Solve Ax=b, where A consist of $x^k$, is my method using RREF on augmented matrix valid?












0












$begingroup$


I have Ax=b,
$$
begin{bmatrix}
1 & x & x^2 \
x & 1 & x \
x^2 & x & 1
end{bmatrix}
begin{bmatrix}
a \ b \ c
end{bmatrix} =
begin{bmatrix}
x \ x^2 \ x^3
end{bmatrix}
$$
and I transform it into an augmented matrix:



$$
begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
x & 1 & x & x^2\
x^2 & x & 1 & x^3
end{array}end{bmatrix}
$$



I tried to do RREF by considering x and function of x as part of constants, for example, $R_2-R_1x$:
$$
begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
x-x & 1-x^2 & x-x^3 & x^2-x^2\
x^2 & x & 1 & x^3
end{array}end{bmatrix}=
begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
0 & 1-x^2 & x-x^3 & 0\
x^2 & x & 1 & x^3
end{array}end{bmatrix}
$$



And finally arrive at,
begin{bmatrix}begin{array}{ccc|c}
1 & 0 & 0 & x \
0 & 1 & 0 & 0\
0 & 0 & 1 & 0
end{array}end{bmatrix}



$$ a = x, b = 0, c = 0$$



Are the operations to reduce the matrix by multiplying by x and function of x valid?






linear-algebra matrices systems-of-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is $x$ a constant or variable?
    $endgroup$
    – Shubham Johri
    Dec 10 '18 at 9:25
















0












$begingroup$


I have Ax=b,
$$
begin{bmatrix}
1 & x & x^2 \
x & 1 & x \
x^2 & x & 1
end{bmatrix}
begin{bmatrix}
a \ b \ c
end{bmatrix} =
begin{bmatrix}
x \ x^2 \ x^3
end{bmatrix}
$$
and I transform it into an augmented matrix:



$$
begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
x & 1 & x & x^2\
x^2 & x & 1 & x^3
end{array}end{bmatrix}
$$



I tried to do RREF by considering x and function of x as part of constants, for example, $R_2-R_1x$:
$$
begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
x-x & 1-x^2 & x-x^3 & x^2-x^2\
x^2 & x & 1 & x^3
end{array}end{bmatrix}=
begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
0 & 1-x^2 & x-x^3 & 0\
x^2 & x & 1 & x^3
end{array}end{bmatrix}
$$



And finally arrive at,
begin{bmatrix}begin{array}{ccc|c}
1 & 0 & 0 & x \
0 & 1 & 0 & 0\
0 & 0 & 1 & 0
end{array}end{bmatrix}



$$ a = x, b = 0, c = 0$$



Are the operations to reduce the matrix by multiplying by x and function of x valid?






linear-algebra matrices systems-of-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is $x$ a constant or variable?
    $endgroup$
    – Shubham Johri
    Dec 10 '18 at 9:25














0












0








0


1



$begingroup$


I have Ax=b,
$$
begin{bmatrix}
1 & x & x^2 \
x & 1 & x \
x^2 & x & 1
end{bmatrix}
begin{bmatrix}
a \ b \ c
end{bmatrix} =
begin{bmatrix}
x \ x^2 \ x^3
end{bmatrix}
$$
and I transform it into an augmented matrix:



$$
begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
x & 1 & x & x^2\
x^2 & x & 1 & x^3
end{array}end{bmatrix}
$$



I tried to do RREF by considering x and function of x as part of constants, for example, $R_2-R_1x$:
$$
begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
x-x & 1-x^2 & x-x^3 & x^2-x^2\
x^2 & x & 1 & x^3
end{array}end{bmatrix}=
begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
0 & 1-x^2 & x-x^3 & 0\
x^2 & x & 1 & x^3
end{array}end{bmatrix}
$$



And finally arrive at,
begin{bmatrix}begin{array}{ccc|c}
1 & 0 & 0 & x \
0 & 1 & 0 & 0\
0 & 0 & 1 & 0
end{array}end{bmatrix}



$$ a = x, b = 0, c = 0$$



Are the operations to reduce the matrix by multiplying by x and function of x valid?






linear-algebra matrices systems-of-equations







share|cite|improve this question











$endgroup$




I have Ax=b,
$$
begin{bmatrix}
1 & x & x^2 \
x & 1 & x \
x^2 & x & 1
end{bmatrix}
begin{bmatrix}
a \ b \ c
end{bmatrix} =
begin{bmatrix}
x \ x^2 \ x^3
end{bmatrix}
$$
and I transform it into an augmented matrix:



$$
begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
x & 1 & x & x^2\
x^2 & x & 1 & x^3
end{array}end{bmatrix}
$$



I tried to do RREF by considering x and function of x as part of constants, for example, $R_2-R_1x$:
$$
begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
x-x & 1-x^2 & x-x^3 & x^2-x^2\
x^2 & x & 1 & x^3
end{array}end{bmatrix}=
begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
0 & 1-x^2 & x-x^3 & 0\
x^2 & x & 1 & x^3
end{array}end{bmatrix}
$$



And finally arrive at,
begin{bmatrix}begin{array}{ccc|c}
1 & 0 & 0 & x \
0 & 1 & 0 & 0\
0 & 0 & 1 & 0
end{array}end{bmatrix}



$$ a = x, b = 0, c = 0$$



Are the operations to reduce the matrix by multiplying by x and function of x valid?






linear-algebra matrices systems-of-equations




linear-algebra matrices systems-of-equations






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 8:58









José Carlos Santos

161k22128232




161k22128232










asked Dec 10 '18 at 8:25









drerDdrerD

1609




1609












  • $begingroup$
    Is $x$ a constant or variable?
    $endgroup$
    – Shubham Johri
    Dec 10 '18 at 9:25


















  • $begingroup$
    Is $x$ a constant or variable?
    $endgroup$
    – Shubham Johri
    Dec 10 '18 at 9:25
















$begingroup$
Is $x$ a constant or variable?
$endgroup$
– Shubham Johri
Dec 10 '18 at 9:25




$begingroup$
Is $x$ a constant or variable?
$endgroup$
– Shubham Johri
Dec 10 '18 at 9:25










2 Answers
2






active

oldest

votes


















1












$begingroup$

$begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
0 & 1-x^2 & x-x^3 & 0\
x^2 & x & 1 & x^3
end{array}end{bmatrix}xrightarrow{R_3to R_3-x^2R_1}begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
0 & 1-x^2 & x-x^3 & 0\
0 & x(1-x^2) & 1-x^4 & 0
end{array}end{bmatrix}\xrightarrow{R_3to R_3-xR_2}begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
0 & 1-x^2 & x-x^3 & 0\
0 & 0 & 1-x^2 & 0
end{array}end{bmatrix}xrightarrow[R_1to R_1+R_3]{R_2to R_2+xR_3}begin{bmatrix}begin{array}{ccc|c}
1 & x & 1 & x \
0 & 1-x^2 & 0 & 0\
0 & 0 & 1-x^2 & 0
end{array}end{bmatrix}$



If $x$ is a constant and $1-x^2=0, b,cinBbb R, a=x(1-b)-c=pm(1-b)-c$.



If $x$ is a constant but $1-x^2ne0, b=c=0,a=xnepm1$.



If $x$ is a variable, then in general, $b=c=0, a=x.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    it makes sense if it's constant, but if it's variable does it make sense to do RREF still?
    $endgroup$
    – drerD
    Dec 11 '18 at 1:31










  • $begingroup$
    Yes, you can still do Gaussian Elimination, but if $x$ is a variable and $a,b,c$ are constants, you don't have a solution to the system as shown above.
    $endgroup$
    – Shubham Johri
    Dec 11 '18 at 6:34



















3












$begingroup$

We only can divide a row by $x$ if $xneq0$. Therefore, the case in which $x=0$ should be treated as a special case. More generaly, you can only divide a row by $f(x)$ if $f(x)neq0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    interesting, the x is actually representing $e^{-t_0}, t_0geq 0$. So it seems to work in general. But there's a case where I had to divide a row by $1-x^2 = 1-e^{-2t_0}$ which is zero if $t_0 = 0 $.
    $endgroup$
    – drerD
    Dec 10 '18 at 8:39











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2 Answers
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2 Answers
2






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

$begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
0 & 1-x^2 & x-x^3 & 0\
x^2 & x & 1 & x^3
end{array}end{bmatrix}xrightarrow{R_3to R_3-x^2R_1}begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
0 & 1-x^2 & x-x^3 & 0\
0 & x(1-x^2) & 1-x^4 & 0
end{array}end{bmatrix}\xrightarrow{R_3to R_3-xR_2}begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
0 & 1-x^2 & x-x^3 & 0\
0 & 0 & 1-x^2 & 0
end{array}end{bmatrix}xrightarrow[R_1to R_1+R_3]{R_2to R_2+xR_3}begin{bmatrix}begin{array}{ccc|c}
1 & x & 1 & x \
0 & 1-x^2 & 0 & 0\
0 & 0 & 1-x^2 & 0
end{array}end{bmatrix}$



If $x$ is a constant and $1-x^2=0, b,cinBbb R, a=x(1-b)-c=pm(1-b)-c$.



If $x$ is a constant but $1-x^2ne0, b=c=0,a=xnepm1$.



If $x$ is a variable, then in general, $b=c=0, a=x.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    it makes sense if it's constant, but if it's variable does it make sense to do RREF still?
    $endgroup$
    – drerD
    Dec 11 '18 at 1:31










  • $begingroup$
    Yes, you can still do Gaussian Elimination, but if $x$ is a variable and $a,b,c$ are constants, you don't have a solution to the system as shown above.
    $endgroup$
    – Shubham Johri
    Dec 11 '18 at 6:34
















1












$begingroup$

$begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
0 & 1-x^2 & x-x^3 & 0\
x^2 & x & 1 & x^3
end{array}end{bmatrix}xrightarrow{R_3to R_3-x^2R_1}begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
0 & 1-x^2 & x-x^3 & 0\
0 & x(1-x^2) & 1-x^4 & 0
end{array}end{bmatrix}\xrightarrow{R_3to R_3-xR_2}begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
0 & 1-x^2 & x-x^3 & 0\
0 & 0 & 1-x^2 & 0
end{array}end{bmatrix}xrightarrow[R_1to R_1+R_3]{R_2to R_2+xR_3}begin{bmatrix}begin{array}{ccc|c}
1 & x & 1 & x \
0 & 1-x^2 & 0 & 0\
0 & 0 & 1-x^2 & 0
end{array}end{bmatrix}$



If $x$ is a constant and $1-x^2=0, b,cinBbb R, a=x(1-b)-c=pm(1-b)-c$.



If $x$ is a constant but $1-x^2ne0, b=c=0,a=xnepm1$.



If $x$ is a variable, then in general, $b=c=0, a=x.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    it makes sense if it's constant, but if it's variable does it make sense to do RREF still?
    $endgroup$
    – drerD
    Dec 11 '18 at 1:31










  • $begingroup$
    Yes, you can still do Gaussian Elimination, but if $x$ is a variable and $a,b,c$ are constants, you don't have a solution to the system as shown above.
    $endgroup$
    – Shubham Johri
    Dec 11 '18 at 6:34














1












1








1





$begingroup$

$begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
0 & 1-x^2 & x-x^3 & 0\
x^2 & x & 1 & x^3
end{array}end{bmatrix}xrightarrow{R_3to R_3-x^2R_1}begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
0 & 1-x^2 & x-x^3 & 0\
0 & x(1-x^2) & 1-x^4 & 0
end{array}end{bmatrix}\xrightarrow{R_3to R_3-xR_2}begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
0 & 1-x^2 & x-x^3 & 0\
0 & 0 & 1-x^2 & 0
end{array}end{bmatrix}xrightarrow[R_1to R_1+R_3]{R_2to R_2+xR_3}begin{bmatrix}begin{array}{ccc|c}
1 & x & 1 & x \
0 & 1-x^2 & 0 & 0\
0 & 0 & 1-x^2 & 0
end{array}end{bmatrix}$



If $x$ is a constant and $1-x^2=0, b,cinBbb R, a=x(1-b)-c=pm(1-b)-c$.



If $x$ is a constant but $1-x^2ne0, b=c=0,a=xnepm1$.



If $x$ is a variable, then in general, $b=c=0, a=x.$






share|cite|improve this answer









$endgroup$



$begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
0 & 1-x^2 & x-x^3 & 0\
x^2 & x & 1 & x^3
end{array}end{bmatrix}xrightarrow{R_3to R_3-x^2R_1}begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
0 & 1-x^2 & x-x^3 & 0\
0 & x(1-x^2) & 1-x^4 & 0
end{array}end{bmatrix}\xrightarrow{R_3to R_3-xR_2}begin{bmatrix}begin{array}{ccc|c}
1 & x & x^2 & x \
0 & 1-x^2 & x-x^3 & 0\
0 & 0 & 1-x^2 & 0
end{array}end{bmatrix}xrightarrow[R_1to R_1+R_3]{R_2to R_2+xR_3}begin{bmatrix}begin{array}{ccc|c}
1 & x & 1 & x \
0 & 1-x^2 & 0 & 0\
0 & 0 & 1-x^2 & 0
end{array}end{bmatrix}$



If $x$ is a constant and $1-x^2=0, b,cinBbb R, a=x(1-b)-c=pm(1-b)-c$.



If $x$ is a constant but $1-x^2ne0, b=c=0,a=xnepm1$.



If $x$ is a variable, then in general, $b=c=0, a=x.$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 9:35









Shubham JohriShubham Johri

5,192717




5,192717












  • $begingroup$
    it makes sense if it's constant, but if it's variable does it make sense to do RREF still?
    $endgroup$
    – drerD
    Dec 11 '18 at 1:31










  • $begingroup$
    Yes, you can still do Gaussian Elimination, but if $x$ is a variable and $a,b,c$ are constants, you don't have a solution to the system as shown above.
    $endgroup$
    – Shubham Johri
    Dec 11 '18 at 6:34


















  • $begingroup$
    it makes sense if it's constant, but if it's variable does it make sense to do RREF still?
    $endgroup$
    – drerD
    Dec 11 '18 at 1:31










  • $begingroup$
    Yes, you can still do Gaussian Elimination, but if $x$ is a variable and $a,b,c$ are constants, you don't have a solution to the system as shown above.
    $endgroup$
    – Shubham Johri
    Dec 11 '18 at 6:34
















$begingroup$
it makes sense if it's constant, but if it's variable does it make sense to do RREF still?
$endgroup$
– drerD
Dec 11 '18 at 1:31




$begingroup$
it makes sense if it's constant, but if it's variable does it make sense to do RREF still?
$endgroup$
– drerD
Dec 11 '18 at 1:31












$begingroup$
Yes, you can still do Gaussian Elimination, but if $x$ is a variable and $a,b,c$ are constants, you don't have a solution to the system as shown above.
$endgroup$
– Shubham Johri
Dec 11 '18 at 6:34




$begingroup$
Yes, you can still do Gaussian Elimination, but if $x$ is a variable and $a,b,c$ are constants, you don't have a solution to the system as shown above.
$endgroup$
– Shubham Johri
Dec 11 '18 at 6:34











3












$begingroup$

We only can divide a row by $x$ if $xneq0$. Therefore, the case in which $x=0$ should be treated as a special case. More generaly, you can only divide a row by $f(x)$ if $f(x)neq0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    interesting, the x is actually representing $e^{-t_0}, t_0geq 0$. So it seems to work in general. But there's a case where I had to divide a row by $1-x^2 = 1-e^{-2t_0}$ which is zero if $t_0 = 0 $.
    $endgroup$
    – drerD
    Dec 10 '18 at 8:39
















3












$begingroup$

We only can divide a row by $x$ if $xneq0$. Therefore, the case in which $x=0$ should be treated as a special case. More generaly, you can only divide a row by $f(x)$ if $f(x)neq0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    interesting, the x is actually representing $e^{-t_0}, t_0geq 0$. So it seems to work in general. But there's a case where I had to divide a row by $1-x^2 = 1-e^{-2t_0}$ which is zero if $t_0 = 0 $.
    $endgroup$
    – drerD
    Dec 10 '18 at 8:39














3












3








3





$begingroup$

We only can divide a row by $x$ if $xneq0$. Therefore, the case in which $x=0$ should be treated as a special case. More generaly, you can only divide a row by $f(x)$ if $f(x)neq0$.






share|cite|improve this answer









$endgroup$



We only can divide a row by $x$ if $xneq0$. Therefore, the case in which $x=0$ should be treated as a special case. More generaly, you can only divide a row by $f(x)$ if $f(x)neq0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 8:29









José Carlos SantosJosé Carlos Santos

161k22128232




161k22128232












  • $begingroup$
    interesting, the x is actually representing $e^{-t_0}, t_0geq 0$. So it seems to work in general. But there's a case where I had to divide a row by $1-x^2 = 1-e^{-2t_0}$ which is zero if $t_0 = 0 $.
    $endgroup$
    – drerD
    Dec 10 '18 at 8:39


















  • $begingroup$
    interesting, the x is actually representing $e^{-t_0}, t_0geq 0$. So it seems to work in general. But there's a case where I had to divide a row by $1-x^2 = 1-e^{-2t_0}$ which is zero if $t_0 = 0 $.
    $endgroup$
    – drerD
    Dec 10 '18 at 8:39
















$begingroup$
interesting, the x is actually representing $e^{-t_0}, t_0geq 0$. So it seems to work in general. But there's a case where I had to divide a row by $1-x^2 = 1-e^{-2t_0}$ which is zero if $t_0 = 0 $.
$endgroup$
– drerD
Dec 10 '18 at 8:39




$begingroup$
interesting, the x is actually representing $e^{-t_0}, t_0geq 0$. So it seems to work in general. But there's a case where I had to divide a row by $1-x^2 = 1-e^{-2t_0}$ which is zero if $t_0 = 0 $.
$endgroup$
– drerD
Dec 10 '18 at 8:39


















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Bundesstraße 106

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Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
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Le Mesnil-Réaume

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Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
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Ida-Boy-Ed-Garten

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Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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