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\begin{document}


\noindent
{\bf Solutions to HW 8.}

\medskip
  
\begin{enumerate}
\item[1.]
Compute the determinant of the following matrix
using each of the methods indicated.
$$
A=
\begin{bmatrix}
1&0&2\\
0&1&2\\
1&1&3\\
\end{bmatrix}
$$  
\begin{enumerate}
\item[(a)] The permutation expansion.
\item[(b)] The Laplace expansion along the second column.
\item[(c)] Gaussian elimination. 
\end{enumerate}
\end{enumerate}

\medskip

\color{blue}
{\bf Solution.}

Part (a): There are three permutation matrices in the support
of $A$ (that is, whose nonzero entries occur in positions
where $A$ has nonzero entries). Using these, we find that
\[
\begin{array}{rl}
\det(A)&=
1\cdot 1\cdot 3\cdot \det\begin{bmatrix}
1&0&0\\
0&1&0\\
0&0&1\\
\end{bmatrix}+
1\cdot 2\cdot 1\cdot
\det\begin{bmatrix}
1&0&0\\
0&0&1\\
0&1&0\\
\end{bmatrix}+
2\cdot 1\cdot 1\cdot \det\begin{bmatrix}
0&0&1\\
0&1&0\\
1&0&0\\
\end{bmatrix}\\
&\vphantom{|}\\
&=3-2-2\\
&=-1.
\end{array}
\]

Part (b): 
\[
\begin{array}{rl}
\det(A)&=
-0\cdot \det\begin{bmatrix}0&2\\1&3\end{bmatrix}
+1\cdot \det\begin{bmatrix}1&2\\1&3\end{bmatrix}
-1\cdot \det\begin{bmatrix}1&2\\0&2\end{bmatrix}\\
&\vphantom{|}\\
&=0+1-2\\
&=-1.
\end{array}\\
\]

Part (c): Use G.E. without scaling, and keeping track of
the number of row interchanges.

\[
\left[
\begin{array}{rrr}
1&0&2\\
0&1&2\\
1&1&3
\end{array}
\right]
{\underset{-R_2+R_3}{\overset{-R_1+R_3}{\xrightarrow{\hspace*{1.2cm}}}}}
\left[
\begin{array}{rrr}
1&0&2\\
0&1&2\\
0&0&-1
\end{array}
\right]
\]
Since there were no row interchanges, the answer
is the product of the diagonal entries: $1\cdot 1\cdot (-1) = -1$.
\color{black}

\medskip

\begin{enumerate}
\item[2.]
Explain why the determinant of an upper triangular matrix
equals the product of the diagonal entries.
\end{enumerate}
\medskip

\color{blue}
{\bf Solution.} 
The only permutation matrix that could be in the support
of an upper triangular matrix $U$ is the identity matrix, so
according to the permutation expansion of the determinant
we get $\det(U) = u_{11}\cdots u_{nn}\cdot \det(I) =$
the product of the diagonal entries of $U$.
\color{black}
  \medskip

\begin{enumerate}
\item[3.]
Let $A$ and $B$ be $n\times n$ matrices
with $B = [{\bf b}_1 \cdots {\bf b}_n]$.
\begin{enumerate}
\item[(a)] Use the fact that $AB = [A{\bf b}_1 \cdots A{\bf b}_n]$
to show that, as a function of the columns of $B$,
$f({\bf b}_1,\ldots,{\bf b}_n):=\det(AB)$
is multilinear and alternating.
\item[(b)] We showed that if $f$ is a multilinear and alternating
function of $n$ columns, then
$$f(B) = \textrm{(perm expansion of $\det(B)$)}\cdot f({\bf e}_1,\ldots,{\bf e}_n).$$
Use this and part (a) to conclude that
$\det(AB)=\det(B)\cdot\det(A)=\det(A)\cdot\det(B)$.
\end{enumerate}
\end{enumerate}
\medskip

\color{blue}
{\bf Solution.}

Part (a): $A$ is a fixed $n\times n$ matrix.

First we argue that the function
$$f({\bf b}_1,\ldots,{\bf b}_n)=\det(A\cdot [{\bf b}_1 \cdots {\bf b}_n)])
=\det([A{\bf b}_1 \cdots A{\bf b}_n])$$
is alternating.
If ${\bf b}_i={\bf b}_j$ for some $i\neq j$, then 
$A{\bf b}_i=A{\bf b}_j$, so
$$f({\bf b}_1,\ldots,{\bf b}_n)=
\det([A{\bf b}_1 \cdots A{\bf b}_n])=0$$
since the determinant is alternating.

Next we argue that $f$ is multilinear.
In fact, we argue more generally that if $F(x_1,\ldots,x_n)$
is any multilinear function and
$T_1(x_1),\ldots,T_n(x_n)$ are linear functions, then
the composition 
$G(x_1,\ldots,x_n):=F(T_1(x_1),\ldots,T_n(x_n))$ is
a multilinear function.
This more general observation can be used here with
$F = \det$ and $T_i(x) = Ax$ for all $i$.

\[
{\scriptsize
\begin{array}{rlcl}
G(\alpha y+\beta z, x_2,\ldots,x_n)&=
F(T_1(\alpha y+\beta z),T_2(x_2),\ldots,T_n(x_n))&\hphantom{AA}&\textrm{Defn of $G$}\\
&=
F(\alpha T_1(y)+\beta T_1(z),T_2(x_2),\ldots,T_n(x_n))
&\hphantom{AA}&\textrm{$T_1$ is linear}\\
&=
\alpha F(T_1(y),T_2(x_2),\ldots,T_n(x_n))+
\beta F(T_1(z),T_2(x_2),\ldots,T_n(x_n))
&\hphantom{AA}&\textrm{$F$ linear in 1st var}\\
&=
\alpha G(y,x_2,\ldots,x_n)+
\beta G(z,x_2,\ldots,x_n)
&\hphantom{AA}&\textrm{Defn of $G$}\\
\end{array}
}
\]
This proves that $G$ is linear in its 1st variable, but the argument
works in an other variable, so $G$ is multilinear.

Part (b): following the hint, 
\[
{\scriptsize
\det(AB)=f(B)=\textrm{(perm expansion of $\det(B)$)}\cdot
f({\bf e}_1,\ldots,{\bf e}_n)=\det(B)\cdot \det(AI)=\det(B)\cdot \det(A),
}
\]
so, since scalars commute, $\det(AB)=\det(B)\cdot \det(A)=\det(A)\cdot \det(B)$.
\color{black}

\end{document}
