Exercises
M
,
P
,
S
{\displaystyle {}M,P,S}
and
T
{\displaystyle {}T}
are the members of one family. In this case,
M
{\displaystyle {}M}
is three times as old as
S
{\displaystyle {}S}
and
T
{\displaystyle {}T}
together,
M
{\displaystyle {}M}
is older than
P
{\displaystyle {}P}
, and
S
{\displaystyle {}S}
is older than
T
{\displaystyle {}T}
, moreover, the age difference between
S
{\displaystyle {}S}
and
T
{\displaystyle {}T}
is twice as large as the difference between
M
{\displaystyle {}M}
and
P
{\displaystyle {}P}
. Furthermore,
P
{\displaystyle {}P}
is seven times as old as
T
{\displaystyle {}T}
, and the sum of the ages of all family members is equal to the paternal grandmother's age, which is
83
{\displaystyle {}83}
.
a) Set up a linear system of equations that expresses the conditions described.
b) Solve this system of equations.
Kevin pays
2.50
{\displaystyle {}2.50}
€ for a winter bunch of flowers with
3
{\displaystyle {}3}
snowdrops and
4
{\displaystyle {}4}
mistletoes, and Jennifer pays
2.30
{\displaystyle {}2.30}
€ for a bunch with
5
{\displaystyle {}5}
snowdrops and
2
{\displaystyle {}2}
mistletoes. How much does a bunch with one snowdrop and
11
{\displaystyle {}11}
mistletoes cost?
Show that the system of linear equations
−
4
x
+
6
y
=
0
5
x
+
8
y
=
0
{\displaystyle {\begin{matrix}-4x+6y&=&0\\5x+8y&=&0\,\end{matrix}}}
has only the trivial solution
(
0
,
0
)
{\displaystyle {}(0,0)}
.
We look at a clock with hour and minute hands. Now it is 6 o'clock, so that both hands have opposite directions. When will the hands have opposite directions again?
Solve the
linear equation
x
+
y
+
z
=
0
.
{\displaystyle {}x+y+z=0\,.}
Solve the system of linear equations
x
=
5
,
2
y
=
3
,
4
z
+
w
=
3.
{\displaystyle x=5,\,2y=3,\,4z+w=3.}
Solve the following system of inhomogeneous linear equations.
3
x
+
z
+
4
w
=
4
2
x
+
2
y
+
w
=
0
4
x
+
6
y
+
w
=
2
x
+
3
y
+
5
z
=
3
.
{\displaystyle {\begin{matrix}3x&\,\,\,\,\,\,\,\,&+z&+4w&=&4\\2x&+2y&\,\,\,\,\,\,\,\,&+w&=&0\\4x&+6y&\,\,\,\,\,\,\,\,&+w&=&2\\x&+3y&+5z&\,\,\,\,\,\,\,\,&=&3\,.\end{matrix}}}
Does there exist a solution
(
a
,
b
,
c
)
∈
Q
3
{\displaystyle {}(a,b,c)\in \mathbb {Q} ^{3}}
for the
system of linear equations
a
(
1
2
11
2
)
+
b
(
2
2
12
3
)
+
c
(
3
1
20
7
)
=
(
1
2
20
5
)
{\displaystyle {}a{\begin{pmatrix}1\\2\\11\\2\end{pmatrix}}+b{\begin{pmatrix}2\\2\\12\\3\end{pmatrix}}+c{\begin{pmatrix}3\\1\\20\\7\end{pmatrix}}={\begin{pmatrix}1\\2\\20\\5\end{pmatrix}}\,}
from
Example 21.1
?
Show that for every
system of linear equations
over
Q
{\displaystyle {}\mathbb {Q} }
, there exists an
equivalent
linear system with the property that all coefficients are integers.
Bring the system of linear equations
3
x
−
4
+
5
y
=
8
z
+
7
x
,
{\displaystyle {}3x-4+5y=8z+7x\,,}
2
−
4
x
+
z
=
2
y
+
3
x
+
6
,
{\displaystyle {}2-4x+z=2y+3x+6\,,}
4
z
−
3
x
+
2
x
+
3
=
5
x
−
11
y
+
2
z
−
8
,
{\displaystyle {}4z-3x+2x+3=5x-11y+2z-8\,,}
into a standard form, and solve it.
Exhibit a linear equation for the straight line in
R
2
{\displaystyle {}\mathbb {R} ^{2}}
, which runs through the two points
(
2
,
3
)
{\displaystyle {}(2,3)}
and
(
5
,
−
7
)
{\displaystyle {}(5,-7)}
.
Before dealing with the next exercise, we recall the concept of a secant, which occurred already in the context of differential calculus.
Determine an equation for the secant of the function
R
⟶
R
,
x
⟼
−
x
3
+
x
2
+
2
,
{\displaystyle \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto -x^{3}+x^{2}+2,}
to the points
3
{\displaystyle {}3}
and
4
{\displaystyle {}4}
.
Determine a linear equation for the plane in
R
3
{\displaystyle {}\mathbb {R} ^{3}}
, where the three points
(
1
,
0
,
0
)
,
(
0
,
1
,
2
)
,
and
(
2
,
3
,
4
)
{\displaystyle (1,0,0),\,(0,1,2),{\text{ and }}(2,3,4)}
lie.
Given a
complex number
z
=
a
+
b
i
≠
0
,
{\displaystyle {}z=a+b{\mathrm {i} }\neq 0\,,}
find its inverse complex number with the help of a real system of linear equations, with two equations in two variables.
Solve, over the
complex numbers ,
the
linear system
of equations
i
x
+
y
+
(
2
−
i
)
z
=
2
7
y
+
2
i
z
=
−
1
+
3
i
(
2
−
5
i
)
z
=
1
.
{\displaystyle {\begin{matrix}{\mathrm {i} }x&+y&+(2-{\mathrm {i} })z&=&2\\&7y&+2{\mathrm {i} }z&=&-1+3{\mathrm {i} }\\&&(2-5{\mathrm {i} })z&=&1\,.\end{matrix}}}
Let
K
{\displaystyle {}K}
be the field with two elements. Solve in
K
{\displaystyle {}K}
the
inhomogeneous linear system
x
+
y
=
1
y
+
z
=
0
x
+
y
+
z
=
0
.
{\displaystyle {\begin{matrix}x&+y&&=&1\\&y&+z&=&0\\x&+y&+z&=&0\,.\end{matrix}}}
Show by an example that the linear system given by three equations I, II, III is not equivalent to the linear system given by the three equations I-II, I-III, II-III.
The following exercises are also about finding appropriate methods to solve the equations.
Solve the system of linear equations
+
7
y
+
3
z
=
4
x
+
4
w
=
9
−
3
y
−
5
z
=
2
−
2
x
+
3
w
=
3
.
{\displaystyle {\begin{matrix}&+7y&+3z&\,\,\,\,\,\,\,\,&=&4\\x&\,\,\,\,\,\,\,\,&\,\,\,\,\,\,\,\,&+4w&=&9\\&-3y&-5z&\,\,\,\,\,\,\,\,&=&2\\-2x&\,\,\,\,\,\,\,\,&\,\,\,\,\,\,\,\,&+3w&=&3\,.\end{matrix}}}
Solve the system of linear equations
7
y
=
5
,
{\displaystyle {}7y=5\,,}
4
z
=
8
,
{\displaystyle {}4z=8\,,}
2
u
−
3
v
=
0
,
{\displaystyle {}2u-3v=0\,,}
5
w
=
0
,
{\displaystyle {}5w=0\,,}
6
x
−
3
y
+
2
z
−
11
u
−
v
+
5
w
=
17
,
{\displaystyle {}6x-3y+2z-11u-v+5w=17\,,}
4
u
−
5
v
=
0
.
{\displaystyle {}4u-5v=0\,.}
Solve the system of linear equations
4
x
−
5
y
+
7
z
=
−
3
,
{\displaystyle {}4x-5y+7z=-3\,,}
−
2
x
+
4
y
+
3
z
=
9
,
{\displaystyle {}-2x+4y+3z=9\,,}
x
=
−
2
.
{\displaystyle {}x=-2\,.}
Solve the system of linear equations
3
x
−
67
y
+
14
z
−
123
u
−
51
w
=
5
,
{\displaystyle {}3x-67y+14z-123u-51w=5\,,}
8
x
−
11
y
+
12
z
−
27
u
−
65
w
=
51
,
{\displaystyle {}8x-11y+12z-27u-65w=51\,,}
66
x
−
67
y
−
77
z
−
u
+
100
w
=
0
,
{\displaystyle {}66x-67y-77z-u+100w=0\,,}
8
x
−
11
y
+
12
z
−
27
u
−
65
w
=
−
15
,
{\displaystyle {}8x-11y+12z-27u-65w=-15\,,}
−
301
x
+
44
y
+
33
z
−
31
u
−
18
w
=
571
.
{\displaystyle {}-301x+44y+33z-31u-18w=571\,.}
Determine, in dependence of the parameter
a
∈
R
{\displaystyle {}a\in \mathbb {R} }
,
the solution space
L
a
⊆
R
3
{\displaystyle {}L_{a}\subseteq \mathbb {R} ^{3}}
of the system of linear equations
5
x
+
a
y
+
(
1
−
a
)
z
=
0
,
{\displaystyle {}5x+ay+(1-a)z=0\,,}
2
a
x
+
a
2
y
+
3
z
=
0
.
{\displaystyle {}2ax+a^{2}y+3z=0\,.}
A system of linear inequalities is given by
x
≥
0
,
{\displaystyle {}x\geq 0\,,}
y
≥
0
,
{\displaystyle {}y\geq 0\,,}
x
+
y
≤
1
.
{\displaystyle {}x+y\leq 1\,.}
Sketch the solution set of this system of inequalities.
Let
a
1
x
+
b
1
y
≥
c
1
,
{\displaystyle {}a_{1}x+b_{1}y\geq c_{1}\,,}
a
2
x
+
b
2
y
≥
c
2
,
{\displaystyle {}a_{2}x+b_{2}y\geq c_{2}\,,}
a
3
x
+
b
3
y
≥
c
3
,
{\displaystyle {}a_{3}x+b3y\geq c_{3}\,,}
be a system of linear inequalities, whose solution set is a triangle. How does the solution set look, when we replace one inequality
≥
{\displaystyle {}\geq }
by
≤
{\displaystyle {}\leq }
?
Hand-in-exercises
Solve the following system of inhomogeneous linear equations.
x
+
2
y
+
3
z
+
4
w
=
1
2
x
+
3
y
+
4
z
+
5
w
=
7
x
+
z
=
9
x
+
5
y
+
5
z
+
w
=
0
.
{\displaystyle {\begin{matrix}x&+2y&+3z&+4w&=&1\\2x&+3y&+4z&+5w&=&7\\x&\,\,\,\,\,\,\,\,&+z&\,\,\,\,\,\,\,\,&=&9\\x&+5y&+5z&+w&=&0\,.\end{matrix}}}
Solve the system of linear equations in the variables
x
1
,
x
2
,
…
,
x
10
{\displaystyle {}x_{1},x_{2},\ldots ,x_{10}}
, which is given by the two equations
x
1
+
x
2
+
x
3
+
x
4
+
x
5
+
x
6
+
x
7
+
x
8
+
x
9
+
x
10
=
0
{\displaystyle {}x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}+x_{7}+x_{8}+x_{9}+x_{10}=0\,}
and
x
1
−
x
2
+
x
3
−
x
4
+
x
5
−
x
6
+
x
7
−
x
8
+
x
9
−
x
10
=
0
.
{\displaystyle {}x_{1}-x_{2}+x_{3}-x_{4}+x_{5}-x_{6}+x_{7}-x_{8}+x_{9}-x_{10}=0\,.}
Consider in
R
3
{\displaystyle {}\mathbb {R} ^{3}}
the two planes
E
=
{
(
x
,
y
,
z
)
∈
R
3
∣
3
x
+
4
y
+
5
z
=
2
}
and
F
=
{
(
x
,
y
,
z
)
∈
R
3
∣
2
x
−
y
+
3
z
=
−
1
}
.
{\displaystyle E={\left\{(x,y,z)\in \mathbb {R} ^{3}\mid 3x+4y+5z=2\right\}}{\text{ and }}F={\left\{(x,y,z)\in \mathbb {R} ^{3}\mid 2x-y+3z=-1\right\}}.}
Determine the intersecting line
E
∩
F
{\displaystyle {}E\cap F}
.
Determine a linear equation for the plane in
R
3
{\displaystyle {}\mathbb {R} ^{3}}
, where the three points
(
1
,
0
,
2
)
,
(
4
,
−
3
,
2
)
,
and
(
2
,
1
,
−
1
)
{\displaystyle (1,0,2),\,\,(4,-3,2),\,{\text{ and }}\,(2,1,-1)}
lie.
We consider the linear system
2
x
−
a
y
=
−
2
a
x
+
3
z
=
3
−
1
3
x
+
y
+
z
=
2
{\displaystyle {\begin{matrix}2x&-ay&&=&-2\\ax&&+3z&=&3\\-{\frac {1}{3}}x&+y&+z&=&2\end{matrix}}}
over the real numbers, depending on the parameter
a
∈
R
{\displaystyle {}a\in \mathbb {R} }
.
For which
a
{\displaystyle {}a}
does the system of equations have no solution, one solution, or infinitely many solutions?
Show that a system of linear equations
a
x
+
b
y
=
0
{\displaystyle {}ax+by=0\,}
c
x
+
d
y
=
0
{\displaystyle {}cx+dy=0\,}
has only the trivial solution
(
0
,
0
)
{\displaystyle {}(0,0)}
if and only if
a
d
−
b
c
≠
0
{\displaystyle {}ad-bc\neq 0}
.
A system of linear inequalities is given by
x
≥
0
,
{\displaystyle {}x\geq 0\,,}
y
+
x
≥
0
,
{\displaystyle {}y+x\geq 0\,,}
−
1
−
y
≤
−
x
,
{\displaystyle {}-1-y\leq -x\,,}
5
y
−
2
x
≥
3
.
{\displaystyle {}5y-2x\geq 3\,.}
a) Sketch the solution set of this system.
b) Determine the corners of this solution set.