{\bf\noindent 10.}

$(0, y_1)$, $(1, y_2)$, and $(2, y_3)$ lie on the same line if $(2, y_3)
= k(1, y_2 - y_1) + (0, y_1) = (k, ky_2 - (k-1)y_1) \to y_3 = 2y_2 - y_1.$

{\bf\noindent 11.}

For $a = 2$ and $a = -2,$ the columns are linearly dependent, and there
is a line of solutions.

{\bf\noindent 15.}

"Lines." "2." "the column vectors."

{\bf\noindent 22.}

If $(a,b)$ is a multiple of $(c,d),$ $c/a = d/b \to c/d = a/b,$ so $(a,
c)$ is a multiple of $(b, d)$.

{\bf\noindent 5.}

$0$ gives no solutions. $20$ gives infinitely many solutions. These
solutions include $(4, -2)$ and $(0, 5).$

{\bf\noindent 8.}

$k=3,$ $k=0,$ and $k=-3$ cause elimination to break down. $k=3$ makes
the system inconsistent, so it has 0 solutions, $k=-3$ causes
infinite solutions, and $k=0$ is consistent with 1 solution but requires
a row exchange.

{\bf\noindent 12.}

If $d=10,$ a row exchange is required, giving a triangular system
$$\pmatrix{2&5&1\cr 0&1&-1\cr 0&0&-1}\pmatrix{x\cr y\cr z} =
\pmatrix{0\cr 3\cr 2}$$

{\bf\noindent 19.}

"Combination." $2x - y = 0$ cannot be solved.

{\bf\noindent 28.}

{\it (a)}

False. If the second row doesn't start with a zero coefficient, then a
multiple of row 1 will be (indirectly) subtracted from row 3 when row 2
is subtracted from row 3.

{\it (b)}

False. After eliminating the $u$ column from the third row, a $v$
``residue'' might remain.

{\it (c)}

True. The third row is already fully ``solved'' for back-substitution.

{\bf\noindent 22.}

{\it (a)}

$$\pmatrix{1&0&0\cr -5&1&0\cr 0&0&1\cr}$$

{\it (b)}

$$\pmatrix{1&0&0\cr 0&1&0\cr 0&-7&1\cr}$$

{\it (c)}

$$\pmatrix{0&1&0\cr 0&0&1\cr 1&0&0\cr}$$

{\bf\noindent 27.}

$R_{31}$ should add 7 times row 1 to row 3. $E_31R_31 = I_3.$

{\bf\noindent 29.}

{\it (a)}

$$E_{13} = \pmatrix{1&0&1\cr 0&1&0\cr 0&0&1}$$

{\it (b)}

$$\pmatrix{1&0&1\cr 0&1&0\cr 1&0&1}$$

{\it (c)}

$$\pmatrix{2&0&1\cr 0&1&0\cr 1&0&1}$$

{\bf\noindent 42.}

{\it (a)}

True.

{\it (b)}

False, they just have to be $m\times n$ and $n\times m.$

{\it (c)}

True, but they don't have the same dimensions.

{\it (d)}

False. This is only true if $B$ is invertible.

{\bf\noindent 51.}

$AX = I_3.$

{\bf\noindent 6.}

$$E^2 = \pmatrix{1&0\cr12&1}$$
$$E^8 = \pmatrix{1&0\cr48&1}$$
$$E^{-1} = \pmatrix{1&0\cr-6&1}$$

{\bf\noindent 9.}

{\it (a)}

If none of $d_1,$ $d_2,$ or $d_3$ are zero, the product is nonsingular.
% Prove it

{\it (b)}

Solving this first system, $c = b,$ by substitution.

Then we have $$Dd = c \to d = \pmatrix{0\cr0\cr 1/d_3}$$
and $$Vx = d \to \pmatrix{1 & -1 & 0\cr 0 & 1 & -1 \cr 0 & 0 &
1}\pmatrix{x_1\cr x_2\cr x_3} = d \to x_3 = x_2 = x_1 = 1/d_3.$$

{\bf\noindent 19.}

In the second matrix, $c=0$ requires a row exchange, and $c=3$ would
make the matrix singular.

In the first matrix, it is singular if $3b = 40-10a.$
And it requires a row exchange if $a=4$ and $b\neq 0.$

{\bf\noindent 31.}

$$\pmatrix{1&1&0\cr 1&2&1\cr 0&1&2} = \pmatrix{1&0&0\cr 1&1&0\cr 0&1&1}
\pmatrix{1&1&0\cr 0&1&1\cr 0&0&1} =
\pmatrix{1&0&0\cr 1&1&0\cr 0&1&1}\pmatrix{1&0&0\cr0&1&0\cr0&0&1}\pmatrix{1&1&0\cr 0&1&1\cr 0&0&1}
$$

$$\pmatrix{a&a&0\cr a&a+b&b\cr 0&b&b+c} =
\pmatrix{1&0&0\cr1&1&0\cr0&1&1}\pmatrix{a&a&0\cr 0&b&b\cr 0&0&c} =
\pmatrix{1&0&0\cr1&1&0\cr0&1&1}\pmatrix{a&0&0\cr0&b&0\cr0&0&c}\pmatrix{1&1&0\cr 0&1&1\cr 0&0&1}
$$

{\bf\noindent 32.}

$$Lc = b \to \pmatrix{1&0\cr4&1}c = \pmatrix{2\cr 11} \to c =
\pmatrix{2\cr 3}.$$
$$Ux = c \to \pmatrix{2&4\cr0&1}x = \pmatrix{2\cr 3} \to x =
\pmatrix{-5\cr 3}.$$

$$A = LU = \pmatrix{1&0\cr4&1}\pmatrix{2&4\cr0&1} =
\pmatrix{2&4\cr8&17}.$$
$$\pmatrix{2&4\cr8&17}x = \pmatrix{2\cr 11} \to \pmatrix{2&4\cr0&1}x =
\underline{\pmatrix{2\cr3}} \to x = \pmatrix{-5\cr 3}$$

\bye
