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\begin{center}\large\bf
Linear Algebra 2270-4 \\
\normalsize\rm
Due in Week 2
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For the second week, the plan is to start chapter 1. Here's the
list of problems, followed by problem notes and a few answers. The Student Study Guide has answers and solutions to the odd problems. The textbook has answers to the odd problems.

\begin{description}
\item[{\bf Problem week2-1}.] State the {\em Three Possibilities} for the solution $\vec{x}$ of a matrix system $A\vec{x}=\vec{b}$. Give an example of each possibility in dimension $3$, then describe the geometry of each example.
\item[{\bf Problem week2-2}.] In section 2.8 the {\em nullspace} of a matrix $A$ is defined to be the set of all solutions $\vec{x}$ to the equation $A\vec{x}=\vec{0}$. Give a $3\times 3$  example in both matrix form and equation form. Then define {\em nullspace} for a system of equations.
%\item[{\bf Problem week2-3}.] The transpose of  matrix $A$ has associated nullspace $A^T\vec{y}=\vec{0}$. Both %$A\vec{x}=\vec{0}$ and $A^T\vec{y}=\vec{0}$ appear in the four-subspace figure on the cover of the textbook. Give an %example of a matrix $A$ for which both nullspaces have a simple geometric realization, but $\vec{x}$ and $\vec{y}$ are %not the same size.
\item[{\bf Section 1.1}.] Exercises  7, 11, 19, 25, 34.
\item[{\bf Section 1.2}.] Exercises 1, 7, 11, 13, 15, 17, 19, 23, 25, 28, 33.
\item[{\bf Section 1.3}.] Exercises 11, 13, 15, 21, 25, 34.
\end{description}

\subsection*{Some Answers}

Exercise 1.1-34: 

(1) Begin by interchanging R1 and R4, then create zeros in the first column.

(2) Scale R1 by –1 and R2 by 1/4, create zeros in the second column, and replace R4 by R4 + R3.

(3) Scale R4 by 1/12, use R4 to create zeros in column 4, and then scale R3 by 1/4.

(4) The last step is to replace R1 by R1 + (–1)R3.

The solution is (20, 27.5, 30, 22.5).

Exercise 1.2-28: Every column in the augmented matrix except the rightmost column is a pivot column, and the rightmost column is not a pivot column.

Exercise 1.3-34: a. For $j = 1,..., n$, $u_j + (-1)u j = (-1)u_j + u_j = 0$, by properties of $\Rr$. By vector equality,
$\vec{u} + (-1)\vec{u} = (-1)\vec{u} + \vec{u} = \vec{0}$.

b. For scalars $c$ and $d$, the $j$th entries of $c(d\vec{u})$ and $(cd )\vec{u}$ are $c(du_j )$ and $(cd )u_j$ , respectively. These
entries in $\Rr$ are equal, so the vectors $c(d\vec{u})$ and $(cd)\vec{u}$ are equal.


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