Gilbert_Strang-Linear_Algebra_and_Its_Applications_4ed_solutions

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Solutions to Selected Exercises

Problem Set 1.2, page 9

1. The lines intersect at (x, y) = (3, 1). Then 3(column 1) + I(column 2) = (4, 4).

3. These "planes" intersect in a line in four-dimensional space. The fourth plane nor-

mally intersects that line in a point. An inconsistent equation like u + w = 5 leaves

no solution (no intersection).

5. The two points on the plane are (1, 0, 0, 0) and (0, 1, 0, 0).

7. Solvable for (3, 5, 8) and (1, 2, 3); not solvable for b = (3, 5, 7) orb = (1, 2, 2).

9. Column 3 = 2(column 2) - column 1.Ifb = (0, 0, 0), then (u, v, w) _ (c, -2c, c).

11. Both a = 2 and a = -2 give a line of solutions. All other a give x = 0, y = 0.

13. The row picture has two lines meeting at (4, 2). The column picture has 4(1, 1) +

2(-2, 1) = 4(column 1) + 2(column 2) = right-hand side (0, 6).

15. The row picture shows four lines. The column picture is infour-dimensional space.

No solution unless the right-hand side is a combination of the two columns.

17. If x, y, z satisfy the first two equations, they also satisfy the third equation. The line

L of solutions contains v = (1, 1, 0), w = (2, 1, 1), and u = 1v + 1w, and all

combinations cv + dw with c + d = 1.

19. Column 3 = column 1; solutions (x, y, z) _ (1, 1, 0) or (0, 1, 1) and you can add

any multiple of (-1, 0, 1); b = (4, 6, c) needs c = 10 for solvability.

21. The second plane and row 2 of the matrix and all columns of the matrix are changed.

The solution is not changed.

23. u = 0, v = 0; w = 1, because 1(column 3) = b.

Problem Set 1.3, page 15

1. Multiply by £ = z = 5, and subtract to find 2x + 3y = 1 and -6y = 6. Pivots

2, -6.

3. Subtract -1 times equation 1 (or add 1 times equation 1). The new second equation

2 2

is 3y = 3. Then y = 1 and x = 5. If the right-hand side changes sign, so does the

solution: (x, y) = (-5, -1).

5. 6x + 4y is 2 times 3x + 2y. There is no solution unless the right-hand side is

2. 10 = 20. Then all points on the line 3x + 2y = 10 are solutions, including (0, 5)

and (4, -1).

7. If a = 2, elimination must fail. The equations have no solution. If a = 0, elimination

stops for a row exchange. Then 3y = -3 gives y = -1 and 4x + 6y = 6 gives

x = 3.

9. 6x - 4y is 2 times (3x - 2y). Therefore, we need b2 = 2b1. Then there will be

infinitely many solutions. The columns (3, 6) and (-2, -4) are on the same line.

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Solutions to Selected Exercises

429

11. 2x - 3y = 3 2x - 3y = 3

x = 3

Subtract 2 x row 1 from row 2

y +

z = 1 gives

y + z = 1 and y= 1

Subtract 1 x row 1 from row 3

2y - 3z = 2

- 5z =0 z = 0

Subtract 2 x row 2 from row 3.

13. The second pivot position will contain -2 - b. If b = -2, we exchange with row

3. If b = -1 (singular case), the second equation is -y - z = 0. A solution is

(1, 1, -1).

15. If row 1 = row 2, then row 2 is zero after the first step; exchange the zero row with

row 3 and there is no third pivot. If column 1 = column 2, there is no second pivot.

17. Row 2 becomes 3y - 4z = 5, then row 3 becomes (q + 4)z = t - 5. If q = -4,

the system is singular - no third pivot. Then, if t = 5, the third equation is 0 = 0.

Choosing z = 1, the equation 3y - 4z = 5 gives y = 3 and equation 1 gives

x=-9.

19. The system is singular if row 3 is a combination of rows 1 and 2. From the end view,

the three planes form a triangle. This happens if rows 1+ 2 = row 3 on the left-hand

side but not the right-hand side: for example, x + y + z = 0, x - 2y - z = 1,

2x - y = 9. No two planes are parallel, but still no solution.

21. The fifth pivot is s . The nth pivot is ("+1)1) .

u+ v+ w= 2

u = 3

23. Triangular system 2v + 2w = -2

Solution v = -2.

2w= 2 w= 1

25. (u, v, w) = (3/2, 1/2,. -3). Change to +1 would make the system singular (2 equal

columns).

27. a = 0 requires a row exchange, but the system is nonsingular: a = 2 makes it singular

(one pivot, infinity of solutions); a = -2 makes it singular (one pivot, no solution).

29. The second term be + ad is (a + b)(c + d) - ac - bd (only 1 additional

multiplication). . ,

31. Elimination fails for a = 2 (equal columns), a = 4 (equal rows), a = 0 (zero

column).

Problem Set 1.4, page 26

[f].

4 , -2 ,

With sides to (2, 1) and (0, 3), the parallelogram goes to (2, 4).

3 5

3. Inner products 54 and 0, column times row gives -6 -10 -2

5. Ax = (0, 0, 0), so x = (2, 1, 1) is a solution; other solutions are cx = (2c, c, c).

7. Examples: Diagonal

0 2

0 0 7

, symmetric

1 3

2 0 , triangular

0 7

1 3

0 0

3 4

skew-symmetric

-3 0 0 .

-4 0 0

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2018-05-09

骗人的，就是书后答案，只有奇数题的

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