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+\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