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Unformatted text preview: ine of intersection,
2
5 (3)
6. fails: 2 + 5 75 6,
:ment 0 = 1. Thus
tically discover.
plutions. When the
1e 0 = 0. Now the
; have a whole line
mes in Figure 1.5b
fferent. The lowest
roblem 1.5c is still
3w. here is no solution
. a whole plane of
.e same. ' It has to go wrong;
the equations, and 1
o 3 =1). (4)
4 "eason is that those
plane (which goes
.ssible (Figure 1.6).
is no solution. But 3 columns
' in a plane of solutions 3e columns. 1.2 The Geometry of Linear Equations 9 there is a chance that b does lie in the plane of the columns. In that case there are too
many solutions; the three columns can be combined in inﬁnitely many ways to produce
b. That column picture in Figure 1.6b corresponds to the row picture in Figure 1.5c. How do we know that the three columns lie in the same plane? One answer is to
ﬁnd a combination of the columns that adds to zero. After some calculation, it is u = 3,
v = — 1, w = —2. Three times column 1 equals column 2 plus twice column 3. Column 1
is in the plane of columns 2 and 3. Only two columns are independent. The vector b = (2, 5,7) is in that plane of the columns—it is column 1 plus
column 3——so (1,0, l) is a solution. We can add any multiple of the combination
(3, —1, ——2) that gives b = 0. So there is a whole line of solutions—as we know from
the row picture. The truth is that we knew the columns would combine to give zero, because the rows
did. That is a fact of mathematics, not of computation—and it remains true in dimension
n. If the n planes have no point in common, or inﬁnitely many points, then the n
columns lie in the same plane. If the row picture breaks down, so does the column picture. That brings out the
difference between Chapter 1 and Chapter 2. This chapter studies the most important
problem—the nonsingular case—where there is one solution and it has to be found.
Chapter 2 studies the general case, where there may be many solutions or none. In
both cases we cannot continue without a decent notation (matrix notation) and a decent
algorithm (elimination). After the exercises, we start with elimination. Problem Set 1.2 1. For the equations x + y = 4, 2x — 2y = 4, draw the row picture (two intersecting
lines) and the column picture (combination of two columns equal to the column
vector (4, 4) on the right side). 2. Solve to ﬁnd a combination of the columns that equals 17:
u — v — w = b1
'Il'iangular system v + w = b2
w = b3. 3. (Recommended) Describe the intersection of the three planes u + v + w + z = 6
and u + w + z = 4 and u + w = 2 (all in fourdimensional space). Is it a line or
a point or an empty set? What is the intersection if the fourth plane u = —l is
included? Find a fourth equation that leaves us with no solution. 4. Sketch these three lines and decide if the equations are solvable: x+2y=2
3by25ystem x— y=2
y=1. What happens if all righthand sides are zero? Is there any nonzero choice of right
hand sides that allows the three lines to intersect at the same point? 5. Find two points on the line of intersection of the three planes t = 0 and z = 0 and
x + y + z + t = 1 in fourdimensional space. 10 Chapteri Matrices and Gaussian Elimination 6. When b = (2, 5, 7), ﬁnd a solution (u, v, w) to equation (4) different from the
solution (1, 0, 1) mentioned in the text. 7. Give two more right—hand sides in addition to b = (2, 5, 7) for which equation (4)
can be solved. Give two more righthand sides in addition to b = (2, 5, 6) for which
it cannot be solved. f8. Explain why the system u + v + w = 2
u + 21) + 3w = 1
v + 2w = 0
is singular by ﬁnding a combination of the three equations that adds up to 0 = 1. What value should replace the last zero on the right side to allow the equations to
have solutions—and what is one of the solutions? 9. The column picture for the previous exercise (singular system) is 1 1 1
u 1 +12 2 +w 3 =b.
0 1 2 Show that the three columns on the left lie in the same plane by expressing the third
column as a combination of the ﬁrst two. What are all the solutions (u, v, w) if b is
the zero vector (0,0, 0)? 10. (Recommended) Under what condition on y1, y2, y3 do the points (0, yl), (1, yz),
(2, y3) lie on a straight line? 11. These equations are certain to have the solution x = y = 0. For which values of a
is there a whole line of solutions? ax+2y=0
2x+ay=0 2. Starting with x + 4y = 7, ﬁnd the equation for the parallel line through x = 0,
y = 0. Find the equation of another line that meets the ﬁrst at x = 3, y = 1. Problems 13—15 are a review of the row and column pictures.
13. Draw the two pictures in two planes for the equations x — 2y = 0, x + y = 6. 14. For two linear equations in three unknowns x, y, z, the row picture will show (2 or 3)
(lines or planes) in (two or three)dimensional space. The column picture is in (two
or three)dimensional space. The solutions normally lie on a _ _. 15. For four linear equations in two unknowns x and y, the row picture shows four
__ . The column picture is in _dimensional space. The equations have
no solution unless the vector on the righthand side is a combination of _ _. 16. Find a point with z = 2 on the intersection line of the planes x + y + 32 = 6 and
x — y + z = 4. Find the point with z = 0 and a third point halfway between. lifferent from the Ihich equation (4)
2, 5, 6) for which adds up to 0 = 1.
v the equations to :pressing the third
ns (u, v, w) ifbis IS (0, yl), (1: 3’2): which values of a e through x = 0,
= 3, y = 1. 0, x + y = 6.
twill show (2 or 3) 1 picture is in (two icture shows four
he equations have
tion of ___ ._. + y + 32 = 6 and
ay between. 1.3 An Example of Gaussian Elimination 11 17. The ﬁrst of these equations plus the second equals the third: x + y + z = 2
x + 2y + z = 3
2x + 3y + 22 = 5.
The ﬁrst two planes meet along a line. The third plane contains that line, because if x, y, z satisfy the ﬁrst two equations then they also . The equations have
inﬁnitely many solutions (the whole line L). Find three solutions. 18. Move the third plane in Problem 17 to a parallel plane 2x + 3y + ZZ = 9. Now the
three equations have no solution—why not? The ﬁrst two planes meet along the line L, but the third plane doesn’t _ _ that line.
19. In Problem 17 the columns are (1, 1, 2) and (1, 2, 3) and (1, 1, 2). This is a “singular
case” because the third column is . Find two combinations of the columns that give b = (2, 3, 5). This is only possible for b = (4, 6, c) if c = 20. Normally 4 “planes” in four—dimensional space meet at a . Normally 4 col
umn vectors in fourdimensional space can combine to produce b. What combination of(1,0,0,0), (1, 1,0,0), (1, l, 1,0), (1, 1, 1, 1) produces b = (3, 3, 3, 2)? What
4 equations for x, y, z, t are you solving? 21. When equation 1 is added to equation 2, which of these are changed: the planes in
the row picture, the column picture, the coefﬁcient matrix, the solution? 22. If (a, b) is a multiple of (c, d) with abcd +‘ 0, show that (a, c) is a multiple of
(b, d). This is surprisingly important: call it a challenge question. You could use
numbers ﬁrst to see how a, b, c, and d are related. The question will lead to: a b IfA=[C d has dependent rows then it has dependent columns. 23. In these equations, the third column (multiplying w) is the same as the right side b.
The column form of the equations immediately gives what solution for (u, v, w)? 6u+7v+8w=8
4u+5v+9w=9
2u—2v+7w=7. 1.3 AN EXAMPLE OF GAUSSIAN ELIMINATION The way to understand elimination is by example. We begin in three dimensions:
214 + v + w = 5
Original system 4u — 61) = —2 (1)
—2u+7v+2w= 9. The problem is to ﬁnd the unknown values of u, v, and w, and we shall apply
Gaussian elimination. (Gauss is recognized as the greatest of all mathematicians, but
certainly not because of this invention, which probably took him ten minutes. Ironically, Chapter 1 Matrices and Gaussian Elimination 7. For which numbers a does elimination break down (a) permanently, and (b) tem—
porarily? ax + 3y = —3
4x + 6y = 6.
Solve for x and y after ﬁxing the second breakdown by a row exchange. 8. For which three numbers k does elimination break down? Which is ﬁxed by a row
exchange? In each case, is the number of solutions 0 or 1 or 00? kx+3y= 6
3x+ky=—6. 9. What test on b1 and b2 decides whether these two equations allow a solution? How
many solutions will they have? Draw the column picture. 3x—2y=b1
6X—‘4y=b2. Problems 10—19 study elimination on 3 by 3 systems (and possible failure). @Reduce this system to upper triangular form by two row operations: ‘ 2x+3y+ Z: 8
4x+7y+5z=20
—2y+22= 0. Circle the pivots. Solve by back—substitution for z, y, x.
. Apply elimination (circle the pivots) and backsubstitution to solve
2x — 3y = 3
4x — 5y + z = 7
2x — y — 3z = 5.
List the three row operations: Subtract _ times row _ from row _. . Which number d forces a row exchange, and what is the triangular system (not
singular) for that d? Which d makes this system singular (no third pivot)?
2x + 5 y + z = 0
4x + dy + z = 2
y — z = 3. . Which number b leads later to a row exchange? Which b leads to a missing pivot?
In that singular case ﬁnd a nonzero solution x, y , z. x+by =0
x—Zy—z—O
y+z=0. 26 Chapteri Matrices and Gaussian Elimination for elimination. But fortunately it is the right order for reversing the elimination steps—
which also comes in the next section.
Notice that the product of lower triangular matrices is again lower triangular. E Problem Set 1.4
1. Compute the products
4 O 1 3 1 0 O 5 2 0 1
0 1 O 4 and O 1 0 ~2 and 1 3 1.
4 0 1 5 O 0 1 3 For the third one, draw the column vectors (2, 1) and (0, 3). Multiplying by (1, 1)
just adds the vectors (do it graphically). 2. Working a column at a time, compute the products 411 1230 431
5 1 [3] and 4 5 6 land 6 6
61 7890 895 3. Find two inner products and a matrix product: 1 3 1
[1 —2 7: —2 and [1 —2 7] 5 and —2 [3 5 1].
7 1 7 . The ﬁrst gives the length of the vector (squared). 4. If an m by n matrix A multiplies an ndimensional vector x, how many separate
multiplications are involved? What if A multiplies an n by p matrix B? 5. Multiply Ax to ﬁnd a solution vector x to the system Ax = zero vector. Can you
ﬁnd more solutions to Ax = 0? 3 —6 0 2
Ax= 0 2 —2 1 .
1 —1 —1 1 6. Write down the 2 by 2 matrices A and B that have entries (1,, = i + j and bi, =
(~1)‘+i, Multiply them to ﬁnd AB and BA. 7. Give 3 by 3 examples (not just the zero matrix) of (a) adiagonal matrix: au = 0 if i 76 j. (b) a symmetric matrix: a;, = a ,7 for all i and j. (c) an upper triangular matrix: a;, = 0 if i > j. (d) a skewsymmetric matrix: au = —a,~,~ for alli and j. 8. Do these subroutines multiply Ax by rows or columns? Start with EU) = O:
D0101=1,N D010J=1,N
D010J=1,N D0101=1,N 10 B(I) = B(I) + A(I,J) * X0) 10 B(I) = B(I) + A(I,J) * X(J) ...
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This note was uploaded on 02/28/2008 for the course EE 441 taught by Professor Neely during the Spring '08 term at USC.
 Spring '08
 Neely

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