\def\bmatrix#1{\left[\matrix{#1}\right]}
\def\fr#1#2{{#1\over #2}}

    {\bf Section 4.2}

\noindent{\bf 12.}

$$\det\bmatrix{1&a&a^2\cr1&b&b^2\cr1&c&c^2} =
\det\bmatrix{1&a&a^2\cr0&b-a&b^2-a^2\cr0&c-a&c^2-a^2} =
(b-a)(c-a)\det\bmatrix{1&a&a^2\cr0&1&b+a\cr0&1&c+a} =$$$$
(b-a)(c-a)\det\bmatrix{1&a&a^2\cr0&1&b+a\cr0&0&c-b}
= (b-a)(c-a)(c-b).
$$

\noindent{\bf 17.}

The determinant of $A$ is $4*3 - 2*1 = 10.$

$$\det(AA^{-1}) = \det I = 1 = \det(A)\det(A^{-1}) \to \det(A^{-1}) =
1/\det(A) = 1/10.$$

$$\det(A-\lambda I) = (4-\lambda)(3-\lambda) - 2 = \lambda^2 - 7\lambda
+ 10 = (\lambda - 2)(\lambda - 5).$$
This gives zeroes (meaning $A-\lambda I$ is singular) of $\lambda = 2, 5.$

\noindent{\bf 34.}

By linearity in each row, the determinant of B is the sum of the
determinants of every choice from the original set. However, if any row
is repeated in the matrix being determined, the determinant is zero, so
\def\row{\mathop{\rm row}}
$$\det B = \left|\matrix{\row 1\cr\row 2\cr\row 3}\right|
+ \left|\matrix{\row 2\cr\row 3\cr\row 1}\right|
= 6 + 6,$$
the determinant of the second matrix being equivalent to $\det A$ by two
row switches.

\iffalse % this is the answer to 35 :'(
Row operations give us
$$\det(I+M) = \det\bmatrix{1+a & b & c & d\cr -1 & 1 & 0 & 0\cr -1 & 0 &
1 & 0\cr -1 & 0 & 0 & 1} =
\det\bmatrix{1+a+b+c+d & 0 & 0 & 0\cr -1 & 1 & 0 & 0\cr -1 & 0 &
1 & 0\cr -1 & 0 & 0 & 1},$$
giving determinant by product of the diagonal entries $1+a+b+c+d.$
\fi

    {\bf Section 4.3}

\noindent{\bf 3.}

{\it (a)}

True.
$$\det(S^{-1}AS) = \det(S^{-1})\det(A)\det(S) = \det(A)\det(S^{-1}S) =
\det(A).$$

{\it (b)}

False, the matrix
$$\bmatrix{1&1\cr 1&1}$$
has determinant $1*1 - 1*1,$ which is a cofactor expansion with every
cofactor either $1$ or $-1,$ but the determinant is still zero.

{\it (c)}

False,
$$\left|\matrix{1&1&0\cr1&0&1\cr0&1&1}\right| = -2.$$

\noindent{\bf 6.}

{\it (a)}

With $a_{ij}$ the entry on $A$ in the $i$th row and $j$th column, and
$c_{ij}$ the determinant of the matrix minor of $A$ (matrix $A$ without
the $i$th row or $j$th column).

$$D_n = a_{11}c_{11} - a_{12}c_{12} + a_{13}c_{13} + \cdots = c_{11} -
c_{12} + 0 + \cdots = D_{n-1} - D_{n-2}.$$
$c_{11}$ is clearly $D_{n-1}$ because the remainder after that elimination
is an $n-1 \times n-1$ tridiagonal matrix with 1s on the diagonals.

$c_{12}$ is an $n-2\times n-2$ tridiagonal with $1,1,0,\ldots$ as the
first row, inserted above it, and $1,0,0,\ldots$ to its left as the
first column (overlapping with the first row). This gives a simple
cofactor expansion (computing on the first column and ignoring the
zeroes) of the determinant of the $n-2\times n-2$ matrix. Thus, $c_{12}
= D_{n-2}.$

{\it (b)}

$D_3 = 0 - 1,$ $D_4 = -1 - 0,$ $D_5 = -1 - (-1) = 0,$ $D_6 = 0 - (-1) =
1,$ $D_7 = 1 - 0,$ $D_8 = 1 - 1 = 0.$
Therefore, the cycle has period 6, and $1000\bmod6 = 4,$ so $D_{1000}
= D_4 = -1.$

\noindent{\bf 15.}

$\det A$ is zero because this is a triangular matrix, so the determinant
is the product of the diagonal entries $x\cdot0\cdot x = 0.$
The rank of $A$ is 2 unless $x = 0,$ in which case it is 0. This is
because the first and second columns are linearly dependent.

\noindent{\bf 34.}

{\it (a)}

Row operations (including permutations) which make $A$ and $D$ diagonal
will be contained within their respective ``block-rows,'' and once we've
got that, the product of the diagonal entries is the same as the product
of the determinants of the new $A$ and $D$ blocks (because those
triangular matrices have determinant equal to the product of diagonal
entries). None of this requires knowledge of $B.$

{\it (b)}

$$
\det
\bmatrix{ 0 & 0 & 1 & 0\cr
          0 & 0 & 0 & 1\cr
          1 & 0 & 0 & 0\cr
          0 & 1 & 0 & 0 }
= 1,
$$
because this can be permuted to give the identity in two operations
(switch rows 3 and 1 and rows 4 and 2).

However, $B = C = I_2,$ which would give our block determinant formula
$|A||D| - |C||B| = 0 - 1 \neq 1.$

{\it (c)}

$$
\det
\bmatrix{ 0 & 1 & 0 & 0\cr
          0 & 0 & 1 & 0\cr
          0 & 0 & 0 & 1\cr
          1 & 0 & 0 & 0 }
= -1,
$$

whereas
$$\det(AD - CB) = \det\bmatrix{0&1\cr-1&0} = 1.$$

    {\bf Section 4.4}

\noindent{\bf 28.}

The volume is easily found as the absolute determinant of

$$\left|\matrix{3&1&1\cr 1&3&1\cr 1&1&3}\right| =
\left|\matrix{3&1&1\cr 0&8/3&2/3\cr 0&-2&2}\right| =
1/3\left|\matrix{3&1&1\cr 0&8&2\cr 0&0&5/2}\right| =
20.
$$

The area of the parallelogram faces formed by each pair is the same for
each pair by symmetry of the xyz coordinates.

If two vectors are orthogonal, the area of the parallelogram they form
(the rectangle) is the product of their norms.
And the area computation is also linear. The area of $\{v_1+v_3, v_2\}$
is the area of $\{v_1, v_2\}$ plus the area of $\{v_3, v_2\},$ and as
was proven for determinants, the area of $\{v_1, av_1\}$ is zero.
This lets us orthogonalize the first two vectors of our set to find
area.
$$\bmatrix{3&1\cr1&3\cr1&1} =
\bmatrix{3&-10/11\cr1&26/11\cr1&4/11}\bmatrix{1&7/11\cr0&1}.$$

This isn't a complete QR decomposition, but the norms of the given
vectors are $\sqrt{11}$ and $\sqrt{792}/11,$ giving an area (by their
product) of $\sqrt{72}$.

\bye