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\begin{document}


\noindent
{\bf Solutions to HW 2.}

\bigskip
  
\begin{enumerate}
\item[1.]
  Let $A = \begin{bmatrix} 1&1\\ 0&1\end{bmatrix}$.
    Show by induction that $A^n = \begin{bmatrix} 1& n\\ 0&1\end{bmatrix}$.
\end{enumerate}

\bigskip

\color{blue}
{\bf Solution.}

(Base Case ($n=1$))

Let $A^1 = A = \begin{bmatrix} 1&1\\ 0&1\end{bmatrix}$.
  \bigskip

  (Inductive Step (Assume true for $n=k$, show true for $n=k+1$))
  
Assume that $A^k = \begin{bmatrix} 1& k\\ 0&1\end{bmatrix}$.
Then
\[  
A^{k+1} = A\cdot A^k =
\begin{bmatrix} 1&1\\ 0&1\end{bmatrix}\begin{bmatrix} 1& k\\ 0&1\end{bmatrix}
=\begin{bmatrix} 1& k+1\\ 0&1\end{bmatrix}.
\]
\color{black}

\bigskip
\bigskip

\begin{enumerate}
\item[2.]
I have three matrices: $A$ is $m\times n$, $B$ is $p\times q$,
and $C$ is $r\times s$. Suppose I tell you that,
because of their dimensions, the triple product
$ABC$ is not defined, even though the triple product in every
other order is defined
(i.e., $ACB, BAC, BCA, CAB, CBA$ are all defined).
Explain how you know that I made a mistake.
\end{enumerate}
\bigskip

\color{blue}
{\bf Solution.}
If $ACB$ is defined
(a product of the form $(m\times n)(r\times s)(p\times q)$
makes sense),
then $n=r$ and $s=p$.
If $BAC$ is defined
(a product of the form $(p\times q)(m\times n)(r\times s)$
makes sense),
then $q=m$ and $n=r$.
If $BCA$ is defined
(a product of the form $(p\times q)(r\times s)(m\times n)$
makes sense),
then $q=r$ and $s=m$.

Even without considering the other products, $CAB$ and $CBA$,
we have the information that $n=r=q=m=s=p$. $A, B$ and $C$
are therefore (square) $n\times n$ matrices, and their triple
product makes sense in any order, so $ABC$ is defined.
\color{black}

\bigskip
\bigskip

\begin{enumerate}
\item[3.]
3. This problem is about multiplicative inverses of matrices.
\begin{enumerate}
\item[(a)]
Find the multiplicative inverse of each type of elementary matrix:
$P_{ij}$, and $E_{ij}(r)$ for the case $i\neq j$ and the case $i=j$.
\end{enumerate}  
\bigskip

\color{blue}
$P_{ij}^{-1}=P_{ij}$.

$E_{ij}(r)^{-1}=E_{ij}(-r)$ when $i\neq j$.

$E_{ii}(r)^{-1}=E_{ij}(r^{-1})$.
\color{black}
\bigskip

\begin{enumerate}
\item[(b)]
Show that if $A^{-1}$ is the multiplicative inverse of $A$ and $B^{-1}$
is the multiplicative
inverse of $B$, then $B^{-1}A^{-1}$ is the
multiplicative inverse of $AB$.  
\end{enumerate}
\bigskip

\color{blue}
To show that $B^{-1}A^{-1}$ is the multiplicative inverse of $AB$,
we must show that the two multiply to the identity in either order.
\[
\begin{array}{rl}
(B^{-1}A^{-1})(AB)&=B^{-1}(A^{-1}A)B\\
&=B^{-1}IB\\
&=B^{-1}B\\
&=I
\end{array}
\]
and
\[
\begin{array}{rl}
  (AB)(B^{-1}A^{-1})&=A(BB^{-1})A^{-1}\\
&=AIA^{-1}\\
&=AA^{-1}\\
&=I
\end{array}
\]
\bigskip

The main point to remember about this exercise is:
``The inverse of a product $AB$ is the product of the
inverses in the reverse order, $B^{-1}A^{-1}$''.
\color{black}
\bigskip

\end{enumerate}
\end{document}
