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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
2
2
2
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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