Mixing
Solve linear equation system with Gauss
$begingroup$
I have the following matrix and have to see if it has solutions depending on $a$. My solution:
$M=
left[ {begin{array}{cc}
a & a^2 &| &1 \
-1 & -1& | & -a \
1 & a & | & a
end{array} } right]
$
My attemp was:
Changing first with third line
$=
left[ {begin{array}{cc}
1 & a & | & a \
-1 & -1& | & -a \
a & a^2 &| &1 \
end{array} } right]
$
Add the first row to the second one
$=
left[ {begin{array}{cc}
1 & a & | & a \
0 & a-1& | & 0 \
a & a^2 &| &1 \
end{array} } right]
$
Add the $-a$ of the first line to the third line
$=
left[ {begin{array}{cc}
1 & a & | & a \
0 & a-1& | & 0 \
0 & 0 &| &1-a^2 \
end{array} } right]
$
From $0=1-a^2$there we can obtain that the LES has a solution if $a=+/-1$
Is this a valid solution to the problem and is there a free variable?
matrices
asked Jan 10 at 13:32
SegonutSegonut
62
$endgroup$
add a comment |
$begingroup$
I have the following matrix and have to see if it has solutions depending on $a$. My solution:
$M=
left[ {begin{array}{cc}
a & a^2 &| &1 \
-1 & -1& | & -a \
1 & a & | & a
end{array} } right]
$
My attemp was:
Changing first with third line
$=
left[ {begin{array}{cc}
1 & a & | & a \
-1 & -1& | & -a \
a & a^2 &| &1 \
end{array} } right]
$
Add the first row to the second one
$=
left[ {begin{array}{cc}
1 & a & | & a \
0 & a-1& | & 0 \
a & a^2 &| &1 \
end{array} } right]
$
Add the $-a$ of the first line to the third line
$=
left[ {begin{array}{cc}
1 & a & | & a \
0 & a-1& | & 0 \
0 & 0 &| &1-a^2 \
end{array} } right]
$
From $0=1-a^2$there we can obtain that the LES has a solution if $a=+/-1$
Is this a valid solution to the problem and is there a free variable?
matrices
asked Jan 10 at 13:32
SegonutSegonut
62
$endgroup$
add a comment |
$begingroup$
I have the following matrix and have to see if it has solutions depending on $a$. My solution:
$M=
left[ {begin{array}{cc}
a & a^2 &| &1 \
-1 & -1& | & -a \
1 & a & | & a
end{array} } right]
$
My attemp was:
Changing first with third line
$=
left[ {begin{array}{cc}
1 & a & | & a \
-1 & -1& | & -a \
a & a^2 &| &1 \
end{array} } right]
$
Add the first row to the second one
$=
left[ {begin{array}{cc}
1 & a & | & a \
0 & a-1& | & 0 \
a & a^2 &| &1 \
end{array} } right]
$
Add the $-a$ of the first line to the third line
$=
left[ {begin{array}{cc}
1 & a & | & a \
0 & a-1& | & 0 \
0 & 0 &| &1-a^2 \
end{array} } right]
$
From $0=1-a^2$there we can obtain that the LES has a solution if $a=+/-1$
Is this a valid solution to the problem and is there a free variable?
matrices
asked Jan 10 at 13:32
SegonutSegonut
62
$endgroup$
I have the following matrix and have to see if it has solutions depending on $a$. My solution:
$M=
left[ {begin{array}{cc}
a & a^2 &| &1 \
-1 & -1& | & -a \
1 & a & | & a
end{array} } right]
$
My attemp was:
Changing first with third line
$=
left[ {begin{array}{cc}
1 & a & | & a \
-1 & -1& | & -a \
a & a^2 &| &1 \
end{array} } right]
$
Add the first row to the second one
$=
left[ {begin{array}{cc}
1 & a & | & a \
0 & a-1& | & 0 \
a & a^2 &| &1 \
end{array} } right]
$
Add the $-a$ of the first line to the third line
$=
left[ {begin{array}{cc}
1 & a & | & a \
0 & a-1& | & 0 \
0 & 0 &| &1-a^2 \
end{array} } right]
$
From $0=1-a^2$there we can obtain that the LES has a solution if $a=+/-1$
Is this a valid solution to the problem and is there a free variable?
matrices
matrices
asked Jan 10 at 13:32
SegonutSegonut
62
asked Jan 10 at 13:32
SegonutSegonut
62
asked Jan 10 at 13:32
SegonutSegonut
62
asked Jan 10 at 13:32
SegonutSegonut
62
asked Jan 10 at 13:32
SegonutSegonut
62
62
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $a=pm 1$, then indeed we have a solution.
Now $a=1$, the second row become the zero row, and hence there is a free variable. Can you write down the general solution?
If $a=-1$, the second row is not a zero row, there is no free variable. It has a unique solution.
answered Jan 10 at 13:37
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
$begingroup$
but for $a=-1$ $x_2$ has to be $0$ so the row becomes a zero row again, doesn't it?
$endgroup$
– Segonut
Jan 10 at 13:48
$begingroup$
$x_2=0$ is true, but the second row is $[0, -2, 0]$ which is not a zero row.
$endgroup$
– Siong Thye Goh
Jan 10 at 13:50
$begingroup$
okay, thank you! And the transformations are correct?
$endgroup$
– Segonut
Jan 10 at 13:50
$begingroup$
And for $a=1$ the general solution should be $L=[1-t, t]$, shouldn't it? Because $x_1=1-x_2$ and $x_2=t$ (or any other name)? And for $a-1$ the solution is $[a,0]$, right?
$endgroup$
– Segonut
Jan 10 at 13:54
$begingroup$
Your Gaussian elimination step looks fine. if $a=1$, the general solution is indeed $(1-t,t)$ where $t$ is a free parameter. If $a=-1$, we have $x_2=0$, and the solution is $(a,0)=(-1,0)$.
$endgroup$
– Siong Thye Goh
Jan 10 at 14:03
add a comment |
$begingroup$
begin{align}
M=
left[begin{array}{cc|c}
a & a^2 & 1 \
-1 & -1 & -a \
1 & a & a
end{array}right]
&to
left[begin{array}{cc|c}
1 & a & a \
-1 & -1 & -a \
a & a^2 & 1
end{array}right]
&&R_1leftrightarrow R_3
\[4px]&to
left[begin{array}{cc|c}
1 & a & a \
0 & a-1 & 0 \
0 & 0 & 1-a^2
end{array}right]
&&begin{aligned}R_2&gets R_2+R_1\R_3&gets R_3-aR_1end{aligned}
end{align}
Your work was pretty good.
Now you have to distinguish the cases $a=1$, $a=-1$, $anepm1$
Case $a=1$
The matrix becomes
$$
left[begin{array}{cc|c}
1 & 1 & 1 \
0 & 0 & 0 \
0 & 0 & 0
end{array}right]
$$
and the system has infinitely many solutions
$$
begin{bmatrix} 1-h \ h end{bmatrix}=
begin{bmatrix} 1 \ 0 end{bmatrix} + hbegin{bmatrix} -1 \ 1 end{bmatrix}
$$
Case $a=-1$
The matrix becomes
$$
left[begin{array}{cc|c}
1 & -1 & -1 \
0 & 2 & 0 \
0 & 0 & 0
end{array}right]
to
left[begin{array}{cc|c}
1 & 0 & -1 \
0 & 1 & 0 \
0 & 0 & 0
end{array}right]
$$
and the system has a single solution $begin{bmatrix} -1 \ 0 end{bmatrix}$.
Case $anepm1$
No solution.
answered Jan 10 at 14:58
egregegreg
181k1485203
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068641%2fsolve-linear-equation-system-with-gauss%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $a=pm 1$, then indeed we have a solution.
Now $a=1$, the second row become the zero row, and hence there is a free variable. Can you write down the general solution?
If $a=-1$, the second row is not a zero row, there is no free variable. It has a unique solution.
answered Jan 10 at 13:37
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
$begingroup$
but for $a=-1$ $x_2$ has to be $0$ so the row becomes a zero row again, doesn't it?
$endgroup$
– Segonut
Jan 10 at 13:48
$begingroup$
$x_2=0$ is true, but the second row is $[0, -2, 0]$ which is not a zero row.
$endgroup$
– Siong Thye Goh
Jan 10 at 13:50
$begingroup$
okay, thank you! And the transformations are correct?
$endgroup$
– Segonut
Jan 10 at 13:50
$begingroup$
And for $a=1$ the general solution should be $L=[1-t, t]$, shouldn't it? Because $x_1=1-x_2$ and $x_2=t$ (or any other name)? And for $a-1$ the solution is $[a,0]$, right?
$endgroup$
– Segonut
Jan 10 at 13:54
$begingroup$
Your Gaussian elimination step looks fine. if $a=1$, the general solution is indeed $(1-t,t)$ where $t$ is a free parameter. If $a=-1$, we have $x_2=0$, and the solution is $(a,0)=(-1,0)$.
$endgroup$
– Siong Thye Goh
Jan 10 at 14:03
add a comment |
$begingroup$
If $a=pm 1$, then indeed we have a solution.
Now $a=1$, the second row become the zero row, and hence there is a free variable. Can you write down the general solution?
If $a=-1$, the second row is not a zero row, there is no free variable. It has a unique solution.
answered Jan 10 at 13:37
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
$begingroup$
but for $a=-1$ $x_2$ has to be $0$ so the row becomes a zero row again, doesn't it?
$endgroup$
– Segonut
Jan 10 at 13:48
$begingroup$
$x_2=0$ is true, but the second row is $[0, -2, 0]$ which is not a zero row.
$endgroup$
– Siong Thye Goh
Jan 10 at 13:50
$begingroup$
okay, thank you! And the transformations are correct?
$endgroup$
– Segonut
Jan 10 at 13:50
$begingroup$
And for $a=1$ the general solution should be $L=[1-t, t]$, shouldn't it? Because $x_1=1-x_2$ and $x_2=t$ (or any other name)? And for $a-1$ the solution is $[a,0]$, right?
$endgroup$
– Segonut
Jan 10 at 13:54
$begingroup$
Your Gaussian elimination step looks fine. if $a=1$, the general solution is indeed $(1-t,t)$ where $t$ is a free parameter. If $a=-1$, we have $x_2=0$, and the solution is $(a,0)=(-1,0)$.
$endgroup$
– Siong Thye Goh
Jan 10 at 14:03
add a comment |
$begingroup$
If $a=pm 1$, then indeed we have a solution.
Now $a=1$, the second row become the zero row, and hence there is a free variable. Can you write down the general solution?
If $a=-1$, the second row is not a zero row, there is no free variable. It has a unique solution.
answered Jan 10 at 13:37
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
If $a=pm 1$, then indeed we have a solution.
Now $a=1$, the second row become the zero row, and hence there is a free variable. Can you write down the general solution?
If $a=-1$, the second row is not a zero row, there is no free variable. It has a unique solution.
answered Jan 10 at 13:37
Siong Thye GohSiong Thye Goh
101k1466118
answered Jan 10 at 13:37
Siong Thye GohSiong Thye Goh
101k1466118
answered Jan 10 at 13:37
Siong Thye GohSiong Thye Goh
101k1466118
answered Jan 10 at 13:37
Siong Thye GohSiong Thye Goh
101k1466118
101k1466118
$begingroup$
but for $a=-1$ $x_2$ has to be $0$ so the row becomes a zero row again, doesn't it?
$endgroup$
– Segonut
Jan 10 at 13:48
$begingroup$
$x_2=0$ is true, but the second row is $[0, -2, 0]$ which is not a zero row.
$endgroup$
– Siong Thye Goh
Jan 10 at 13:50
$begingroup$
okay, thank you! And the transformations are correct?
$endgroup$
– Segonut
Jan 10 at 13:50
$begingroup$
And for $a=1$ the general solution should be $L=[1-t, t]$, shouldn't it? Because $x_1=1-x_2$ and $x_2=t$ (or any other name)? And for $a-1$ the solution is $[a,0]$, right?
$endgroup$
– Segonut
Jan 10 at 13:54
$begingroup$
Your Gaussian elimination step looks fine. if $a=1$, the general solution is indeed $(1-t,t)$ where $t$ is a free parameter. If $a=-1$, we have $x_2=0$, and the solution is $(a,0)=(-1,0)$.
$endgroup$
– Siong Thye Goh
Jan 10 at 14:03
add a comment |
$begingroup$
but for $a=-1$ $x_2$ has to be $0$ so the row becomes a zero row again, doesn't it?
$endgroup$
– Segonut
Jan 10 at 13:48
$begingroup$
$x_2=0$ is true, but the second row is $[0, -2, 0]$ which is not a zero row.
$endgroup$
– Siong Thye Goh
Jan 10 at 13:50
$begingroup$
okay, thank you! And the transformations are correct?
$endgroup$
– Segonut
Jan 10 at 13:50
$begingroup$
And for $a=1$ the general solution should be $L=[1-t, t]$, shouldn't it? Because $x_1=1-x_2$ and $x_2=t$ (or any other name)? And for $a-1$ the solution is $[a,0]$, right?
$endgroup$
– Segonut
Jan 10 at 13:54
$begingroup$
Your Gaussian elimination step looks fine. if $a=1$, the general solution is indeed $(1-t,t)$ where $t$ is a free parameter. If $a=-1$, we have $x_2=0$, and the solution is $(a,0)=(-1,0)$.
$endgroup$
– Siong Thye Goh
Jan 10 at 14:03
$begingroup$
but for $a=-1$ $x_2$ has to be $0$ so the row becomes a zero row again, doesn't it?
$endgroup$
– Segonut
Jan 10 at 13:48
$begingroup$
but for $a=-1$ $x_2$ has to be $0$ so the row becomes a zero row again, doesn't it?
$endgroup$
– Segonut
Jan 10 at 13:48
$begingroup$
$x_2=0$ is true, but the second row is $[0, -2, 0]$ which is not a zero row.
$endgroup$
– Siong Thye Goh
Jan 10 at 13:50
$begingroup$
$x_2=0$ is true, but the second row is $[0, -2, 0]$ which is not a zero row.
$endgroup$
– Siong Thye Goh
Jan 10 at 13:50
$begingroup$
okay, thank you! And the transformations are correct?
$endgroup$
– Segonut
Jan 10 at 13:50
$begingroup$
okay, thank you! And the transformations are correct?
$endgroup$
– Segonut
Jan 10 at 13:50
$begingroup$
And for $a=1$ the general solution should be $L=[1-t, t]$, shouldn't it? Because $x_1=1-x_2$ and $x_2=t$ (or any other name)? And for $a-1$ the solution is $[a,0]$, right?
$endgroup$
– Segonut
Jan 10 at 13:54
$begingroup$
And for $a=1$ the general solution should be $L=[1-t, t]$, shouldn't it? Because $x_1=1-x_2$ and $x_2=t$ (or any other name)? And for $a-1$ the solution is $[a,0]$, right?
$endgroup$
– Segonut
Jan 10 at 13:54
$begingroup$
Your Gaussian elimination step looks fine. if $a=1$, the general solution is indeed $(1-t,t)$ where $t$ is a free parameter. If $a=-1$, we have $x_2=0$, and the solution is $(a,0)=(-1,0)$.
$endgroup$
– Siong Thye Goh
Jan 10 at 14:03
$begingroup$
Your Gaussian elimination step looks fine. if $a=1$, the general solution is indeed $(1-t,t)$ where $t$ is a free parameter. If $a=-1$, we have $x_2=0$, and the solution is $(a,0)=(-1,0)$.
$endgroup$
– Siong Thye Goh
Jan 10 at 14:03
add a comment |
$begingroup$
begin{align}
M=
left[begin{array}{cc|c}
a & a^2 & 1 \
-1 & -1 & -a \
1 & a & a
end{array}right]
&to
left[begin{array}{cc|c}
1 & a & a \
-1 & -1 & -a \
a & a^2 & 1
end{array}right]
&&R_1leftrightarrow R_3
\[4px]&to
left[begin{array}{cc|c}
1 & a & a \
0 & a-1 & 0 \
0 & 0 & 1-a^2
end{array}right]
&&begin{aligned}R_2&gets R_2+R_1\R_3&gets R_3-aR_1end{aligned}
end{align}
Your work was pretty good.
Now you have to distinguish the cases $a=1$, $a=-1$, $anepm1$
Case $a=1$
The matrix becomes
$$
left[begin{array}{cc|c}
1 & 1 & 1 \
0 & 0 & 0 \
0 & 0 & 0
end{array}right]
$$
and the system has infinitely many solutions
$$
begin{bmatrix} 1-h \ h end{bmatrix}=
begin{bmatrix} 1 \ 0 end{bmatrix} + hbegin{bmatrix} -1 \ 1 end{bmatrix}
$$
Case $a=-1$
The matrix becomes
$$
left[begin{array}{cc|c}
1 & -1 & -1 \
0 & 2 & 0 \
0 & 0 & 0
end{array}right]
to
left[begin{array}{cc|c}
1 & 0 & -1 \
0 & 1 & 0 \
0 & 0 & 0
end{array}right]
$$
and the system has a single solution $begin{bmatrix} -1 \ 0 end{bmatrix}$.
Case $anepm1$
No solution.
answered Jan 10 at 14:58
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
begin{align}
M=
left[begin{array}{cc|c}
a & a^2 & 1 \
-1 & -1 & -a \
1 & a & a
end{array}right]
&to
left[begin{array}{cc|c}
1 & a & a \
-1 & -1 & -a \
a & a^2 & 1
end{array}right]
&&R_1leftrightarrow R_3
\[4px]&to
left[begin{array}{cc|c}
1 & a & a \
0 & a-1 & 0 \
0 & 0 & 1-a^2
end{array}right]
&&begin{aligned}R_2&gets R_2+R_1\R_3&gets R_3-aR_1end{aligned}
end{align}
Your work was pretty good.
Now you have to distinguish the cases $a=1$, $a=-1$, $anepm1$
Case $a=1$
The matrix becomes
$$
left[begin{array}{cc|c}
1 & 1 & 1 \
0 & 0 & 0 \
0 & 0 & 0
end{array}right]
$$
and the system has infinitely many solutions
$$
begin{bmatrix} 1-h \ h end{bmatrix}=
begin{bmatrix} 1 \ 0 end{bmatrix} + hbegin{bmatrix} -1 \ 1 end{bmatrix}
$$
Case $a=-1$
The matrix becomes
$$
left[begin{array}{cc|c}
1 & -1 & -1 \
0 & 2 & 0 \
0 & 0 & 0
end{array}right]
to
left[begin{array}{cc|c}
1 & 0 & -1 \
0 & 1 & 0 \
0 & 0 & 0
end{array}right]
$$
and the system has a single solution $begin{bmatrix} -1 \ 0 end{bmatrix}$.
Case $anepm1$
No solution.
answered Jan 10 at 14:58
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
begin{align}
M=
left[begin{array}{cc|c}
a & a^2 & 1 \
-1 & -1 & -a \
1 & a & a
end{array}right]
&to
left[begin{array}{cc|c}
1 & a & a \
-1 & -1 & -a \
a & a^2 & 1
end{array}right]
&&R_1leftrightarrow R_3
\[4px]&to
left[begin{array}{cc|c}
1 & a & a \
0 & a-1 & 0 \
0 & 0 & 1-a^2
end{array}right]
&&begin{aligned}R_2&gets R_2+R_1\R_3&gets R_3-aR_1end{aligned}
end{align}
Your work was pretty good.
Now you have to distinguish the cases $a=1$, $a=-1$, $anepm1$
Case $a=1$
The matrix becomes
$$
left[begin{array}{cc|c}
1 & 1 & 1 \
0 & 0 & 0 \
0 & 0 & 0
end{array}right]
$$
and the system has infinitely many solutions
$$
begin{bmatrix} 1-h \ h end{bmatrix}=
begin{bmatrix} 1 \ 0 end{bmatrix} + hbegin{bmatrix} -1 \ 1 end{bmatrix}
$$
Case $a=-1$
The matrix becomes
$$
left[begin{array}{cc|c}
1 & -1 & -1 \
0 & 2 & 0 \
0 & 0 & 0
end{array}right]
to
left[begin{array}{cc|c}
1 & 0 & -1 \
0 & 1 & 0 \
0 & 0 & 0
end{array}right]
$$
and the system has a single solution $begin{bmatrix} -1 \ 0 end{bmatrix}$.
Case $anepm1$
No solution.
answered Jan 10 at 14:58
egregegreg
181k1485203
$endgroup$
begin{align}
M=
left[begin{array}{cc|c}
a & a^2 & 1 \
-1 & -1 & -a \
1 & a & a
end{array}right]
&to
left[begin{array}{cc|c}
1 & a & a \
-1 & -1 & -a \
a & a^2 & 1
end{array}right]
&&R_1leftrightarrow R_3
\[4px]&to
left[begin{array}{cc|c}
1 & a & a \
0 & a-1 & 0 \
0 & 0 & 1-a^2
end{array}right]
&&begin{aligned}R_2&gets R_2+R_1\R_3&gets R_3-aR_1end{aligned}
end{align}
Your work was pretty good.
Now you have to distinguish the cases $a=1$, $a=-1$, $anepm1$
Case $a=1$
The matrix becomes
$$
left[begin{array}{cc|c}
1 & 1 & 1 \
0 & 0 & 0 \
0 & 0 & 0
end{array}right]
$$
and the system has infinitely many solutions
$$
begin{bmatrix} 1-h \ h end{bmatrix}=
begin{bmatrix} 1 \ 0 end{bmatrix} + hbegin{bmatrix} -1 \ 1 end{bmatrix}
$$
Case $a=-1$
The matrix becomes
$$
left[begin{array}{cc|c}
1 & -1 & -1 \
0 & 2 & 0 \
0 & 0 & 0
end{array}right]
to
left[begin{array}{cc|c}
1 & 0 & -1 \
0 & 1 & 0 \
0 & 0 & 0
end{array}right]
$$
and the system has a single solution $begin{bmatrix} -1 \ 0 end{bmatrix}$.
Case $anepm1$
No solution.
answered Jan 10 at 14:58
egregegreg
181k1485203
answered Jan 10 at 14:58
egregegreg
181k1485203
answered Jan 10 at 14:58
egregegreg
181k1485203
answered Jan 10 at 14:58
egregegreg
181k1485203
181k1485203
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068641%2fsolve-linear-equation-system-with-gauss%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
