\magnification = \magstep1
\def\soln{\bigskip\item{}{\it Solution}:\ }
\nopagenumbers
\centerline{Mathematics 243, section 1 -- Algebraic Structures}
\centerline{Solutions for Midterm Exam 1 -- September 29, 2006}
\bigskip
\noindent
I.   For this problem, 
$$\eqalign{U &= {\bf Z}\cr
           A &= \{ x \in {\bf Z} : x {\rm\ is\ odd \ and\ } -10 \le x \le -4\}\cr
           B &= \{-2,-1,0,1\}\cr
           C &= \{x \in {\bf Z} : x < 0 \}\cr}$$
and $f: {\bf Z} \to {\bf Z}$ is the mapping defined by 
$$f(x) = \cases{x+3 & if $x$ is even \cr
                x-1 & if $x$ is odd\cr}.$$
\item{A)} (10) What is the set $A \cup (B \cap C')$?
\soln $C' = \{x \in {\bf Z} : x \ge 0\}$, so $B \cap C' = \{0,1\}$, and 
$$A \cup (B \cap C') = \{-9,-7,-5,0,1\}.$$
\item{B)} (10) What is the set $f(B)$?
\soln $f(B) = \{f(-2),f(-1),f(0),f(1)\} = \{1,-2,3,0\}$ (using the 
first part of the definition to find $f(-2),f(0)$ and the second to
get $f(-1),f(1)$).
\medskip
\item{C)} (10) Given: $f$ is a permutation of ${\bf Z}$ (you don't need
to show this).  Find the inverse mapping $f^{-1} : {\bf Z} \to {\bf Z}$.  
\soln  Since $x+3$ is odd when $x$ is even, and $x-1$ is even when 
$x$ is odd:
$$f^{-1}(x) = \cases{x-3 & when $x$ is odd\cr
                     x+1 & when $x$ is even\cr}$$ 
\bigskip
\noindent
II.  Let $f : A \to B$ be any mapping.
\item{A)}  (15) Prove that for any subsets $T_1$ and $T_2$ of $B$,
$f^{-1}(T_1 \cup T_2) = f^{-1}(T_1) \cup f^{-1}(T_2)$.  
\soln by the definitions of the inverse image and union of sets we have
$$\eqalign{
x \in f^{-1}(T_1 \cup T_2) &\Leftrightarrow f(x) \in T_1 \cup T_2\cr
                           &\Leftrightarrow f(x) \in T_1 {\rm\ or\ } f(x) \in T_2\cr
                           &\Leftrightarrow x \in f^{-1}(T_1) {\rm\ or\ } x \in f^{-1}(T_2)\cr
                           &\Leftrightarrow x \in f^{-1}(T_1) \cup f^{-1}(T_2)\cr}$$
Hence $f^{-1}(T_1\cup T_2) = f^{-1}(T_1) \cup f^{-1}(T_2)$ because each of these
sets is contained in the other.
\medskip
\item{B)}  (10) Is $f^{-1}(f(S))$ always equal to $S$ for 
all subsets $S \subseteq A$?  If you say no, give a counterexample;
if you say yes, say why.
\soln This is {\it not} always true.  It can fail when the mapping $f$
is not injective (one-to-one).  For instance, let $f : {\bf R} \to {\bf R}$
be defined by $f(x) = x^2$.  Then if $S = \{1\}$, we have $f(S) = \{1\}$.
But $f^{-1}(f(S)) = \{-1,1\}$, since both $f(-1) = 1$ and $f(1) = 1$.
\bigskip
\noindent
III.  Let $*$ be the binary operation on ${\bf R}$
defined by $x*y = x y - 2 x$. 
\item{A)} (10) Show with specific counterexamples that 
$*$ is {\it not associative}, and {\it not commutative}.
\soln $*$ is not associative because for instance
$(1*2)*1 = (2-2)*1 = 0*1 = 0$, but $1*(2*1) = 1*(2-4) = 1*(-2) = -2 - 2 = -4$.
Similarly, $*$ is not commutative:  $1*2 = 0$ but $2*1 = -2$.
\medskip
\item{B)} (15) An element $e$ is said to be a 
{\it left identity} for a binary operation if 
$e * x = x$ for all $x$.  Similarly, $e$ is a 
{\it right identity} if $x * e = x$ for all $x$.  
Does the $*$ operation have a left identity?
Does it have a right identity?  If so, say what $e$ is
in each case.  If not, explain how you reach your conclusions.  
\soln $*$ {\it does have a right identity} $e = 3$ since 
$x*3 = 3x - 2x = x$ for all $x$.  On the other hand, it 
does not have a left identity because if 
$e*y = y$ for all $y$, then we have $ey - 2e = y$ for all $y \in {\bf R}$
This equation says $e = {y \over y - 2}$, so while there is some $e$
that works for each $y$ individually, there is no single $e$ that works
for all $y$.   
\bigskip
\noindent
IV.  
In both parts of this problem, 
$$A = \pmatrix{1 & 1 \cr 3 & 4} \qquad  B = \pmatrix{4 & -1 \cr -3 & 1}.$$
\item{A)} (10) Compute the matrix product $AB$ or say why it doesn't exist.
\soln $A$ and $B$ are $2\times 2$ matrices, so the product is defined, and
$$AB = \pmatrix{1 & 1 \cr 3 & 4}\pmatrix{4 & -1 \cr -3 & 4} = \pmatrix{1 & 0 \cr 0 & 1\cr}$$
(the $2\times 2$ identity matrix).  Note that 
$$BA = \pmatrix{4 & -1 \cr -3 & 1}\pmatrix{1 & 1 \cr 3 & 4} = \pmatrix{1 & 0 \cr 0 & 1\cr}$$
also.
\medskip
\item{B)} (10) Define a mapping 
$$F : M_{2\times 2}({\bf R}) \to  M_{2\times 2}({\bf R})$$
for $X$ a $2\times 2$ matrix by by $F(X) = BX$ (matrix product).  
Prove that $F$ is one-to-one (injective) and give its left inverse mapping.
\soln  We show $F$ is injective first.  Assume $F(X) = F(Y)$ for two matrices $X$, $Y$.
Then $BX = BY$.  Multiply $A$ on the left of both sides in this equation:
$A(BX) = A(BY)$.  By associativity of matrix multiplication, this says
$(AB)X = (AB)Y$.  Since $AB = I_2$ (the $2\times 2$ identity matrix), 
this implies $I_2X = I_2Y$, so $X = Y$. 
\medskip
\item{} The left inverse mapping of $F$ is 
$G : M_{2\times 2}({\bf R}) \to M_{2\times 2}({\bf R})$
defined by $G(X) = AX$.  This follows because for all $X$
$$(G\circ F)(X) = G(F(X)) = A(BX) = (AB)X = I_2 X = X$$
(The fact that $F$ is injective also follows from the fact that there 
is a left inverse.)  
\bye

 
