\magnification=\magstep1
\def\sol{{\it Solution}\ :\ }
\centerline{Mathematics 244, section 1 -- Linear Algebra}
\centerline{Solutions for Exam 1 -- February 16, 2007}
If $A,B$ are $n \times n$ matrices and 
$A$ is invertible, then there is a unique $n\times n$ matrix $X$
with $AXA = B$.
\bigskip
\noindent
I. 
\item{A)} (10)
For which value(s) of $c \in {\bf R}$ is the system of linear equations
represented
by the following augmented matrix {\it consistent}? 
$$\pmatrix{-1 & 2 & 0 & c &|&  1\cr
            0 & 1 & 2 & 0 &|& -1\cr
            2 & -4 & 0 & 8 &|&  5\cr}$$
\bigskip
\noindent
\sol
Perform the row operation $R_3 \mapsto R_3 + 2R_1$:
$$\pmatrix{-1 & 2 & 0 & c &|&  1\cr
            0 & 1 & 2 & 0 &|& -1\cr
            0 & 0 & 0 & 8+2c &|&  7\cr}$$
If $8+2c = 0$, then the system is {\it inconsistent}, otherwise
it is consistent.  So the answer here is that the system is 
consistent for all $c \in {\bf R}$, $c \ne -4$.  
\bigskip
\item{B)}  (10) Give a parametrization
of the set of solutions for a value of $c$ (your choice)
that makes the system consistent.
\bigskip
\noindent
\sol
We will take $c = 0$ for simplicity.  Continuing from the matrix in the 
solution for part A, we obtain the row-reduced echelon form
$$\pmatrix{ 1 & 0 & 4 & 0 &|& -3\cr
            0 & 1 & 2 & 0 &|& -1\cr
            0 & 0 & 0 & 1 &|&  7/8\cr}$$
This gives a system where $x_3$ is a free variable. Set of solutions is
$$\left\{
\pmatrix{x_1 \cr x_2 \cr x_3 \cr x_4} = \pmatrix{-3 \cr -1 \cr 0 \cr 7/8}
+ t \pmatrix{-4 \cr -2 \cr 1\cr 0} : t \in {\bf R}\right\}$$
\bigskip
\noindent
II.  
\item{A)}  (5) {\it Define:}  The set $S$ of vectors in ${\bf R}^n$
is {\it linearly independent}.
\bigskip
\noindent
$S$ is linearly independent if whenever a linear combination 
$$c_1 v_1 + c_2 v_2 + \cdots + c_kv_k = 0$$
for $v_i \in S$, then $c_1 = c_2 = \cdots = c_k = 0$.  
\bigskip
\item{B)}  (10) For which value of $a \in {\bf R}$ is
$$S = \left\{\pmatrix{3\cr 2\cr 1 \cr},\pmatrix{-1\cr 1\cr 0\cr},
\pmatrix{1\cr 0\cr a\cr}\right\}$$
a linearly {\it dependent} subset of ${\bf R}^3$?  
\bigskip
\noindent
\sol
We do row operations on the matrix with columns equal to the vectors
in $S$:
$$\eqalign{
\pmatrix{3 & -1 & 1\cr 2 & 1 & 0 \cr 1 & 0 & a} 
&\mapsto 
\pmatrix{1 & 0 & a\cr 2 & 1 & 0\cr 3 &-1 &1\cr}\cr
&\mapsto
\pmatrix{1 & 0 & a\cr 0 & 1 & -2a\cr 0 & -1 & 1-3a\cr}\cr 
&\mapsto
\pmatrix{1 & 0 & a\cr 0 & 1 & -2a\cr 0 & 0 & 1-5a\cr} }$$
In order for the set to be linearly dependent, the matrix must
have fewer than 3 pivot columns, which means $1 - 5a = 0$, so 
$a = 1/5$.   
\item{C)}  (15) Let $b = \pmatrix{3\cr 3\cr 4\cr}$.  Using your 
value of $a$ from part B, is $b \in {\rm Span}(S)$?  Why or why not?
\bigskip
\noindent
\sol
We do the row operations as in the previous part, but on the augmented 
matrix:
$$\eqalign{
\pmatrix{3 & -1 & 1 &| &3 \cr 2 & 1 & 0 &| & 3\cr 1 & 0 & 1/5 &| & 4} 
&\mapsto 
\pmatrix{1 & 0 & 1/5 &| & 4\cr 2 & 1 & 0 &| & 3\cr 3 &-1 &1 &| & 3\cr}\cr
&\mapsto
\pmatrix{1 & 0 & 1/5 &| & 4\cr 0 & 1 & -2/5 &| & -5\cr 0 & -1 & 2/5 &| &-9\cr}\cr 
&\mapsto
\pmatrix{1 & 0 & 1/5 &| & 4\cr 0 & 1 & -2/5&| &-5\cr 0 & 0 & 0 &| &-14\cr} }$$
This says the system is inconsistent, so $b$ is {\it not} in the span of 
$S$.  
\bigskip
\noindent
III.  Let $T : {\bf R}^3 \to {\bf R}^3$ be the mapping defined by
$$T(x) = \pmatrix{1 & 2 & 1\cr
                 -1 & 4 & 0\cr
                  1 & 0 & 1\cr}x$$
\item{A)} (10) Show that $T$ is a linear mapping by verifying the 
properties in the definition.
\bigskip
\noindent
\sol
For {\it any} $3\times 3$ (or $m\times n$) matrix, $A(x+y) = Ax + Ay$
and $A(cx) = cAx$.  This says $T(x+y) = T(x) + T(y)$ and $T(cx) = cT(x)$, 
so $T$ is linear.  
\bigskip
\item{B)} (10) Is $T$ an onto mapping?  Why or why not?  
\bigskip
\noindent
\sol
There are a number of ways to answer this.  One of the most direct is
just to see whether the matrix is {\it invertible}.  If so, the 
Invertible Matrix Theorem says $T$ is onto.  We follow our procedure
for computing the matrix inverse:
$$\eqalign{
\pmatrix{1 & 2 & 1 & 1 & 0 & 0\cr
         -1 & 4 & 0 & 0 & 1 & 0\cr
         1 & 0 & 1 & 0 & 0 & 1\cr}
&\mapsto
\pmatrix{1 & 2 & 1 & 1 & 0 & 0\cr
         0 & 6 & 1 & 1 & 1 & 0\cr
         0 & -2 & 0 & -1 & 0 & 1\cr}\cr
&\mapsto
\pmatrix{1 & 2 & 1 & 1 & 0 & 0\cr
         0 & -2 & 0 & -1 & 0 & 1\cr
         0 & 6 & 1 & 1 & 1 & 0\cr
         }\cr
&\mapsto
\pmatrix{1 & 0 & 1 & 0 & 0 & 1\cr
         0 & -2 & 0 & -1 & 0 & 1\cr
         0 & 6 & 1 & 1 & 1 & 0\cr
         }\cr
&\mapsto
\pmatrix{1 & 0 & 1 & 0 & 0 & 1\cr
         0 & -2 & 0 & -1 & 0 & 1\cr
         0 & 0 & 1 & -2 & 1 & 3\cr
         }\cr
&\mapsto
\pmatrix{1 & 0 & 0 & 2 & -1 & -2\cr
         0 & 1 & 0 & 1/2 & 0 & -1/2\cr
         0 & 0 & 1 & -2 & 1 & 3\cr
         }\cr}
$$
Since $A$ is invertible the answer is yes -- the mapping $T$ is onto.
\bigskip
\noindent
IV.  (True-False)  Determine whether each of the following statements
is true or false.  For the ones that are true, give short proofs;
for those that are false, give counterexamples.
\bigskip
\item{A)} (10) If the homogeneous system $Ax = 0$ has two free variables,
then the set of solutions of $Ax = b$ is a plane for all $b$. 
\bigskip
\noindent
\sol
This is FALSE because the system $Ax = b$ could also
be inconsistent for some $b$.  An example is
given by 
$$[A|b] = \pmatrix{1 & 2 & 3 &| & 1\cr 2 & 4 & 6 &| & -1}$$
\bigskip
\item{B)} (10)  If $x \in {\rm Span}(S)$ ({\it and $x\notin S$}), then the set
$S \cup \{x\}$ is linearly dependent.  
\bigskip
\noindent
\sol
This is TRUE.  If $x = c_1 v_1 + \cdots + c_k v_k$, where $v_i \in S$
for all $i$, then 
$$0 = (-1)x + c_1 v_1 + \cdots + c_k v_k.$$
Since the coefficient $-1\ne 0$ in this linear combination, the 
set $S \cup \{x\}$ is linearly dependent.
\bigskip
\noindent
{\it Comment:} Without the added statement $x \notin S$, this
would actually be FALSE(!)  Counterexample:  $S = \{e_1,e_2\}$
in ${\bf R}^n$ for any $n \ge 2$.  $e_1 = 1\cdot e_1 + 0\cdot e_2$
is in ${\rm Span}(S)$, but $S \cup \{e_1\} = \{e_1,e_2\}$ is still
linearly independent. 
\bigskip
\item{C)} (10)  Let $A$ be an $m\times n$ matrix that defines an {\it onto}
mapping $T_A : {\bf R}^n \to {\bf R}^m$ ($T_A(x) = Ax$ for $x \in {\bf R}^n$).
Then the span of the set of columns of $A$ is ${\bf R}^m$.  
\bigskip
\noindent
\sol
This is TRUE.  Since $T$ is onto, for every $y \in {\bf R}^m$, there
is some $x \in {\bf R}^n$ such that $T(x) = Ax = y$.  If 
$x = \pmatrix{x_1 \cr \vdots \cr x_n\cr}$, then 
$$y = Ax = x_1 a_1 + \cdots + x_n a_n$$
where $a_1, \ldots, a_n$ are the columns of $A$ .  This shows
that every $y \in {\bf R}^m$ is a linear combination of the columns of $A$.  
\bye
