Homework #1: Solutions to select problems
Linear Algebra
Summer 2011
Math S2010X (3)
Corrin Clarkson
June 6, 2011
Section 1.1
Exercise (42). Find a system of linear equations with three unknowns whose
solutions are the points on the line through (1,1,1) and (3,5,0).
Solution. The line through (1,1,1) and (3,5,0) can be parametrized bu the
following three equations:
x = 2t + 1
y = 4t + 1
z = −t + 1
Solving the last equation for t gives t = −z + 1. We can then plug this in for
t in the first two equations to get
x = 2(−z + 1) + 1
y = 4(−z + 1) + 1
Finally, we put this linear system in standard form
x + 2z = 3
y + 4z = 5
We check this solution by first noting the both of the given points satisfy
these equations. Then we check that the two equations describe two different
planes in R3 . This is is shown by finding a point that satisfies one of the
1
equations, but not the other; (1, 0, 1) is such a point.Thus we have two
distinct plains intersecting in R3 so we know their intersection is a line.
Because (1,1,1) and (3,5,0) are on both planes, the intersection line must be
the line through the given points.
Section 1.2
Exercise (48). For an arbitrary positive integer n ≥ 3, find all solutions
x1 , x2 , . . . , xn of the simultaneous equations
x2 =
x3 =
..
.
xn−1 =
1
(x1
2
1
(x2
2
+ x3 )
+ x4 )
1
(xn−2
2
+ xn )
Solution. We begin by considering the case when n = 3. This will not give
a complete solution, but will provide insight into the general case. When
n = 3, there is only one equation in the linear system
1
x2 = (x1 + x3 ).
2
In standard form, this equations becomes
1
1
x1 − x2 + x3 = 0
2
2
and solving for x1 gives
x1 = 2x2 − x3 .
The parametrized solution is then
x1 = 2t − s
x2 = t
x3 = s
The solution has one leading variable and two free variables.
Before solving the system for a general n, we will consider one more special
case: n = 4. In this case the linear system consists of the two equations
1
1
x2 = (x1 + x3 ) and x3 = (x2 + x4 )
2
2
2
In standard form, these equations become
1
1
1
1
x1 − x2 + x3 = 0 and x2 − x3 + x4 = 0.
2
2
2
2
From these equations, we can determine the augmented matrix of the linear
system
1
−1 12 0 0
2
0 12 −1 21 0
To row reduce this matrix, multiply both rows by 2 and then add twice the
second row to the first row. The reduce row echelon form is then
1 0 −3 2 0
0 1 −2 1 0
This gives the following parametrized solution:
x1
x2
x3
x4
=
=
=
=
3t + 2s
2t + s
t
s
Now we will consider the system for a general n. The basic strategy will
be the same: find the augmented matrix of the system, row reduce it and
then find the parametrized solution. First, we put the linear equations in
standard form.
1
x − x2 + 12 x3 = 0
2 1
1
x − x3 + 12 x4 = 0
2 2
..
.
1
x
2 n−2
− xn−1 + 12 xn = 0
The augmented matrix of this system is then
 1
−1 12
0 ··· 0 0
2
 0 1 −1 1 · · · 0 0

2
2

.. .. ..

.
.
.
0
0
0
0
···
3
1
2
−1

0 0
0 0 

.. 
. 
1
2
0
(1)
We begin row reducing this
is a leading one in each row

1 −2
 0 1



0 0
matrix by multiplying all the rows by 2. There
and in the first n − 2 columns.

1
0 ··· 0 0 0 0
−2 1 · · · 0 0 0 0 

.. 
.. .. ..
.
.
.
. 
0
0 · · · 1 −2 1 0
We begin to clear the columns with leading ones by adding twice the second
row to the first row. At this point the first two columns are done; they each
contain a single leading one and no other nonzero entries.


1 0 −3 2 · · · 0 0 0 0
 0 1 −2 1 · · · 0 0 0 0 



.. 
.
.
.
.
.
.

.
.
.
. 
0 0 0
0 · · · 1 −2 1 0
The third column can be cleared by adding the thrice
first row and twice the third row to the second row.

1 0 0 −4 · · · 0
0 0 0
 0 1 0 −3 · · · 0
0 0 0


..
.
.
.
.. .. ..

.
0 0 0 0 · · · 1 −2 1 0
the third row to the





Continuing to apply row operations in this manner will put the matrix in
reduced row echelon form. The complete sequence of row operations can be
summarized as follows:
row 1 = 2(r1 + 2r2 + 3r4 + · · · + (n − 2)rn−2 )
row 2 = 2(r2 + 2r3 + 3r4 + · · · + (n − 3)rn−2 )
..
.
row n − 3 = 2(rn−3 + 2rn−2 )
row n − 2 = 2(rn−2 )
Here ri indicates the ith row. We can check that these row operations have
the desired affect on the middle columns by considering their affect on three
consecutive rows.
m
[· · · 1
−2
1
0
0
···]
(m + 1) [· · · 0
1
−2
1
0
···]
+ (m + 2) [· · · 0
0
1
−2
1
···]
[· · · m −m + 1 0 −m − 3 m + 2 · · · ]
4
The resulting row reduced

1 0 0
 0 1 0



0 0 0
matrix is then.
0 ···
0 ···
... ..
.
0 ···
0 1−n n−2
0
−n n − 1
..
.
1
−2
0
0
..
.





(2)
0
1
The following parametrized solution describes all possible solutions to the
general linear system.
x1 = (n − 1)t + (2 − n)s
x2 = nt + (1 − n)s
..
.
xn−3
xn−2
xn−1
xn
=
=
=
=
3t − 2s
2t − s
t
s
Exercise (56). Find the conics through the points (0,0), (1,0), (0, 1) and
(1, -1). Draw a rough sketch of your solutions curve(s).
Solution. We are given that a conic is described by an equation of the form
c1 + c2 x + c3 y + c4 x2 + c5 xy + c6 y 2 = 0
Plugging each of the points into this equation gives the linear system
c1
c1 + c2 + c4
c1 + c3 + c6
c1 + c2 − c3 + c4 − c5 + c6
The augmented matrix of this linear

1 0 0
 1 1 0

 1 0 1
1 1 −1
=
=
=
=
system is
0 0
1 0
0 0
1 −1
5
0
0
1
1

0
0 

0 
0
0
0
0
0
We begin to row reduce this matrix
of the rows below it.

1 0 0
 0 1 0

 0 0 1
0 1 −1
Next, we subtract the second from
the forth row.

1 0 0
 0 1 0

 0 0 1
0 0 0
Finally, we multiply the fourth
form.

1 0
 0 1

 0 0
0 0
by subtracting the first row from each
0 0
1 0
0 0
1 −1
0
0
1
1

0
0 

0 
0
the forth row and add the third row to

0 0 0 0
1 0 0 0 

0 0 1 0 
0 −1 2 0
row by −1 to achieve reduced row echelon

0 0 0 0 0
0 1 0 0 0 

1 0 0 1 0 
0 0 1 −2 0
The parameterized solution of the linear system is
c1
c2
c3
c4
c5
c6
=
=
=
=
=
=
0
−t
−s
t
2s
s
Thus the conics that pass through the given points are described by equations
of the form
−tx − sy + tx2 + 2sxy + sy 2 = 0
Figure 1 is a graph of a few of these conics.
6
Figure 1: Conics through (0,0), (1,0), (0, 1) and (1, -1)
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