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Math 314/814: Matrix Theory Dr. S. Cooper, Fall 2008
Homework Solutions – Week of August 26
Section 1.1:
(3) A complete solution to this exercise can be found at the back of your text (page
(5) All of the drawings for this exercise can be found at the back of your text (pages
671 - 672). We compute each of the vectors as follows:
(a)
−→ AB= [4 − 1 , 2 − (−1)] = [3, 3]
(b)
−→
AB= [2 − 0 , − 1 − (−2)] = [2, 1]
(c)
−→ AB=
[
1 2
3 2
]
[
3 2
3 2
]
(d)
−→
AB=
[
1 6
1 3
1 2
1 3
]
[
1 6
1 6
]
2 a + 3c = 2[0, 2 , 0] + 3[1, − 2 , 1]
= [0, 4 , 0] + [3, − 6 , 3]
= [0 + 3, 4 − 6 , 3 + 0]
= [3, − 2 , 3]
2 c − 3 b − d = 2[1, − 2 , 1] − 3[3, 2 , 1] − [− 1 , − 1 , −2]
= [2, − 4 , 2] + [− 9 , − 6 , −3] + [1, 1 , 2]
= [2 − 9 + 1, − 4 − 6 + 1, 2 − 3 + 2]
= [− 6 , − 9 , 1]
2(a − 3 b) + 3(2b + a) = (2a − 6 b) + (6b + 3a) (Theorem 1.1, part e)
= ((2a − 6 b) + 6b) + 3a (Theorem 1.1, part b)
= (2a + (− 6 b + 6b)) + 3a (Theorem 1.1, part b)
= (2a + (−6 + 6)b) + 3a (Theorem 1.1, part f)
= (0b + 2a) + 3a (Theorem 1.1, part a)
= 0 b + (2a + 3a) (Theorem 1.1, part b)
= (2 + 3)a (Theorem 1.1, part f)
= 5 a
x − a = 2(x − 2 a) =⇒ x − a = 2x − 4 a =⇒ x = 3a
x + 2a − b = 3(x + a) − 2(2a − b)
=⇒ x + 2a − b = 3 x + 3a − 4 a + 2b
=⇒ x + 2a − b = 3 x − a + 2b
=⇒ − 2 x = − 3 a + 3b
=⇒ x =
a −
b =
(a − b)
(21) A complete solution to this exercise can be found at the back of your text (page
Section 1.3:
(3) A complete solution to this exercise can be found at the back of your text (page
(7) 2x + y = 7 − 3 y ⇐⇒ 2 x + 4y = 7. The linear equation 2x + 4y = 7 has the
same solution set as the given equation.
(9) If x 6 = 0 and y 6 = 0, then
x
y
xy
y + x
xy
xy
⇐⇒ x + y = 4.
So, the linear equation x + y = 4 where x, y 6 = 0 has the same solution set as
the given equation.
3 x − 6 y = 0 ⇐⇒ 3 x = 6y ⇐⇒ x = 2y.
Thus, for any real number t, we see that [x, y] = [2t, t] is a solution. Observe
that the complete set of solutions corresponds to the set of points on the line
determined by the given equation. Setting y = t, a parametric solution is given
by
x = 2 t
y = t.
So, the solution set to the equation 3x − 6 y = 0 is
{[
2 t
t
]
: t ∈ R
(13) The complete set of solutions corresponds to the set of points in the plane
determined by the equation x + 2y + 3z = 4. Setting y = s and z = t, a
parametric solution is given by
x = 4 − 2 s − 3 t
y = s
z = t.
So, the solution set to the equation x + 2y + 3z = 4 is
4 − 2 s − 3 t
s
t
: s, t ∈ R
(14) The complete set of solutions corresponds to the set of points in the plane
determined by the equation 4x 1 + 3x 2 + 2x 3 = 1. Setting x 2 = s and x 3 = t, a
parametric solution is given by
x 1 =
s −
t
x 2 = s
x 3 = t.
Thus, the solution set to the equation 4x 1 + 3x 2 + 2x 3 = 1 is
1 4
3 4
s −
1 2
t
s
t
: s, t, ∈ R
(15) The geometric solution to this exercise can be found at the back of your text
(page 674). Algebraically, we first subtract 2 times the first equation from the
second:
x + y = 0
−y = 3.
This gives y = −3. Using back substitution, we see that x − 3 = 0 =⇒ x = 3.
Thus, given system has the unique solution
[3, −3].
(16) The graphs of the lines x − 2 y = 7 and 3x + y = 7 show that there is a unique
solution (i.e. there is one and only one point of intersection between these two
lines). Algebraically, we subtract 3 times the first equation from the second:
x − 2 y = 7
7 y = − 14.
(29) A complete solution to this exercise can be found at the back of your text (page
(35) We solve the system using the technique of Example 2.6 from the text. The
augmented matrix is:
We subtract row one from the third row:
We now divide row three by -3:
Finally, we subtract row two from row three:
This gives us the system
x + 5y = − 1
y = − 1
0 x + 0y = 0
which has the same solution set as the given system. Using back substitution
we see that y = −1 and so x = 4. So, the solution to the given system is
[4, −1].
