Exercise for the break Determine an equation for the line in
R
2
{\displaystyle {}\mathbb {R} ^{2}}
(
−
4
,
3
)
{\displaystyle {}(-4,3)}
(
5
,
−
6
)
{\displaystyle {}(5,-6)}
Exercises 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}}}
Determine a linear equation for the plane in
R
3
{\displaystyle {}\mathbb {R} ^{3}}
(
1
,
0
,
0
)
,
(
0
,
1
,
2
)
,
and
(
2
,
3
,
4
)
{\displaystyle (1,0,0),\,(0,1,2),{\text{ and }}(2,3,4)}
lie.
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.
Solve, over the
complex numbers ,MDLD/complex numbers
the
linear system MDLD/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}
K
{\displaystyle {}K}
inhomogeneous linear system MDLD/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}}}
Given a
complex number MDLD/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.
Q
[
3
]
{\displaystyle {}\mathbb {Q} [{\sqrt {3}}]}
a
+
b
3
{\displaystyle {}a+b{\sqrt {3}}}
a
,
b
∈
Q
{\displaystyle {}a,b\in \mathbb {Q} }
a
+
b
3
≠
0
{\displaystyle {}a+b{\sqrt {3}}\neq 0}
a
a
2
+
3
b
2
−
b
a
2
+
3
b
2
3
{\displaystyle {}{\frac {a}{a^{2}+3b^{2}}}-{\frac {b}{a^{2}+3b^{2}}}{\sqrt {3}}}
Solve the following linear system over the field
K
=
Q
[
3
]
{\displaystyle {}K=\mathbb {Q} [{\sqrt {3}}]}
(
2
+
3
−
3
1
2
−
2
−
3
3
)
(
x
y
)
=
(
1
−
3
4
−
2
3
)
.
{\displaystyle {}{\begin{pmatrix}2+{\sqrt {3}}&-{\sqrt {3}}\\{\frac {1}{2}}&-2-3{\sqrt {3}}\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}1-{\sqrt {3}}\\4-2{\sqrt {3}}\end{pmatrix}}\,.}
Solve the linear system
4
y
+
7
z
+
3
=
4
11
y
+
9
z
+
13
=
9
6
y
+
8
z
+
5
=
2
{\displaystyle {\begin{matrix}4y&+7z&+3&=&4\\11y&+9z&+13&=&9\\6y&+8z&+5&=&2\,\end{matrix}}}
with the
substitution method .MDLD/substitution method
Solve the linear system
5
y
+
6
z
+
2
=
6
4
y
+
8
z
+
9
=
5
11
y
+
5
z
+
7
=
8
{\displaystyle {\begin{matrix}5y&+6z&+2&=&6\\4y&+8z&+9&=&5\\11y&+5z&+7&=&8\,\end{matrix}}}
with the
equating method .MDLD/equating method
We consider the linear system over
Q
{\displaystyle {}\mathbb {Q} }
3
7
x
+
4
9
y
−
3
15
z
=
1
35
{\displaystyle {}{\frac {3}{7}}x+{\frac {4}{9}}y-{\frac {3}{15}}z={\frac {1}{35}}\,}
and
−
5
3
x
+
1
4
y
−
6
7
z
=
11
10
.
{\displaystyle {}-{\frac {5}{3}}x+{\frac {1}{4}}y-{\frac {6}{7}}z={\frac {11}{10}}\,.}
Determine a linear system that is equivalent to the given system and has the property that all coefficients are integers.
Show that for every
system of linear equations MDLD/system of linear equations
over
Q
{\displaystyle {}\mathbb {Q} }
equivalent MDLD/equivalent (linear system)
linear system with the property that all coefficients are integers.
Show that, for every
linear system MDLD/linear system
over
Q
{\displaystyle {}\mathbb {Q} }
equivalent MDLD/equivalent (linear system)
system with the property that the modulus of all coefficients is smaller than
1
{\displaystyle {}1}
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.
Out of the resources
R
1
,
R
2
{\displaystyle {}R_{1},R_{2}}
R
3
{\displaystyle {}R_{3}}
P
1
,
P
2
,
P
3
,
P
4
{\displaystyle {}P_{1},P_{2},P_{3},P_{4}}
R
1
{\displaystyle {}R_{1}}
R
2
{\displaystyle {}R_{2}}
R
3
{\displaystyle {}R_{3}}
P
1
{\displaystyle {}P_{1}}
6
{\displaystyle {}6}
2
{\displaystyle {}2}
3
{\displaystyle {}3}
P
2
{\displaystyle {}P_{2}}
4
{\displaystyle {}4}
1
{\displaystyle {}1}
2
{\displaystyle {}2}
P
3
{\displaystyle {}P_{3}}
0
{\displaystyle {}0}
5
{\displaystyle {}5}
2
{\displaystyle {}2}
P
4
{\displaystyle {}P_{4}}
2
{\displaystyle {}2}
1
{\displaystyle {}1}
5
{\displaystyle {}5}
a) Establish a matrix that computes, applied to a four-tuple of commodities, the required resources.
P
1
{\displaystyle {}P_{1}}
P
2
{\displaystyle {}P_{2}}
P
3
{\displaystyle {}P_{3}}
P
4
{\displaystyle {}P_{4}}
6
{\displaystyle {}6}
4
{\displaystyle {}4}
7
{\displaystyle {}7}
5
{\displaystyle {}5}
What resources are necessary?
R
1
{\displaystyle {}R_{1}}
R
2
{\displaystyle {}R_{2}}
R
3
{\displaystyle {}R_{3}}
12
{\displaystyle {}12}
9
{\displaystyle {}9}
13
{\displaystyle {}13}
What tuples of commodities can be produced from this without waste?
Let
D
=
(
d
11
0
⋯
⋯
0
0
d
22
0
⋯
0
⋮
⋱
⋱
⋱
⋮
0
⋯
0
d
n
−
1
n
−
1
0
0
⋯
⋯
0
d
n
n
)
{\displaystyle {}D={\begin{pmatrix}d_{11}&0&\cdots &\cdots &0\\0&d_{22}&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&d_{n-1\,n-1}&0\\0&\cdots &\cdots &0&d_{nn}\end{pmatrix}}\,}
be a
diagonal matrix ,MDLD/diagonal matrix
and let
c
=
(
c
1
⋮
c
n
)
{\displaystyle {}c={\begin{pmatrix}c_{1}\\\vdots \\c_{n}\end{pmatrix}}}
n
{\displaystyle {}n}
field MDLD/field
K
{\displaystyle {}K}
x
=
(
x
1
⋮
x
n
)
{\displaystyle {}x={\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}}
D
x
=
c
,
{\displaystyle {}Dx=c\,,}
and how can you solve it?
Solve the
linear systems MDLD/linear systems
(
7
4
1
2
)
(
x
1
y
1
)
=
(
4
9
)
,
(
7
4
1
2
)
(
x
2
y
2
)
=
(
6
5
)
,
(
7
4
1
2
)
(
x
3
y
3
)
=
(
−
2
5
)
{\displaystyle {\begin{pmatrix}7&4\\1&2\end{pmatrix}}{\begin{pmatrix}x_{1}\\y_{1}\end{pmatrix}}={\begin{pmatrix}4\\9\end{pmatrix}},\,{\begin{pmatrix}7&4\\1&2\end{pmatrix}}{\begin{pmatrix}x_{2}\\y_{2}\end{pmatrix}}={\begin{pmatrix}6\\5\end{pmatrix}},\,{\begin{pmatrix}7&4\\1&2\end{pmatrix}}{\begin{pmatrix}x_{3}\\y_{3}\end{pmatrix}}={\begin{pmatrix}-2\\5\end{pmatrix}}}
simultaneously.
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 }
≤
{\displaystyle {}\leq }
Exercise Exercise 5.19
change ===
Prove the
superposition principle
for systems of linear equations.
Determine, in dependence of the parameter
a
∈
R
{\displaystyle {}a\in \mathbb {R} }
L
a
⊆
R
3
{\displaystyle {}L_{a}\subseteq \mathbb {R} ^{3}}
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\,.}
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}}}
Consider in
R
3
{\displaystyle {}\mathbb {R} ^{3}}
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}}
(
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} }
a
{\displaystyle {}a}
Solve the linear system
5
y
−
z
+
7
=
6
3
y
+
6
z
+
3
=
−
2
8
y
+
8
z
+
7
=
3
{\displaystyle {\begin{matrix}5y&\,\,\,\,-z&+7&=&6\\3y&+6z&+3&=&-2\\8y&+8z&+7&=&3\,\end{matrix}}}
with the
substitution method .MDLD/substitution method
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.
![{\displaystyle {}\mathbb {Q} [{\sqrt {3}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d5c04e759fea7ec9feb40cb4f3a2bcc738b1eb0f)
![{\displaystyle {}K=\mathbb {Q} [{\sqrt {3}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67da0fe9dffae746547f718f932a39acc1a422e2)
