EE441_Homework_01_012208

EE441_Homework_01_012208 - ine of intersection, 2 5 (3) 6....

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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 four-dimensional 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 right-hand 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 four-dimensional 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 right-hand 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 right-hand 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 four-dimensional 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 back-substitution 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 n-dimensional 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 skew-symmetric 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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EE441_Homework_01_012208 - ine of intersection, 2 5 (3) 6....

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