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\begin{document}
\examtitle{MATH 2270-2}{Exam 1}{Spring 2016}

\studentinfo
\instructions{No books or notes. No electronic devices, please.\\
The questions have credits 10 to 30, which reflects the time required to
write the solution.\\ If you must write a solution out of order or on
the back side, then supply a road map.
\\~
}
\finishfirstpage
% Problems Start Here % ----------------------------------------------------- %


\problem{10}{%
~\\
(a) Give a counter example or explain why it is true. If $A$ and $B$ are $n\times n$
invertible, and $C^T$ denotes the transpose of a matrix $C$, then
$((A+B)^{-1})^{T}=(B^T)^{-1} +  (A^{T})^{-1}$.\\
\medskip
(b) Give a counter example or explain why it is true. If $A$ is a square matrix and
$A^TA=I$, then both $A$ and $A^T$ are invertible.
}{}{%
~\\
(a) FALSE. If $A=A^T$ and $B=B^T$, then the claim is $(A+B)^{-1}=A^{-1} + B^{-1}$. If $B=-A$, then the left side is ill-defined (has no inverse). A counter example is $A=-B=I$.
\\
(b) By the determinant product theorem, $1=|I|=|A^TA|=|A^T||A|$. Then both $|A|$ and $|A^T|$ are nonzero, hence both $A$ and $A^T$ have an inverse.
}
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\problem{10}{%
Let $A$ be a $3\times 4$ matrix. Find the elimination matrix $E$ which
under left multiplication against $A$ performs both (1) and (2) with one matrix multiply.
\begin{itemize}
\item[{\bf (1)}] Replace Row 2 of $A$ with Row 2 minus Row 1.
\item[{\bf (2)}] Replace Row 3 of $A$ by Row 3 minus 5 times Row 2.
\end{itemize}
}{}{%
Perform combo(1,2,-1) on $I$ then combo(2,3,-5) on the result. The elimination matrix is
\[ E=\begin{pmatrix} 1 & 0 &  0\\
                     -1 & 1 & 0 \\
                    5 & -5 &  1 \end{pmatrix}\]
}
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\problem{30}{
Let $a$, $b$ and $c$ denote
constants and consider the system of equations
$$
\left(\begin{array}{ccc}
1  &  -b  & c \\
1  &   c   & a \\
2  &  -b+c  & -a
\end{array}\right)
\left(\begin{array}{r} x \\ y \\ z
\end{array}\right)
= \left(\begin{array}{r} a \\ -a \\ -a
\end{array}\right)
$$
Use techniques learned in this course to briefly explain the following facts. Only write what is needed to justify a statement.
\begin{itemize}
\item[{\bf (a)}.] The system has a unique solution for $(b+c)(2a+c)\ne 0$.
\newline
\item[{\bf (b)}.] The system has no solution if $2a+c=0$ and $a\ne 0$ (don't explain the other possibilities).
\newline
\item[{\bf (c)}.] The system has infinitely many solutions if $a=c=0$ (don't explain the other possibilities).
\end{itemize}

}{}{
% A4 := [[1, -b, c, a], [0, b+c, -c+a, -2*a], [0, 0, -c-2*a, -a]]
Combo, swap and mult are used to obtain in 3 combo steps the matrix
$$A_3=\left(\begin{array}{rrrr}1& -b& c& a\\ 0& b+c& -c+a& -2a\\ 0& 0& -c-2a& -a \end{array}\right)$$
(a) Uniqueness requires zeros free variables. Then the diagonal entries of the last frame must be nonzero, written simply as
$-(c+b)(2a+c)\ne 0$, which is equivalent to the determinant of $A$ not equal to zero.

(b) No solution: The last row of $A_3$ is a signal equation if $c+2a=0$ and $a\ne 0$. 

(c) Infinitely many solutions: If $a=c=0$, then $A_3$ has last row zero. If $a=c=0$ and $b=0$, then there is one lead variable and two free variables, because the last two rows of $A_3$ are zero. If $a=c=0$ and $b\ne 0$, then there are two lead variables and one free variable. The homogeneous problem has infinitely many solutions, because of at least one free variable and no signal equation.

The sequence of steps are documented below for maple.
%\begin{verbatim}
% with(LinearAlgebra):\newline
% combo:=(A,s,t,m)->LinearAlgebra[RowOperation](A,[t,s],m);\newline
% mult:=(A,t,m)->LinearAlgebra[RowOperation](A,t,m);\newline
% swap:=(A,s,t)->LinearAlgebra[RowOperation](A,[s,t]);\newline
% A:=(a,b,c)->Matrix([[1,b,c,-a],[1,c,-a,a],[2,b+c,a,a]]);\newline
% A0:=A(a,b,c);\newline
% A1:=combo(A(a,b,c),1,2,-1);\newline
% A2:=combo(A1,1,3,-2);\newline
% A3:=combo(A2,2,3,-1);\newline
% A4:=convert(A3,list,nested=true);\newline
% A4 := [[1, -b, c, a], [0, b+c, -c+a, -2*a], [0, 0, -c-2*a, -a]];
%\end{verbatim}
}
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%Continued\vfill\anotherpage
\problem{20}{%
{\bf Definition}. {\small\em Vectors $\vec{v}_1, \ldots, \vec{v}_k$ are
called {\bf independent} provided solving the equation $c_1 \vec{v}_1 +
\cdots + c_k \vec{v}_k = \vec{0}$ for constants $c_1,\ldots, c_k$ has
the unique solution $c_1=\cdots=c_k=0$. Otherwise the vectors are called
{\bf dependent}.}
\newline
Find a largest set of independent vectors from the following set of vectors,
using the Pivot Theorem or the definition of independence (above).
\[
\begin{pmatrix}
  1\\ 0 \\ 0 \\ 0 \\ 0
\end{pmatrix},
\begin{pmatrix}
  1\\ 2 \\ 0 \\ 0 \\ 0
\end{pmatrix},
\begin{pmatrix}
  0\\ 0 \\ 0 \\ 0 \\ 0
\end{pmatrix},
\begin{pmatrix}
  1\\ 0 \\ 0 \\ 0 \\ 1
\end{pmatrix},
\begin{pmatrix}
  3\\ 2 \\ 0 \\ 0 \\ 1
\end{pmatrix},
\begin{pmatrix}
  2\\ 0 \\ 0 \\ 0 \\ 1
\end{pmatrix}
\]
}{}{
The set of vectors is dependent. The augmented matrix of the six vectors
has pivot columns 1,2,4. A largest set of independent vectors are vectors 1, 2, 4.
}

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\problem{20}{
Find the vector general solution $\vec{x}$ to the equation $A\vec{x}=\vec{b}$ for
\[A=
\begin{pmatrix}
  1 & 0 & 0 &4\\
  3 & 0 & 1 &0 \\
  4 & 0 & 0 & 1
\end{pmatrix}
,\quad
\vec{b}=
\begin{pmatrix}
  0\\4 \\ 0
\end{pmatrix}
\]
}{}{
The augmented matrix for this system of equations is
\[\begin{pmatrix}
  1 & 0 & 0 &4 & 0\\
  3 & 0 & 1 &0 & 4 \\
  4 & 0 & 0 & 1 & 0
\end{pmatrix}\]
The reduced row echelon form is found as follows:
\[\begin{pmatrix}
  1 & 0 & 0 &4 & 0\\
  0 & 0 & 1 &-12 & 4 \\
  4 & 0 & 0 & 1 & 0
\end{pmatrix}\quad\mbox{combo(1,2,-3)}
\]
\[\begin{pmatrix}
  1 & 0 & 0 &4 & 0\\
  0 & 0 & 1 &-12 & 4 \\
  0 & 0 & 0 &-16 & 0
\end{pmatrix}\quad\mbox{combo(1,3,-4)}
\]
\[\begin{pmatrix}
  1 & 0 & 0 &4 & 0\\
  0 & 0 & 1 &-12 & 4 \\
  0 & 0 & 0 &1 & 0
\end{pmatrix}\quad\mbox{mult(3,-1/16)}
\]
\[\begin{pmatrix}
  1 & 0 & 0 &0 & 0\\
  0 & 0 & 1 &0 & 4 \\
  0 & 0 & 0 &1 & 0
\end{pmatrix}\quad\mbox{last frame}
\]
The last frame, or RREF, implies the system
$$\begin{array}{rrrrrrrrrrrr}
x_1 && &&     &&     &=& 0 \\
    && && x_3 &&     &=& 4\\
    && &&     && x_4 &=& 0
\end{array}
$$
The lead variables are $x_1,x_3,x_4$ and the free variable is $x_2$.
The last frame algorithm introduces invented symbol $t_1$. The free
variable is set to this symbol, then back-substitute into the lead
variable equations of the last frame to obtain the general solution
$$
\begin{array}{rrrrrrr}
x_1 &=& 0,\\
x_2 &=& t_1,\\
x_3 &=& 4,\\
x_4 &=& 0.
\end{array}
$$
Strang's {\em special solution} $\vec{s}_1$ is the partial of $\vec{x}$
on the invented symbol $t_1$. A particular solution $\vec{x}_p$ is
obtained by setting all invented symbols to zero. Then
$$
\vec{x} = \vec{x}_p + t_1\vec{s}_1 =
\left(\begin{array}{r}0\\ 0 \\ 4\\ 0\end{array}\right)
+t_1\left(\begin{array}{r}0\\ 1 \\ 0\\ 0\end{array}\right)
$$
}

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\problem{20}{
 Determinant problem, chapter 3. Parts reduced on Exam 1.\\
 (a) [10\%] True or False? The value of a determinant is the product of
 the diagonal elements.
\newline
 (b) [10\%] True or False? The determinant of the negative of the
 $n\times n$ identity matrix is $-1$.
\newline
 (c) [30\%] Assume given $3\times 3$ matrices $A$, $B$. Suppose
 $E_2E_1A^2=AB$ and $E_1$, $E_2$ are
 elementary matrices representing respectively a combination
and a multiply by $3$. Assume $\det(B)=27$.
 Let $C=-A$. Find all possible values of $\det(C)$.
\newline
(d) [20\%] Determine all values of $x$ for which $(2I+C)^{-1}$
fails to exist, where $I$ is the $3\times 3$ identity and $C=\left(
\begin{array}{cccc}
2 &   x  & -1 \\
3x &  0 & 1  \\
1 & 0 & -1
\end{array}\right)
$.
% A:=<2,x,-1|3*x,0,1|1,0,-1>^+ + 2*<1,0,0|0,1,0|0,0,1>;
% # ans = -5/3, 2
\newline
 (e) [30\%] Let symbols $a, b, c$ denote constants and define
$$ A=\left(\begin{array}{rrrr}
 1 & -1 &  0 & 0  \\
 1 & 0 &  0 & 0 \\
 a & b &  0 & 1 \\
 1 & c &  1 & \frac12
\end{array}\right)
$$
Apply the  adjugate [adjoint] formula for the inverse
$$A^{-1} = \frac{\adj(A)}{|A|}$$
to find the value of the entry in row
 $4$, column $2$ of $A^{-1}$.

}{}{
 ~ \\
(a)  FALSE. True only if the matrix is triangular.
\\
(b)  FALSE. It equals $1$ when $n$ is even.
\\
(c)  Start with the determinant product theorem $|FG|=|F||G|$. Apply it to obtain
$|E_2||E_1||A|^2=|A||B|$. Let $x=|A|$ in this equation and solve for $x$. You will need to know that
$|E_1|=1$ and $|E_2|=3$. Then $|C|=|(-I)A|=|-I||A|=(-1)^3x$. The answer is $|C|=0$ or $|C|=-9$.
(d)  Find $C+I=\left( \begin {array}{ccc} 4&x&-1\\ \noalign{\medskip}3\,x&2&1
\\ \noalign{\medskip}1&0&1\end {array} \right)
$, then evaluate its determinant, to eventually solve for $x=-5/3$ and $x= 2$. Used here is $F^{-1}$ exists if and only if $|F|\ne 0$.
\\
(e) Find the cross-out determinant in row 2, column 4 (no mistake, the transpose swaps rows and columns). Form the fraction, top=checkboard sign times cross-out determinant, bottom=$|A|$. The value is $-b-a$. A maple check:

\begin{verbatim}
 C4:=Matrix([[1,-1,0,0],[1,0,0,0],[a,b,0,1],[1,c,1,1/2]]);\newline
 1/C4; # The inverse matrix\newline
 C5:=linalg[minor](C4,2,4);\newline
 (-1)$^$(2+4)*linalg[det](C5)/linalg[det](C4);\newline
 # ans = -b-a
\end{verbatim}
}



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\begin{center} End Exam 1.\end{center}

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