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\section*{Strang: Chapter 3}
\begin{description}

\item[{\bf Section 3.1}.] Exercises 1, 3, 11, 12, 17, 18, 22, 23

\item[{\bf Section 3.2}.] Exercises 4, 10, 15, 17, 18, 19, 24, 37

\item[{\bf Section 3.3}.] Exercises 1, 12, 21

\item[{\bf Section 3.4}.] Exercises 1, 3, 4, 6, 13, 16, 33

\item[{\bf Section 3.5}.] Exercises 2, 5, 9, 10, 11, 15, 18, 23, 26, 31, 45

\item[{\bf Section 3.6}.] Exercises 1, 2, 5, 7, 11, 24

\end{description}



\subsection*{Some Answers}


{\bf 3.1}. Exercises 1, 3, 11, 23 are in the textbook.

{\bf 3.1-12}. For the plane $x+y-2z=4$, the sum of $(4,0,0)$ and $(0,4,0)$
is not on the plane. (The key is that this plane does not go through $(0,0,0)$.)

{\bf 3.1-17}.  (a) The invertible matrices do not include the zero
matrix, so they are not a subspace

(b) The sum of singular matrices
$\twobytwo{1& 0}{0& 0}+\twobytwo{0& 0}{ 0& 1}$
is not singular: not a subspace.

{\bf 3.1-18}. (a) True: The symmetric matrices do form a subspace (b)
True: The matrices with $A^T=-A$ do form a subspace (c) False: The sum
of two unsymmetrical matrices could be symmetric.


{\bf 3.1-22}. (a) Solution for every b (b) Solvable only if $b_3=0$. (c)
Solvable only if $b_3=b_2$.

\bigskip

{\bf 3.2}. Exercises  4, 15, 24, 37 are in the textbook.

{\bf 3.2-10}. (a) Impossible row 1 (b) $A=$ invertible (c) $A=$ all ones
(d) $A=2I$, $R=I$.

{\bf 3.2-17}. $A=\colvector{1\\ -3\\ -1}$ gives the plane $x-3y-z=12$;
$y$ and $z$ are free variables. The special solutions are $(3,1,0)$ and
$(1,0,1)$.

{\bf 3.2-18}. Fill in 12 then 4 then 1 to get the complete solution to $x-3y-z=12$:
$$\displaystyle \colvector{x\\ y\\ z} = \colvector{12\\ 0\\ 0}+y\colvector{4\\ 1\\ 0} + z\colvector{1\\ 0 \\ 1} = x_{\mbox{particular}} + x_{\mbox{nullspace}}$$.

{\bf 3.2-19}. If $LUx=0$, multiply by $L^{-1}$ to find $Ux=0$. Then $U$
and $LU$ have the same nullspace.

\bigskip

{\bf 3.3}. Exercises 1, 21 have a textbook answer.

{\bf 3.3-12}.
Invertible $r$ by $r$ submatrices
use pivot rows and columns
$S= \twobytwo{1& 3}{ 1& 4}$
and $S=[1]$  and $S= \twobytwo{1& 0}{ 0& 1}$.

\bigskip

{\bf 3.4}. Exercises 4, 6, 13, 16 have a textbook answer.

{\bf 3.4-1}. Row reduce the augmented matrix to upper triangular form
$$\left(\begin{array}{rrrr|l}
2 & 4 & 6 & 4 & b_1 \\
0 & 1 & 1 & 2 & b_2 - b_1 \\
0 & 0 & 0 & 0 &b_3 + b_2 -2b_1
\end{array}\right)
$$
Then $Ax=b$ has a solution when the last row is all zeros. This is the plane
given by the equation $b_3+b_2-2b_1=0$. The nullspace is obtained by solving
$Ax=0$, which is a step away by back-substitution. The answer is
$\vec{x}_{\mbox{nullspace}}=c_1\vec{s}_1+c_2\vec{s}_2$ where $\vec{s}_1=\colvector{-1\\ -1\\ 1\\ 0}$, $\vec{s}_2=\colvector{2 \\ -2 \\ 0 \\ 1}$.
Then the complete solution is
$c_1\vec{s}_1+c_2\vec{s}_2+\colvectorC{b_1\\ b_2-b_1\\ 0}$,
subject to the restraint $b_3+b_2-2b_1=0$ ($b_1, b_2$ unrestrained).
Choosing $b_1=4$ and $b_2=3$ with $c_1=c_2=0$ gives particular solution
$\vec{x}_{\mbox{particular}}=\colvector{4\\ -1\\ 0\\ 0}$.

{\bf 3.4-3}. $\vec{x}_{\mbox{complete}} = \colvector{-2\\ 0\\ 1} + x_2\colvector{-3\\ 1 \\ 0}$.
The matrix is singular but the equations are still
solvable; $b$ is in the column space. Our particular solution has free
variable $y=0$.


{\bf 3.4-33}.
If the complete solution to $Ax=\colvector{1\\ 3}$ is $x=\colvector{1\\
0}+\colvector{0\\ c}$ then $A=\twobytwo{ 1& 0}{ 3& 0}$.

\bigskip



{\bf 3.5}. Exercises 2, 11, 15, 18 have textbook answers.

{\bf 3.5-5}. (a) Reduce to upper triangular
$\threebythree{1&2&3}{0&-5&-7}{0&0&-18/5}$. Invertible implies independent columns.

(b) Reduce to upper triangular
$\threebythree{1&2&-3}{0&7&-7}{0&0&0}$. Then $A\colvector{1\\ 1\\ 1}=\colvector{0\\ 0\\ 0}$,
columns add to zero.


{\bf 3.5-9}. a) The four vectors in $\Rr^3$ are the columns of a $3$ by
$4$ matrix $A$. There is a nonzero solution to $Ax=0$ because there is
at least one free variable

(b) Two vectors are dependent if $[v_1| v_2]$ has rank $0$ or $1$. (OK to say "they are on the same line" or "one is
a multiple of the other" but not $v_2$ is a multiple of $v_1$ -- since
$v_1$ might be $0$.)

(c) A nontrivial combination of $v_1$ and $0$ gives $0$:
$0\vec{v}_1+3\vec{0}=\vec{0}$.


{\bf 3.5-10}. The plane is the nullspace of $1$ by $4$ matrix $A=\left[\begin{array}{rrrr} 1 &2& -3& -1\end{array}\right]$. Three free
variables give three solutions $(x, y, z, t)= (2,-1,0,0)$, $(3,0,1,0)$
and $(1,0,0,1)$. Combinations of those special solutions give more
solutions (all solutions).

{\bf 3.5-23}. Columns 1 and 2 are bases for the (different) column
spaces of $A$ and $U$; rows 1 and 2 are bases for the (equal) row spaces
of $A$ and $U$; $(1,-1,1)$ is a basis for the (equal) nullspaces

{\bf 3.5-26}.

\def\AA{\threebythree{1& 0& 0}{0& 0& 0}{0& 0& 0}}
\def\BB{\threebythree{0& 0& 0}{0& 1& 0}{0& 0& 0}}
\def\CC{\threebythree{0& 0& 0}{0& 0& 0}{0& 0& 1}}
(a) $\AA,\BB,\CC$

\def\AA{\threebythree{0& 1& 0}{1& 0& 0}{0& 0& 0}}
\def\BB{\threebythree{0& 0& 1}{0& 0& 0}{1& 0& 0}}
\def\CC{\threebythree{0& 0& 0}{0& 0& 1}{0& 1& 0}}
(b) Add $\AA,\BB,\CC$

\def\AA{\threebythree{0& 1& 0}{-1& 0& 0}{0& 0& 0}}
\def\BB{\threebythree{0& 0& 1}{0& 0& 0}{-1& 0& 0}}
\def\CC{\threebythree{0& 0& 0}{0& 0& 1}{0& -1& 0}}
(c) $\AA,\BB,\CC$

These are simple bases (among many others) for (a) diagonal matrices (b)
symmetric matrices (c) skew-symmetric matrices. The dimensions are $3,
6, 3$.


{\bf 3.5-31}.  (a) $y(x)$ = constant $C$ (b) $y(x)=3x$ (c) $y(x)=3x+C=y_p+y_n$ solves $dy/dx=3$.

{\bf 3.5-45}.  If the left side of $\dim(V)+\dim(W)=\dim(V\cap
W)+\dim(V+W)$ is greater than $n$, then $\dim(V+W)\le n$ implies $\dim(V\cap W)$ must be
greater than zero. So $V\cap W$ contains nonzero vectors.

\bigskip

{\bf 3.6}. Exercises 1, 11, 24 have textbook answers.

{\bf 3.6-2}.

A: Row space basis = row 1 =$(1,2,4)$; nullspace $(-2,1,0)$ and $(-4,0,1)$; column
space basis = column 1 = $(1,2)$; left nullspace $(-2,1)$.

B: Row space basis = both rows = $(1,2,4)$ and $(2,5,8)$;
 column space basis = two columns = $(1,2)$ and $(2,5)$; nullspace $(-4,0,1)$; left nullspace basis is empty because the space contains
only $y= 0$.

{\bf 3.6-5}. $A=\left(\begin{array}{rrr} 1& 1& 1\\ 2& 1& 0 \end{array}\right)$ has those rows spanning its row space;
$B=\left[\begin{array}{rrr}1 & -2& 1 \end{array}\right]$ has the same rows spanning its nullspace and $BA^T= 0$.

{\bf 3.6-7}. Invertible $3$ by $3$ matrix $A$: row space basis = column space basis = $(1,0,0)$, $(0,1,0)$, $(0,0,1)$;
nullspace basis and left nullspace basis are empty. Matrix $B=\left[\begin{array}{rrr}A & A\end{array}\right]$: row
space basis $(1,0,0,1,0,0)$, $(0,1,0,0,1,0)$,  and $(0,0,1,0,0,1)$; column space basis
$(1,0,0)$, $(0,1,0)$, $(0,0,1)$; nullspace basis $(-1, 0, 0, 1, 0, 0)$ and $(0,-1,0,0,1,0)$ and
$(0,0,-1,0,0,1)$; left nullspace basis is empty.

{\bf 3.6-28}. $B$ and $C$ (checkers and chess) both have rank $2$ if $p\ne 0$. Row $1$ and row $2$ are a basis for the
row space of $C$, $B^Ty= 0$ has 6 special solutions with $-1$ and $1$ separated by a zero;
$N(C^T)$ has $(-1,0,0,0,0,0,0,1)$ and $(0,-1,0,0,0,0,1,0)$ and columns $3,4,5,6$ of
the identity $I$; $N(C)$ is a challenge.



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