## SUGGESTIONS FOR HOMEWORK #9

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Homework suggestions: **
5.2

#32(b) Use the fact that when all the entries of a matrix except
for the second column are held fixed, the determinant is a linear
function of the second column. Use also that

3 1 1
5 = 1 + 2 2
7 1 3

5.3
#4 (not bracketed) Let u,v,w be the vertices of the triangle.
Thus, for example, u =
a_1

a_2

The sides of the triangle at the vertex u are given by
v-u, w-u, emanating from the tip of u. The area of the
triangle is therefore half the absolute value of the
determinant of the matrix whose columns are v-u, w-u.

Now look at the 3 by 3 determinant in the book. If you
subtract the first column from the each of the other two
the third row becomes 1 0 0. Expanding with respect
to this row gives 1 det B, where B is the matrix with
columns v-u, w-u. Thus, the area of the triangle is
also half the determinant of the given 3 by 3 matrix.

#5 (not bracketed) The cone with vertex P over a region
R in a plane (P is not in the plane) consists of all points
that lie on a line joining P to a point of R. When R
is a disk and P is on a line perpendicular to the plane
through the center of the disk, this gives an ordinary cone.
The volume of a cone is (1/3)(altitude)(area of the base),
no matter what shape the base is. The reason is that if the
altitude is h and we let x be the distance of a variable
point on the altitude to the vertex (x runs from 0 to h) then
the area of a cross-section through a point on the altitude
at distance x from the vertex is A(x/h)^2. One integrates
this as x goes from 0 to h to get the formula Ah/3.

Pick two of the three vectors determining the parallelepiped:
they determuine a parallelogram. Say that parallelogram has
area A. Say the height of the parallepiped is h. The
volume of the parallelepiped is hA. The tetrahedron can be
thought of as a cone of height h over a triangle that has
half the area of the parallegram, i.e., A/2. Thus, the volume
of the tetrahedron is (1/3)h(A/2) = Ah/6, or one sixth the
volume of the parallepiped.

#6 (to be turned in) Referring to the analyses above, you should
be able to calculate the ratio of the area of the specified
triangle to the volume of the specified tetrahedron. The answer
is a completely specific real number that does not depend in any
way on the values of the a's, b's, and c's.

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