Math 308 Solutions to Conceptual Problems - Chapter 1

(1) The linear equations are:

x = 0.05(103, 000 − y − z) = 5150 − 0.05y − 0.05z
y = 0.05(103, 000 − x) = 5150 − 0.05x
z = 0.4(103, 000 − x − y) = 41200 − 0.4x − 0.4y
Solving the above system of equations we get the unique solution:
x = $3000, y = $5000, z = $38, 000
(2) (a) Any two parallel lines will work. For example, x + 2y = 3 and x + 2y = 4.
(b) Two lines that are not parallel, which intersect at this point (both go through
that point). For example, x + y = −3 and 2x + y = −2.
(c) You need two equations that describe the same line. One equation will
necessarily be a constant multiple of the other. When you put
(x, y) = (1 + 2s, 3 − s) into both equations you should get 0 = 0 after simplifying.
For example, x + 2y = 7 and 10x + 20y = 70.
(i) You add the equation of a line that does not contain that point. For example,
x + y = 1.
(ii) Remove either equation.
(iii) Add the equation of a line (it could be the same as one of the others) that goes
through that point. For example, 3x + y = −1 (different slope), 10x + 10y = −30 or
20x + 10y = −2.
(iv) You can remove one of them and add a third. You cannot do it by removing or
adding just one equation.
(3) (a) Here you have options. You can do three parallel planes for example
x + y + z = 1, x + y + z = 2, x + y + z = 2, or you can do two parallel planes and a
third one that is not, or you can do two equations that describe the same plane and
a third for a parallel plane.
(b) You need three distinct non-parallel planes that contain this point, whose
intersection is not a line. Solve your system to verify it has the given unique
solution.
(c) Three non-parallel planes (their normals should not be parallel) which all
contain this line. For example, −x + y − z = 3, 2x + y − 4z = −3 and
3x − y − z = −7.
(c) The three equations must describe the same plane and the solution given must
satisfy them for any value of s1 and s2 . For example, −x + y + z = 0,
−2x + 2y + 2z = 0, −3x + 3y + 3z = 0..
(4) For an extra challenge (or if you plan to major in math) do this problem first with
arbitrary points (a, b, c) and (d, e, f ) in R3 , and then with arbitrary points (a1 , ..., an
and (b1 , ..., bn ) in Rn .
(a) The answer here is not unique. It depends on which point and which direction
vector you used. One answer is (x1 , x2 , x3 ) = (2 + 7t, 3 − 3t, −1 + 4t). (Most
likely in your equation when t = 1 you get one point when t = 0 you get the
other point.)
(b) The points p and q satisfy the linear equation means:
9a1 + 0a2 + 3a3 = d and 2a1 + 3a2 − a3 = d
Now we check (x1 , x2 , x3 ) = (2 + 7t, 3 − 3t, −1 + 4t):
a1 (2 + 7t) + a2 (3 − 3t) + a3 (−1 + 4t) = 2a1 + 3a2 − a3 + t(7a1 − 3a2 + 4a3 )
 At this point you have to remember where the numbers 7, -3 and 4 came from
(the direction vector for the line) so we have
= 2a1 + 3a2 − a3 + t [(9 − 2)a1 + (0 − 3)3a2 + (3 − (−1))a3 ]
after some reorganization
= 2a1 + 3a2 − a3 + t [(9a1 + 0a2 + 3a3 ) − (2a1 + 3a2 − a3 )]
finally using the fact that the initial points were solutions we get
= d + t(d − d) = d
So, any point on the line is also a solution.
(c) If we were given an system of linear equations, we would do this process above
of every one of the equations and get the same result so any point on the line
would be a solution to the system.
(5) (a) We basically have f (−1) = 7, f (1) = 0, f (2) = −17 and f (0) = 2. So the
system is:
a0 − a1 + a2 − a3 = 7
a0 + a1 + a2 + a3 = 0
a0 + 2a1 + 4a2 + 8a3 = −17
a0 = 2
with augmented matrix  
1 −1 1 −1 7
1 1 1 1 0
 
1 2 4 8 −17
1 0 0 0 2
and solution f (x) = 2 − 0.5x + 1.5x2 − 3x3 .
(b) This says f (1) = 7, f 0 (1) = 2, f (2) = 5 and f 0 (2) = −7. The answer is
f (x) = 3 + 5x − x3 .
(6) (a) We’ll use row operations, starting from the augmented matrix. Remember to
work left-to-right (during Gaussian elimination, until reaching echelon form), then
right-to-left (during the “Jordan” part, to reach reduced echelon form).
     
0 1 3 −1 0 −1 −1 −1 1 0 −1 −1 −1 1 0
−1 −1 −1 1 0 ∼R1 ↔R2  0 1 3 −1 0 ∼R3 −2R1 →R3 0
  1 3 −1 0
−2 −4 4 −2 0 −2 −4 4 −2 0 0 −2 6 −4 0
   
−1 −1 −1 1 0 1 1 1 −1 0
∼R3 +2R2 →R3 0
 1 3 −1 0 (echelon form!) ∼ −R1 → R1 0 1 3 −1
  0
0 0 12 −6 0 1
R → R2
12 2
0 0 1 − 21 0
     
1 1 1 −1 0 1 1 0 − 21 0 1 0 0 −1 0
∼R2 −3R3 →R2 0 1 0 2 0 ∼R1 −R3 →R1 0 1 0 2 0 ∼R1 −R2 →R1 0 1 0 12
 1   1   0.
0 0 1 − 12 0 0 0 1 − 12 0 0 0 1 − 21 0
Whew. So, the final equations are z1 − z4 = 0, z2 + 21 z4 = 0, z3 − 12 z4 = 0. The
leading variables are z1 , z2 , z3 ; the free variable is z4 , so we set z4 = t, a free
parameter. The general solution is: z = (z1 , z2 , z3 , z4 ) = (t, − 21 t, 12 t, t).
(b) Since the system is homogeneous, we know immediately that there’s the
trivial solution: (0, 0, 0, 0). For a nontrivial solution, we could set t = 2 to get
z = (2, −1, 1, 2).
(c) We can plug in the three points to get equations in terms of a1 , a2 , a3 , b:
 a1 + 3a3 − b = 0
a1 + a2 + a3 − b = 0
−2a1 − a2 + 2a3 − b = 0
These are almost the same equations! We have to swap the first two variables:
z1 = a2 and z2 = a1 (and z3 = a3 , z4 = b). Then each equation will be proportional
to the corresponding z equation above – so the solution sets are the same. So, our
solution z = (2, −1, 1, 2) tells us (a1 , a2 , a3 , b) = (−1, 2, 1, 2), giving the plane:
−x1 + 2x2 + x3 = 2.
Notice that we needed to have a nontrivial solution to part (b). We could not use
(a1 , a2 , a3 , b) = (0, 0, 0, 0) – that doesn’t give an equation of a plane.
(Comment: implicitly, swapping the first two variables was a column
operation, and it changed the solution vector. By contrast, row operations do not
change the set of solutions.)
(7) (a) The augmented matrix is
 
2 3 −5 b1
 7 2 8 b2 
−1 1 −5 b3
Reducing this matrix to echelon form gives
 
1 −1 5 −b3
0 5 −15 b1 + 2b3 
0 0 0 9b1 − 5b2 − 17b3

(b) The system has a solution exactly when 9b1 − 5b2 − 17b3 = 0
(c) The points (b1 , b2 , b3 ) for which 9b1 − 5b2 − 17b3 = 0 is the plane in R3 passing
through the origin and with normal vector (9, −5, −17).
(d) No. A random point in R3 would not lie on the plane 9b1 − 5b2 − 17b3 = 0. (If
you pick a random point on the plane, how likely is it that it would lie on the
line y = 2x + 3? If you pick a random point on the number line, how likely is it
that it will be the point x = 3? Do you see them similarities between these
questions?

(8) The echelon form of the system is

 
1 −3 1 4
 0 3 −5 −5  .
0 0 a − 14 b + 4
(a) The system has infinitely many solutions when the last row is all zero.
Therefore, a = 14 and b = −4.
(b) For one solution it suffices to take a 6= 14, say a = 15. Then for a given b, the
unique solution is x3 = b + 4, x2 = −5+5(b+4)
3
= 15+5b
3
and
x1 = 4 − (b + 4) + (−5 + 5(b + 4)) = 4b + 15. If we choose b = 0 for example, then
the unique solution with a = 15 and b = 0 is (15, 5, 4).
(c) The system has no solution if a = 14 but b 6= −4.