\def\bmatrix#1{\left[\matrix{#1}\right]}

    {\noindent\bf Section 2.4}

{\noindent\bf 6.}

{\it (a)}

It has a two-sided inverse if $r = m = n.$

{\it (b)}

It has infinitely many solutions if $r = m < n.$

{\noindent\bf 12.}

{\it (a)}

Matrix $A$ is of rank 1 and equal to
$$\bmatrix{1\cr 0\cr 2}\bmatrix{1&0&0&3}$$

{\it (b)}

Matrix $A$ is of rank 1 and equal to
$$\bmatrix{2\cr 6}\bmatrix{1&-1}$$

{\noindent\bf 18.}

The row space has basis:
$$\{\bmatrix{0\cr1\cr2\cr3\cr4}, \bmatrix{0\cr0\cr0\cr1\cr2}\},$$
the null space has basis:
$$\{\bmatrix{1\cr0\cr0\cr0\cr0}, \bmatrix{0\cr-2\cr1\cr0\cr0},
\bmatrix{0\cr0\cr0\cr-2\cr0}\},$$
the column space has basis:
$$\{\bmatrix{1\cr1\cr0},\bmatrix{3\cr4\cr1}\},$$
and the left null space has basis:
$$\{\bmatrix{1\cr-1\cr1}\}.$$

{\noindent\bf 32.}

$A$ has column space of the xy-plane, and left null space of the z-axis.
It also has row space of the yz-plane, and null space of the x-axis.

$I+A$ has full rank, so its column space and row space are ${\bf R}^3,$
and its null space and left null space are the zero vector.

\iffalse % practice problems

{\noindent\bf 2.}

{\noindent\bf 3.}

{\noindent\bf 8.}

{\noindent\bf 9.}

{\noindent\bf 10.}

{\noindent\bf 16.}

{\noindent\bf 17.}

{\noindent\bf 21.}

{\noindent\bf 25.}

{\noindent\bf 27.}

{\noindent\bf 35.}

{\noindent\bf 37.}

\fi

    {\noindent\bf Section 2.6}

{\noindent\bf 16.}

$$\bmatrix{0&1&0&0\cr 0&0&1&0\cr 0&0&0&1\cr 1&0&0&0}$$
If $A$ maps $(x_1, x_2, x_3, x_4)$ to $(x_2, x_3, x_4, x_1),$ $A^2$ maps
$x$ to $(x_3, x_4, x_1, x_2),$ and $A^3$ takes $x$ to $(x_4, x_1, x_2,
x_3),$ and $AA^3 = I = A^4,$ so $A^3 = A^{-1}$ by definition of the
identity.

{\noindent\bf 28.}

{\it (a)}

Range is $V^2,$ and kernel is $0.$

{\it (b)}

Range is $V^2,$ and kernel has basis $(0, 0, 1).$

{\it (c)}

Range is $0,$ and kernel is $V^2$

{\it (d)}

Range is the subspace with basis $(1, 1)$ and kernel has basis $(0, 1).$

{\noindent\bf 36.}

{\it (a)}

$$\bmatrix{2&5\cr 1&3}$$

{\it (b)}

$$\bmatrix{3&-5\cr -1&2}$$

{\it (c)}

Because, by linearity, if $(2, 6) \mapsto (1, 0),$ $.5(2,6) = (1, 3)
\mapsto (.5, 0).$

{\noindent\bf 44.}

This is equivalent to a 180$^\circ$ rotation.

\iffalse % practice problems

{\noindent\bf 6.}

{\noindent\bf 7.}

{\noindent\bf 8.}

{\noindent\bf 9.}

{\noindent\bf 17.}

{\noindent\bf 40.}

{\noindent\bf 45.}

\fi

\bye
