variable y = z + 2v. Iterate: on the next row
up, substitute expressions found 26 Chapter
One. Linear Systems in lower rows x + (z + 2v)
+ 2z + (v) + v = 0 and solve for the leading
variable x = z 2v.
bn an Vectors are equal if they have the
same representation. We arent too careful
about distinguishing between a point and the
vector whose canonical representation ends at
that point. R n = cfw_ v1
the homogeneous system has a non-~0
solution, and thus by the prior paragraph has
infinitely many solutions). QED This table
summarizes the factors affecting the size of a
general solution. number of
left side gives (u1 v1) 2 + (u2 v2) 2 + (u3
v3) 2 = (u 2 1 2u1v1 + v 2 1 ) + (u 2 2 2u2v2
+ v 2 2 ) + (u 2 3 2u3v3 + v 2 3 ) while the
right side gives this. (u 2 1 + u 2 2 + u 2 3 ) + (v
2 1 + v 2 2
miles at 89 degrees east of north, and 6.5
miles at 31 degrees east of north. Find the
distance between starting and ending
positions. (Ignore the earths curvature.) 2.14
Find k so that these two vect
unknowns and solve it. (b) Solutions to the
system are choices that the farmer can make.
Write down two reasonable solutions. (c)
Suppose that in the fall when the crops
mature, the farmer can bring i
classes we can think of the matrices in a
class as derived by row operations from the
unique reduced echelon form matrix in that
class. Put in more operational terms,
uniqueness of reduced echelon for
interaction between the left and right. For
some intuition about that interaction, consider
this system with one of the coefficients left
unspecified, as the variable c. x + 2y + 3z = 1 x
+ y + z = 1
unknowns system having (a) a one-parameter
solution set; (b) a two-parameter solution set;
(c) a three-parameter solution set. ? 2.32
[Shepelev] This puzzle is from a Russian website http:/www.arbuz.u
in the plane they generate. (Remark. They
could generate a degenerate plane a line or
a point but the statement remains true.)
2.36 Prove that, where ~u,~v R n are
nonzero vectors, the vector ~u |~u |
the prior section we defined vectors and
vector operations with an algebraic
motivation; r v1 v2 ! = rv1 rv2 ! v1 v2 ! + w1
w2 ! = v1 + w1 v2 + w2 ! we can now
understand those operations geometricall
third truck driver pay for a sandwich, a cup of
coffee, and a doughnut? X 2.21 The Linear
Combination Lemma says which equations can
be gotten from Gaussian reduction of a given
linear system. (1) Pro
b c = 2 2a + c = 3 a b = 0 (e) x + 2y z = 3 2x
+ y + w = 4 x y + z + w = 1 (f) x + z + w = 4 2x +
y w = 2 3x + y + z = 7 X 2.19 Solve each
system using matrix notation. Give each
solution set in vecto
row of the second is a linear combination of
the rows of the first. Proof For any two
interreducible matrices A and B there is some
minimum number of row operations that will
take one to the other. We
are straight and planes are flat. We can easily
check from the definition that linear surfaces
have the property that for any two points in
that surface, the line segment between them
is contained in
step so by the principle of mathematical
induction the proposition is true. QED This
shows, as discussed between the lemma and
its proof, that we can parametrize solution sets
using the free variables
2 1 0 1 3 5 ! (1/2)1+2 2 1 0 0 5/2 5 !
(1/2)1 (2/5)2 1 1/2 0 0 1 2 ! (1/2)2+1
1 0 1 0 1 2 ! Denoting those matrices A
D G B and writing the rows of A as 1
and 2, etc., we have this. 1 2 ! (1/2)1+2
are equal. How can things that are in different
places be equal? Think of a vector as
representing a displacement (the word vector
is Latin for carrier or traveler). These two
squares undergo displace
have the zero vector as their only solution. 3x
+ 2y + z = 0 6x + 4y = 0 y + z = 0 21+2 3x
+ 2y + z = 0 2z = 0 y + z = 0 23 3x + 2y
+ z = 0 y + z = 0 2z = 0 3.5 Example Some
homogeneous systems have m
has after a Gaussian reduction. The solution
description has two parts, the particular
solution ~p and the unrestricted linear
combination of the ~ s. We shall prove the
theorem with two corresponding
equations conflict. But the associated
homogeneous system does have a solution, as
do all homogeneous systems. x + z + w = 0 2x
y + w = 0 x + y + 3z + 2w = 0 21+2
1+3 2+3 x + z + w = 0 y 2z w = 0
0
1 1 8 = 2 1 3 1 0 5
3 4 4.5 = 2 4 0.5
1 0 5 As with the line, note that we
describe some points in this plane with
negative ts or negative ss or both. Calculus
books often describe a plane by using
other rows by moving ci~i to the left and
dividing by ci. Therefore we will have proved
the theorem if we show that in () all of the
coefficients are 0. For that we use induction on
the row number i.
base case is that the matrix has n = 1 column.
If this is the zero matrix then its echelon form
is the zero matrix. If instead it has any nonzero
entries then when the matrix is brought to
reduced ech
solves the given system since for any equation
index i, ai,1(p1 + h1) + + ai,n(pn + hn) =
(ai,1p1 + + ai,npn) + (ai,1h1 + + ai,nhn)
= di + 0 = di where as earlier pj and hj are the jth components of ~
10 (d) 1 1 1 1 2 2 , 0 3 1 2 2 5 (e) 1 1 1 0 0 3 ,
0 1 2 1 1 1 2.11 Describe the matrices in each
of the classes represented in Example 2.9. 2.12
Describe all matrices in the row equivalence
class of
adding a multiple of one row to another ri +j
then only row j of B differs from the matching
row of G, and ~ j = ri + j, which is indeed a
linear combinations of the rows of G. Because
we have proved
the picture, including the parallelogram in the
plane that shows the sum of the vectors
ending at (1.5, 0, 1) and (1.5, 1, 0). The
endpoint of the sum, on the diagonal, is not (3,
1, 1); what is it? 1
equations? X 3.24 Prove that if ~s and ~t
satisfy a homogeneous system then so do
these vectors. (a) ~s +~t (b) 3~s (c) k~s + m~t
for k, m R Whats wrong with this argument:
These three show that if a