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\def\Sol{\bigskip\item{}{\it Solution\/:}  }
\centerline{Mathematics 244, section 1 -- Linear Algebra}
\centerline{Retest Exam 2 -- March 28, 2007}
\bigskip
\noindent
I.  All parts of this question refer to the matrix
$$A = \pmatrix{ 1 & 4 & 2 & 1 & 2t\cr
               -3 & 2 & 2 & 3 & t^2\cr
                2 & 1 & 0 & -1 & -t\cr}$$
\item{A)} (15)  What is $\dim{\rm Nul}(A)$ if $t = 2$?  Find a 
basis for ${\rm Nul}(A)$ in this case. 
\Sol
The row-reduced echelon form with $t = 2$ is
$$E = \pmatrix{1 & 0 & -2/7 & -5/7 & 0\cr
               0 & 1 & 4/7  & 3/7  & 0\cr
               0 & 0 &  0   &  0   & 1\cr}$$
This shows $A x = 0$ has two free variables, so $\dim{\rm Nul}(A) = 2$.
A basis is
$$\left\{ 
\pmatrix{2/7 \cr -4/7 \cr 1 \cr 0\cr 0\cr}, 
\pmatrix{5/7 \cr -3/7 \cr 0 \cr 1 \cr 0\cr} \right\}$$
\item{B)} (10)  What is $\dim{\rm Col}(A)$ if $t = 4$.  Find
a basis for ${\rm Col}(A)$ in this case.  
\Sol  With $t = 4$, the row-reduced echelon form has a row of zeroes,
and pivots in columns 1,2.  A basis for ${\rm Col}(A)$ consists of the 
first two columns of $A$.  
\bigskip
\noindent
II.  
\item{A)} (15) Give the general proof of Cramer's Rule
for systems $Ax = b$ with $A \in M_{n\times n}$ an invertible
matrix.
\Sol See the class notes or the proof of Theorem 7 in section 3.3 of the text.
\bigskip
\item{B)} (10) Give an example of a system $Ax = b$, where
$A$ is an invertible $3 \times 3$ matrix, but the second component
of the solution vector $x$ is $x_2 = 0$.
\Sol  
Here's the way I was thinking of this one.  By Cramer's rule, 
$x_2 = {\det(A_2(b))\over \det(A)}$.  So one way to get $x_2 = 0$ is
to make $b$ a linear combination of columns $1$ and $3$ of $A$.  Then
$\det(A_2(b)) = 0$.  An example is $A x = b$ where
$$A = \pmatrix{1 & 1 & 1\cr
               1 & 1 & 0\cr
               1 & 0 & 0\cr}, b = \pmatrix{1 \cr 1 \cr 1\cr}$$
Note that $\det(A_2(b))$ will have two equal columns in this case,
so its determinant is automatically zero. 
\bigskip\noindent
III.  Let $V = P_2({\bf R})$, the vector space of polynomials
of degree $\le 2$, and let $W = M_{2\times 2}({\bf R})$, the vector space of all 
$2\times 2$ matrices with real entries.
\bigskip
\item{A)} (10) Show that $T: V \to W$ defined by 
$T(p) = \pmatrix{p(1) & p(2) \cr p'(1) & p'(2)\cr}$
($p' = $ derivative of $p$) is a linear mapping.  
\Sol  
If $p,q \in V$, then 
$$T(p+q) = \pmatrix{(p+q)(1) & (p+q)(2) \cr (p+q)'(1) & (p+q)'(2)\cr}
= \pmatrix{p(1) + q(1) & p(2) + q(1) \cr p'(1) + q'(1) & p'(2) + q'(2)\cr}
=T(p) + T(q).$$
Similarly, if $r \in {\bf R}$,
$$T(rp) = \pmatrix{(rp)(1) & (rp)(2) \cr (rp)'(1) & (rp)'(2)\cr}
= r\pmatrix{p(1) & p(2) \cr p'(1) & p'(2)\cr}
=r T(p).$$
Hence $T$ is linear.
\bigskip
\item{B)} (10) Let
$K = \left\{A = \pmatrix{a & b \cr c & d \cr} \in W\ |\ \dim {\rm Col}(A) \le 1\right\}$
Is $K$ a vector subspace of $W$?  Why or why not?
\Sol
The answer is No.  $K$ is not closed under matrix sums.  The condition
$\dim {\rm Col}(A) \le 1$ means that the columns of $A$ are scalar
multiples (possibly both the zero vector).  If we take 
$$A = \pmatrix{1 & 0 \cr 0 & 0 \cr}, \quad B = \pmatrix{0 & 0 \cr 0 & 1 \cr}$$
Then $A,B \in W$ since each has column space of dimension 1.
But $A + B = \pmatrix{1 & 0 \cr 0 & 1 \cr}$ which is not in $K$, since
its column space is ${\bf R}^2$.  
(The other two properties for subspaces do hold in this case.)
\bigskip
\item{C)} (5 {\bf Extra Credit})  Let $T$ be the mapping from 
part A.   What is ${\rm Ker}(T)$?  Justify your answer.
\Sol
${\rm Ker}(T)$ is the set of polynomials $p = ax^2 + bx + c \in P_2({\bf R})$
such that $T(p) = \pmatrix{0 & 0 \cr 0 & 0 \cr}$, that is,
$p(1) = p(2) = 0$ and $p'(1) = p'(2) = 0$.  From the equation
on the derivatives we see that $y = p(x)$ must have two critical points
at $x = 1,2$.
This implies $a = 0$ since $a \ne 0$ implies $p'(x) = 2ax + b= 0$
at only one point $x = -b/(2a)$ (an ``honest-to-God'' parabola
has only one turning point).  But then the graph $y = p(x)$ is 
a straight line.  The conditions $p(1) = p(2)$ say that line
passes through $(1,0)$ and $(2,0)$.  Hence $b = c = 0$ as well.
The kernel contains only the zero polynomial.
  
\bigskip
\noindent
IV.  (True-False)  For each true statement, give a short proof.
For each false statement, give a counterexample.  (Note:
As always, ``true'' means ``true in every case.'' Partial credit {\it will}
be given for definitions or theorems that apply even if you don't find a 
complete solution.)
\bigskip
\item{A)} (10) If $a,b,c$ are distinct real numbers, then the matrix
$V = \pmatrix{1 & a & a^2\cr
           1 & b & b^2\cr
           1 & c & c^2\cr}$ 
is invertible.
\Sol TRUE.  Perform row operations to compute $\det(V)$:
$$\det \pmatrix{1 & a & a^2\cr
           1 & b & b^2\cr
           1 & c & c^2\cr}  = \det \pmatrix{1 & a & a^2\cr
                                            0 & b-a & b^2-a^2\cr
                                            0 & c-a & c^2-a^2\cr}.$$
We can pull out factors of $b-a$ from row 2 and $c - a$ from row 3:
$$\det(V) = (b-a)(c-a)\det \pmatrix{1 & a & a^2\cr
                                    0 & 1 & b+a\cr
                                    0 & 1 & c+a\cr} = (b-a)(c-a)(c-b).$$
This is zero only when $a = b$, $a = c$, or $b = c$.  If
$a,b,c$ are distinct, then $\det(V) \ne 0$ and $V$ is invertible.
\bigskip
\item{B)} (10)  If $A$ is a $5\times 7$ matrix,
then $\dim {\rm Nul} (A) \cdot \dim {\rm Col}(A) $ can equal 9.
\Sol FALSE.  $\dim {\rm Nul} (A) + \dim {\rm Col}(A) = 7$
and $\dim {\rm Col}(A) \le 5$.  None of the possible pairs
of numbers adding up to 7 have product = 9.
\bigskip
\item{C)} (10)  Let 
$$A = \pmatrix{1 & -2 & 0 \cr
               -2 & -1 & -1 \cr}$$
Any $3\times 2$ matrix $X$ such that $AX = 0_{2\times 2}$
has linearly dependent columns.  
\Sol
TRUE.  If $AX = 0$, the columns of $X$ are in ${\rm Nul}(A)$.
But note that the echelon form of $A$ will have two pivot columns
so $\dim {\rm Nul}(A) = 1$.  A set of two vectors in a one-dimensional
space must be linearly dependent (by the second ``theorem to know'').
\bye
