}$,
$B=\twobytwo{0&1}{1&0}$,
$AB - BA=\twobytwo{0&-1}{1&0}$
is invertible.


{\bf 5.1-33}. I now know that maximizing the determinant for $1$, $-1$ matrices is Hadamard's problem
(1893): see Brenner in American Math. Monthly volume 79 (1972) 626-630. Neil
Sloane's wonderful On-Line Encyclopedia of Integer Sequences ({\tt research.att.com/\~{}njas})
includes the solution for small $n$ (and more references) when the problem is
changed to $0$, $1$ matrices. That sequence {\tt A003432} starts from $n=0$ with $1, 1, 1, 2, 3,
5, 9$. Then the $1$, $-1$ maximum for size $n$ is $2^{n-1}$ times the $0$, $1$ maximum for size $n-1$
(so $(32)(5)=160$ for $n=6$ in sequence {\tt A003433}).
To reduce the $1$, $-1$ problem from $6$ by $6$ to the $0$, $1$ problem for $5$ by $5$, multiply the
six rows by $\pm 1$ to put $+1$ in column $1$. Then subtract row $1$ from rows $2$ to $6$ to get a $5$
by $5$ submatrix $S$ of $-2$, $0$ and divide $S$ by $-2$.
Here is an advanced MATLAB code and a $1$, $-1$ matrix with largest $\det(A)=48$ for
$n=5$:
\begin{verbatim}
 n=5; p=(n-1)^2; A0=ones(n); maxdet=0;
for k=0 : 2^p - 1
 Asub=rem(floor(k. * 2.^(-p + 1: 0)),2);
 A=A0;
 A(2:n,2:n)= 1-2*reshape(Asub,n-1,n-1);
 if abs(det(A))>maxdet, maxdet=abs(det(A)); maxA=A;
 end
end
\end{verbatim}
$$\mbox{Output:}\quad\quad
maxA =
\left[\begin{array}{rrrrrr}
1& 1&  1& 1 &  1\\
1& 1&  1&-1 & -1 \\
1& 1& -1& 1 & -1 \\
1&-1&  1& 1 & -1\\
1&-1& -1& -1&  1
\end{array}\right],\quad
maxdet=48
$$

\bigskip

{\bf 5.2}. Exercises 2, 11, 12, 16, 20, 32 have textbook answers.

{\bf 5.2-5}. Four zeros in the same row guarantee $\det=0$. $A=I$ has $12$ zeros (maximum with
$\det\ne 0$).

{\bf 5.2-7}. $5!/2=60$ permutation matrices have $\det= +1$. Move row $5$ of $I$ to the top; starting
from rows in the order $(5,1,2,3,4)$, elimination to reach $I$ will take four row exchanges.

{\bf 5.2-23}.

(a) If we choose an entry from $B$ we must choose an entry from the zero block; result
zero. This leaves entries from $A$ times entries from $D$ leading to $\det(A)\det(D)$.

(b) and (c) Take $A=\twobytwo{1&0}{0&0}$, $B=\twobytwo{0&0}{1&0}$, $C=\twobytwo{0&1}{0&0}$, $D=\twobytwo{0&0}{0&1}$.
 See the solution to problem 25.

\bigskip

{\bf Problem week10-1}. Answer: $c=3/5$.
Use $\int_{-T}^T (\mbox{odd function})dx = 0$ in the report.


{\bf Problem week10-2}. The $-1$, $1$ odd square wave is $f(x)= x/|x|$ for $0 < |x| < \pi$. Its Fourier series in
equation (8), Section 8.5 of Strang's book, is $4/\i$ times $[\sin x + \frac13\sin 3x +\frac15\sin 5x \cdots]$. The sum of the first $N$
terms has an interesting shape, close to the square wave except where the wave jumps
between $-1$ and $1$. At those jumps, the Fourier sum spikes the wrong way to $\pm 1.09$
(the Gibbs phenomenon) before it takes the jump with the true values of $f(x)$.

This happens for the Fourier sums of all functions with jumps. It makes shock
waves hard to compute. You can see it clearly in a graph of the sum of 10 terms.

\bigskip

{\bf 5.3}. Exercises 2, 4, 6, 9, 17, 36  have textbook answers.

{\bf 5.3-28}.Let $x=\rho\sin\phi\cos\theta$, $y=\rho\sin\phi\sin\theta$, $z=\rho\cos\phi$ and $\vec{u}=(x,y,z)$. Then
 $$J=\det[\vec{u}_\rho | \vec{u}_\phi | \vec{u}_\theta]=
\threebythree
{\sin\phi\,\cos\theta & \rho\cos\phi\,\sin\theta & -\rho\sin\phi\,\sin\theta}
{\sin\phi\,\sin\theta & \rho\cos\phi\,\sin\theta & \rho\sin\phi\,\cos\theta}
{\cos\phi             & -\rho\sin\phi            & 0}$$
Expand by cofactors along column 1, then simplify with trig identities, obtaining the answer $=\rho^2 \sin\phi$.

{\bf 5.3-33}. Find the components of $\vec{v}\times\vec{w}$ from determinant expansion
$\det\threebythree{\vec{\imath} & \vec{\jmath} & \vec{k}}{v_1 & v_2 & v_3}{w_1 & w_2 & w_3}$.
These are exactly the cofactors along row one of the determinant
$\det\threebythree{u_1 & u_2 & u_3}{v_1 & v_2 & v_3}{w_1 & w_2 & w_3}$. Then the determinant equals its cofactor expansion = dot product of $\vec{u}$ and $\vec{v}\times\vec{w}$.

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