\def\bmatrix#1{\left[\matrix{#1}\right]}
{\bf\noindent Section 1.6}
{\bf\noindent 4.}
{\it (a)}
That means $A^{-1}$ exists s.t. $A^{-1}A = I.$
$$AB = AC \to A^{-1}AB = A^{-1}AC \to B = C$$

{\it (b)}
$$B = \pmatrix{1 & 1\cr 1 & 0}$$
$$C = \pmatrix{1 & 2\cr 1 & 1}$$

$$AB = AC.$$ % double check

{\bf\noindent 10.}

$$A_1^{-1} = \bmatrix{0 & 0 & 0 & 1\cr 0 & 0 & 1/2 & 0\cr 0 & 1/3 & 0 &
0\cr 1/4 & 0 & 0 & 0}$$

$$A_2^{-1} = \bmatrix{1 & 0 & 0 & 0\cr 1/2 & 1 & 0 & 0\cr 0 & 2/3 & 1 &
0\cr 0 & 0 & 3/4 & 1}$$

\def\rowone{{d\over ad-bc}&{-b\over ad-bc}}
\def\rowtwo{{-c\over ad-bc}&{a\over ad-bc}}
$$A_3^{-1} = \bmatrix{\rowone & 0 & 0\cr \rowtwo & 0 & 0\cr 0 & 0 &
\rowone \cr 0 & 0 & \rowtwo}$$

{\bf\noindent 11.}

If $A = I$ and $B = -I,$ both are invertible, but $A + B$ is not.

If $A = \pmatrix{1&0\cr 0&0}$ and $B = \pmatrix{0&0\cr 0&1},$ $A + B =
I$ is invertible, but neither $A$ nor $B$ are.

$A = B = I$ gives $A + B,$ $A,$ and $B$ invertibility.
$A^{-1} = B^{-1} = I \to A^{-1}(A+B)B^{-1} = B^{-1} + A^{-1} \to 2I =
2I.$

{\bf\noindent 12.}

{\it (a)}

This one remains.

{\it (b)}

This one remains.

{\it (c)}

This one remains.

{\it (d)}

This one doesn't remain.

{\it (e)}

This one remains.

% I don't really have any coherent reasons for this.
% But these all make a lot of sense to me.
% This isn't graded, but I would like to understand it better. Office
% hours?

{\bf\noindent 23.}

$$\bmatrix{10&20\cr 20&50}\bmatrix{x\cr y} = \bmatrix{1\cr 0} \to
\bmatrix{10&20\cr 0&10}\bmatrix{x \cr y} = \bmatrix{1\cr -2}$$$$\to
\bmatrix{1&2\cr 0&1}\bmatrix{x \cr y} = \bmatrix{1/10\cr -2/10} \to
\bmatrix{1&0\cr 0&1}\bmatrix{x \cr y} = \bmatrix{5/10\cr -2/10}.
$$

$$\bmatrix{10&20\cr 20&50}\bmatrix{x\cr y} = \bmatrix{0\cr 1} \to
\bmatrix{1&2\cr 0&1}\bmatrix{x\cr y} = \bmatrix{0\cr 1/10} \to
\bmatrix{1&0\cr 0&1}\bmatrix{x\cr y} = \bmatrix{-2/10\cr 1/10}.
$$

$$A^{-1} = {1\over10}\bmatrix{5&-2\cr -2&1}$$

{\bf\noindent 25.}

{\it (a)}
$$\bmatrix{\hbox{row 1}\cr\hbox{row 2}\cr\hbox{row 1 + row
2}}\bmatrix{x\cr y\cr z} = \bmatrix{1\cr0\cr0}
\to \bmatrix{\hbox{row 1}\cr\hbox{row 2}\cr0}\bmatrix{x\cr y\cr z} =
\bmatrix{1\cr0\cr-1}.$$
$0 \neq -1,$ so this is inconsistent.

{\it (b)}

If $b_3 = b_2 + b_1,$ this matrix admits a solution.

{\it (c)}

As already shown, row 3 becomes a zero row during elimination.

{\bf\noindent 27.}

$B = E_12A,$
and the elementary matrix which switches the first two rows is by
definition invertible ($E_12E_12 = I$) so if $A$ is invertible, $B$ is
also invertible, and $B^{-1} = A^{-1}E_12$ (I think this is a switch of
columns).

{\bf\noindent 36.}

$$\bmatrix{2 & 1 & 0 & 1 & 0 & 0\cr
           1 & 2 & 1 & 0 & 1 & 0\cr
           0 & 1 & 2 & 0 & 0 & 1}
\to
\bmatrix{2 & 1 & 0 & 1 & 0 & 0\cr
           0 & 1.5 & 1 & .5 & 1 & 0\cr
           0 & 1 & 2 & 0 & 0 & 1}
\to
\bmatrix{2 & 1 & 0 & 1 & 0 & 0\cr
           0 & 1 & 0 & .5 & 1 & .5\cr
           0 & 1 & 2 & 0 & 0 & 1}
$$$$\to
\bmatrix{2 & 0 & 0 & .5 & -1 & -.5\cr
           0 & 1 & 0 & .5 & 1 & .5\cr
           0 & 1 & 2 & 0 & 0 & 1}
\to
\bmatrix{2 & 0 & 0 & .5 & -1 & -.5\cr
           0 & 1 & 0 & .5 & 1 & .5\cr
           0 & 0 & 2 & -.5 & -1 & .5}
$$

Inverse is
$$
\bmatrix{.25 & -.5 & -.25\cr
         .5 & 1 & .5\cr
         -.25 & -.5 & .25}
$$

% {\bf\noindent 40.}

{\bf\noindent Section 2.1}
{\bf\noindent 2.}

{\it (a)}

This is. $k(0, b_2, b_3) = (0, kb_2, kb_3),$ and similar for addition.

{\it (b)}

This isn't. $2(0, 1, 0) = (0, 2, 0),$ which isn't in the subset.

{\it (c)}

This isn't. $(1, 1, 0) + (1, 0, 1) = (2, 1, 1),$ which isn't in the
subset, even though its two starting components are.

{\it (d)}

This is, by definition.

{\it (e)}

Yes, this is the null space of a homogenous matrix with one row. It is a
subspace.

{\bf\noindent 7.}

{\it (a)}

This is not a subspace. The given sequence plus the given sequence with
a zero prepended would sum to a sequence without a zero, meaning it is
not closed under addition.

{\it (b)}

This is a subspace because the sums and multiples are closed.

{\it (c)}

$x_{j+1} \leq x_j \land y_{j+1} \leq y_j \to x_{j+1} + y_{j+1} \leq x_j
+ y_j,$
and similar for multiplication.

{\it (d)}

Yes, the sum of two convergent sequences will tend to the sum of their
limits, and each can be multiplied by a real.

{\it (e)}

Yes, the sum of two arithmetic series will be an arithmetic series, and
they work with multiplications by a real.

{\it (f)}

This isn't because $x_1 = 1,$ $k = 2,$ and $x_1 = 1,$ $k = 3$ sum to
$$2 + 5 + 14 + \cdots,$$ which is not a geometric sequence.

{\bf\noindent 8.} % required

The two rows are linearly independent, so with 3 unknows, this forms a
line (intersection of two equations/planes).

{\bf\noindent 18.}

{\it (a)}

It's probably a line, but it could be a plane.

{\it (b)}

It's probably $(0, 0, 0)$, but it could be a line.

% {\it (c)}

% I don't know.

{\bf\noindent 22.} % required

{\it (a)}

This matrix has a column space with basis $(1, 2, -1)^T,$ so it is only
solvable if $b = (b_1, b_2, b_3)^T$ is a multiple of that vector.

{\it (b)}

These two columns are linearly independent, so any b within the span of
them has a solution.

{\bf\noindent 25.}

Unless $b$ is in the span of $A.$ If A is the zero $3\times 1$ matrix,
and $b$ is the first column of $I_3,$ then the column space increases.
If $A = b,$ then it would not extend the column space. Iff $b$ is
already in the column space, then $Ax = b$ has a solution.

{\bf\noindent 26.}

They are not equal only if $B$ is not invertible (it is a necessary but
not sufficient condition). If $A$ is invertible, like $A = I,$ then
non-invertible $B$ gives $AB = B$ with smaller column space.

{\bf\noindent 28.} % required

{\it (a)}

False. Counterexample: The complement of $(0, 0, 0)$ does not pass the
test $0v \in V.$

{\it (b)}

True. The space with only the zero vector has a basis of only zero
vectors.

{\it (c)}

True. Each of $2A$'s columns can be transformed into the columns of $A$
by division by 2.

{\it (d)}

False. The column space of $-I$ is the full space, unlike the zero
matrix it's ``based on.''

{\bf\noindent 30.} % required

$${\bf C}(A) = {\bf R}^9.$$

\bye
