文档介绍：EXERCISE SET
1. (b) Not linear because of the term x1x3.
–2
(d) Not linear because of the term x1 .
3/5
(e) Not linear because of the term x1 .
6. (b) Substituting the given expressions for x and y into the equation x = 5 + 2y yields
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tt=5+2−−=5+ tt 5=
2 2
Since this equation is valid for all values of t, the proposed solution is, indeed, the
general one.
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Alternatively, we let x = t and solve for y, obtaining yt= –
2 2
7. Since each of the three given points must satisfy the equation of the curve, we have the
system of equations
2
ax1 + bx1 + c = y1
2
ax2 + bx2 + c = y2
2
ax3 + bx3 + c = y3
If we consider this to be a system of equations in the three unknowns a, b, and c, the
augmented matrix is clearly the one given in the exercise.
8. If the system is consistent, then we can, for instance, subtract the first two equations from
the last. This yields c – a – b = 0 or c = a + b. Unless this equation holds, the system cannot
have a solution, and hence cannot be consistent.
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2 Exercise Set
9. The solutions of x1 + kx2 = c are x1 = c – kt, x2 = t where t is any real number. If these
ᐉᐉᐉ
satisfy x1 + x2 = d, then c – kt + t = d, or ( – k)t = d – c for all real numbers t. In
particular, if t = 0, then d = c, and if t = 1, then ᐉ= k.
10. If x – y = 3, then 2x – 2y = 6. Therefore, the equations are consistent if and only if k = 6;
that is, there are no solutions if k ≠ 6. If k = 6, then the equations represent the same line,
in which case, there are infinitely many solutions. Since this covers all of the possibilities,
there is never a unique solution.
12. (a) If the system of equations fails to have a solution, then there are three possibilities:
The three lines intersect in (i) distinct points, (ii) 2 distinct points, or (iii) not at all.
(i) (ii) (iii)
(b) If the system of equations has exactly one solution, then all of the lines must pass
through mon point. Moreover, at le