Mixing
When do the solutions to the linear system $Ax=b$ form a vector subspace?
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When do the solutions to the linear system $Ax=b$ form a vector subspace?
A) If and only if $A$ is invertible;
B) if and only if $b=0$;
C) if and only if $A$ is not invertible;
D) if and only if $bne0$.
Is the answer b? As homogeneous solution are closed under addition and multiplication.
linear-algebra systems-of-equations
edited Oct 7 '15 at 13:27
Martin Sleziak
44.8k10119272
asked Oct 7 '15 at 10:27
El Psy CongrooEl Psy Congroo
11
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add a comment |
$begingroup$
When do the solutions to the linear system $Ax=b$ form a vector subspace?
A) If and only if $A$ is invertible;
B) if and only if $b=0$;
C) if and only if $A$ is not invertible;
D) if and only if $bne0$.
Is the answer b? As homogeneous solution are closed under addition and multiplication.
linear-algebra systems-of-equations
edited Oct 7 '15 at 13:27
Martin Sleziak
44.8k10119272
asked Oct 7 '15 at 10:27
El Psy CongrooEl Psy Congroo
11
$endgroup$
$begingroup$
Welcome to our community! Would be nice if you want to get value from the members that you invest a minimum... Editing the question rather than posting an image, describing what you think...
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– mathcounterexamples.net
Oct 7 '15 at 10:32
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You can see that other users have transcribed the text from the picture you posted. There is still long way to go in order to make this a good question, but it is definitely a start.
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– Martin Sleziak
Oct 7 '15 at 13:31
add a comment |
$begingroup$
When do the solutions to the linear system $Ax=b$ form a vector subspace?
A) If and only if $A$ is invertible;
B) if and only if $b=0$;
C) if and only if $A$ is not invertible;
D) if and only if $bne0$.
Is the answer b? As homogeneous solution are closed under addition and multiplication.
linear-algebra systems-of-equations
edited Oct 7 '15 at 13:27
Martin Sleziak
44.8k10119272
asked Oct 7 '15 at 10:27
El Psy CongrooEl Psy Congroo
11
$endgroup$
When do the solutions to the linear system $Ax=b$ form a vector subspace?
A) If and only if $A$ is invertible;
B) if and only if $b=0$;
C) if and only if $A$ is not invertible;
D) if and only if $bne0$.
Is the answer b? As homogeneous solution are closed under addition and multiplication.
linear-algebra systems-of-equations
linear-algebra systems-of-equations
edited Oct 7 '15 at 13:27
Martin Sleziak
44.8k10119272
asked Oct 7 '15 at 10:27
El Psy CongrooEl Psy Congroo
11
edited Oct 7 '15 at 13:27
Martin Sleziak
44.8k10119272
asked Oct 7 '15 at 10:27
El Psy CongrooEl Psy Congroo
11
edited Oct 7 '15 at 13:27
Martin Sleziak
44.8k10119272
edited Oct 7 '15 at 13:27
Martin Sleziak
44.8k10119272
edited Oct 7 '15 at 13:27
Martin Sleziak
44.8k10119272
44.8k10119272
asked Oct 7 '15 at 10:27
El Psy CongrooEl Psy Congroo
11
asked Oct 7 '15 at 10:27
El Psy CongrooEl Psy Congroo
11
asked Oct 7 '15 at 10:27
El Psy CongrooEl Psy Congroo
11
11
$begingroup$
Welcome to our community! Would be nice if you want to get value from the members that you invest a minimum... Editing the question rather than posting an image, describing what you think...
$endgroup$
– mathcounterexamples.net
Oct 7 '15 at 10:32
$begingroup$
You can see that other users have transcribed the text from the picture you posted. There is still long way to go in order to make this a good question, but it is definitely a start.
$endgroup$
– Martin Sleziak
Oct 7 '15 at 13:31
add a comment |
$begingroup$
Welcome to our community! Would be nice if you want to get value from the members that you invest a minimum... Editing the question rather than posting an image, describing what you think...
$endgroup$
– mathcounterexamples.net
Oct 7 '15 at 10:32
$begingroup$
You can see that other users have transcribed the text from the picture you posted. There is still long way to go in order to make this a good question, but it is definitely a start.
$endgroup$
– Martin Sleziak
Oct 7 '15 at 13:31
$begingroup$
Welcome to our community! Would be nice if you want to get value from the members that you invest a minimum... Editing the question rather than posting an image, describing what you think...
$endgroup$
– mathcounterexamples.net
Oct 7 '15 at 10:32
$begingroup$
Welcome to our community! Would be nice if you want to get value from the members that you invest a minimum... Editing the question rather than posting an image, describing what you think...
$endgroup$
– mathcounterexamples.net
Oct 7 '15 at 10:32
$begingroup$
You can see that other users have transcribed the text from the picture you posted. There is still long way to go in order to make this a good question, but it is definitely a start.
$endgroup$
– Martin Sleziak
Oct 7 '15 at 13:31
$begingroup$
You can see that other users have transcribed the text from the picture you posted. There is still long way to go in order to make this a good question, but it is definitely a start.
$endgroup$
– Martin Sleziak
Oct 7 '15 at 13:31
add a comment |
1 Answer
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Yes, almost. The catch that doesn't make the b alternative entirely correct is that if $A$ is not a square matrix then the solutions does not need to be in $mathbb R^n$. Otherwise it's true:
If $bne 0$ you will not have $0$ in the set of solutions which is required in a vector space. If $b=0$ on the other hand you will have that $cx+y$ are solutions to if $x$ and $y$ are solutions.
The other alternatives either prohibits $b=0$ or at least is not equivalent to it.
answered Oct 7 '15 at 10:32
skykingskyking
14.3k1929
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1 Answer
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1 Answer
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oldest
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active
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votes
$begingroup$
Yes, almost. The catch that doesn't make the b alternative entirely correct is that if $A$ is not a square matrix then the solutions does not need to be in $mathbb R^n$. Otherwise it's true:
If $bne 0$ you will not have $0$ in the set of solutions which is required in a vector space. If $b=0$ on the other hand you will have that $cx+y$ are solutions to if $x$ and $y$ are solutions.
The other alternatives either prohibits $b=0$ or at least is not equivalent to it.
answered Oct 7 '15 at 10:32
skykingskyking
14.3k1929
$endgroup$
add a comment |
$begingroup$
Yes, almost. The catch that doesn't make the b alternative entirely correct is that if $A$ is not a square matrix then the solutions does not need to be in $mathbb R^n$. Otherwise it's true:
If $bne 0$ you will not have $0$ in the set of solutions which is required in a vector space. If $b=0$ on the other hand you will have that $cx+y$ are solutions to if $x$ and $y$ are solutions.
The other alternatives either prohibits $b=0$ or at least is not equivalent to it.
answered Oct 7 '15 at 10:32
skykingskyking
14.3k1929
$endgroup$
add a comment |
$begingroup$
Yes, almost. The catch that doesn't make the b alternative entirely correct is that if $A$ is not a square matrix then the solutions does not need to be in $mathbb R^n$. Otherwise it's true:
If $bne 0$ you will not have $0$ in the set of solutions which is required in a vector space. If $b=0$ on the other hand you will have that $cx+y$ are solutions to if $x$ and $y$ are solutions.
The other alternatives either prohibits $b=0$ or at least is not equivalent to it.
answered Oct 7 '15 at 10:32
skykingskyking
14.3k1929
$endgroup$
Yes, almost. The catch that doesn't make the b alternative entirely correct is that if $A$ is not a square matrix then the solutions does not need to be in $mathbb R^n$. Otherwise it's true:
If $bne 0$ you will not have $0$ in the set of solutions which is required in a vector space. If $b=0$ on the other hand you will have that $cx+y$ are solutions to if $x$ and $y$ are solutions.
The other alternatives either prohibits $b=0$ or at least is not equivalent to it.
answered Oct 7 '15 at 10:32
skykingskyking
14.3k1929
answered Oct 7 '15 at 10:32
skykingskyking
14.3k1929
answered Oct 7 '15 at 10:32
skykingskyking
14.3k1929
answered Oct 7 '15 at 10:32
skykingskyking
14.3k1929
14.3k1929
add a comment |
add a comment |
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$begingroup$
Welcome to our community! Would be nice if you want to get value from the members that you invest a minimum... Editing the question rather than posting an image, describing what you think...
$endgroup$
– mathcounterexamples.net
Oct 7 '15 at 10:32
$begingroup$
You can see that other users have transcribed the text from the picture you posted. There is still long way to go in order to make this a good question, but it is definitely a start.
$endgroup$
– Martin Sleziak
Oct 7 '15 at 13:31