Statements about a system of equations true/false
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I have the following question here:
Consider the homogeneous system $Ax=0$ of $m$ equations for $n>m$ unknowns. Which is the following statements is false?
$(A)$ $x=0$ is the only solution
$(B)$ There are always infinitely many distinct solutions.
$(C)$ The general solutions has at least $n-m$ free variables (free parameters).
$(D)$ If $x_1$ and $x_2$ are two solutions, so is $x_1+tx_2$ for any $t epsilon mathbb{R}. $
$(E)$ If $x_1$ solves $Ax=0$ and $x_2$ solves $Ax=b$, then $x_2+tx_1$ solves $Ax=b$, $t epsilon mathbb{R}$.
The answer is supposed to be $(A)$. It makes sense since a homogeneous system does not require the trivial solution for there to be a solution to the homogeneous system of equations (I assume my reasoning is right?) but why are the other choices true?
I think $(B)$ is true because solutions to a system of equations can be parametrized in infinitely many ways so hence it is true. Is that correct?
I am not sure why $(C)$ and $(D)$ are true though. Can someone please explain? Thanks!
linear-algebra systems-of-equations
asked Dec 17 '18 at 6:51
Future Math personFuture Math person
977817
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show 5 more comments
$begingroup$
I have the following question here:
Consider the homogeneous system $Ax=0$ of $m$ equations for $n>m$ unknowns. Which is the following statements is false?
$(A)$ $x=0$ is the only solution
$(B)$ There are always infinitely many distinct solutions.
$(C)$ The general solutions has at least $n-m$ free variables (free parameters).
$(D)$ If $x_1$ and $x_2$ are two solutions, so is $x_1+tx_2$ for any $t epsilon mathbb{R}. $
$(E)$ If $x_1$ solves $Ax=0$ and $x_2$ solves $Ax=b$, then $x_2+tx_1$ solves $Ax=b$, $t epsilon mathbb{R}$.
The answer is supposed to be $(A)$. It makes sense since a homogeneous system does not require the trivial solution for there to be a solution to the homogeneous system of equations (I assume my reasoning is right?) but why are the other choices true?
I think $(B)$ is true because solutions to a system of equations can be parametrized in infinitely many ways so hence it is true. Is that correct?
I am not sure why $(C)$ and $(D)$ are true though. Can someone please explain? Thanks!
linear-algebra systems-of-equations
asked Dec 17 '18 at 6:51
Future Math personFuture Math person
977817
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($D$) follows from ($E$) considering $b=0$. $(A)$ is false because otherwise $A$ were invertible.
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– Yadati Kiran
Dec 17 '18 at 6:55
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But why is $(E)$ true? How are you able to express the solution as a linear combination like that?
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– Future Math person
Dec 17 '18 at 7:06
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Knowing the fact that matrices are always linear, $A(x_1+tx_2)=Ax_1+tcdot Ax_2=?$
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– Yadati Kiran
Dec 17 '18 at 7:10
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$0+b$? Because those are the solutions to the given system?
$endgroup$
– Future Math person
Dec 17 '18 at 7:14
$begingroup$
$0+tcdot b$. Yes according to the if statement in the proposition $E$.
$endgroup$
– Yadati Kiran
Dec 17 '18 at 7:16
|
show 5 more comments
$begingroup$
I have the following question here:
Consider the homogeneous system $Ax=0$ of $m$ equations for $n>m$ unknowns. Which is the following statements is false?
$(A)$ $x=0$ is the only solution
$(B)$ There are always infinitely many distinct solutions.
$(C)$ The general solutions has at least $n-m$ free variables (free parameters).
$(D)$ If $x_1$ and $x_2$ are two solutions, so is $x_1+tx_2$ for any $t epsilon mathbb{R}. $
$(E)$ If $x_1$ solves $Ax=0$ and $x_2$ solves $Ax=b$, then $x_2+tx_1$ solves $Ax=b$, $t epsilon mathbb{R}$.
The answer is supposed to be $(A)$. It makes sense since a homogeneous system does not require the trivial solution for there to be a solution to the homogeneous system of equations (I assume my reasoning is right?) but why are the other choices true?
I think $(B)$ is true because solutions to a system of equations can be parametrized in infinitely many ways so hence it is true. Is that correct?
I am not sure why $(C)$ and $(D)$ are true though. Can someone please explain? Thanks!
linear-algebra systems-of-equations
asked Dec 17 '18 at 6:51
Future Math personFuture Math person
977817
$endgroup$
I have the following question here:
Consider the homogeneous system $Ax=0$ of $m$ equations for $n>m$ unknowns. Which is the following statements is false?
$(A)$ $x=0$ is the only solution
$(B)$ There are always infinitely many distinct solutions.
$(C)$ The general solutions has at least $n-m$ free variables (free parameters).
$(D)$ If $x_1$ and $x_2$ are two solutions, so is $x_1+tx_2$ for any $t epsilon mathbb{R}. $
$(E)$ If $x_1$ solves $Ax=0$ and $x_2$ solves $Ax=b$, then $x_2+tx_1$ solves $Ax=b$, $t epsilon mathbb{R}$.
The answer is supposed to be $(A)$. It makes sense since a homogeneous system does not require the trivial solution for there to be a solution to the homogeneous system of equations (I assume my reasoning is right?) but why are the other choices true?
I think $(B)$ is true because solutions to a system of equations can be parametrized in infinitely many ways so hence it is true. Is that correct?
I am not sure why $(C)$ and $(D)$ are true though. Can someone please explain? Thanks!
linear-algebra systems-of-equations
linear-algebra systems-of-equations
asked Dec 17 '18 at 6:51
Future Math personFuture Math person
977817
asked Dec 17 '18 at 6:51
Future Math personFuture Math person
977817
asked Dec 17 '18 at 6:51
Future Math personFuture Math person
977817
asked Dec 17 '18 at 6:51
Future Math personFuture Math person
977817
asked Dec 17 '18 at 6:51
Future Math personFuture Math person
977817
977817
$begingroup$
($D$) follows from ($E$) considering $b=0$. $(A)$ is false because otherwise $A$ were invertible.
$endgroup$
– Yadati Kiran
Dec 17 '18 at 6:55
$begingroup$
But why is $(E)$ true? How are you able to express the solution as a linear combination like that?
$endgroup$
– Future Math person
Dec 17 '18 at 7:06
$begingroup$
Knowing the fact that matrices are always linear, $A(x_1+tx_2)=Ax_1+tcdot Ax_2=?$
$endgroup$
– Yadati Kiran
Dec 17 '18 at 7:10
$begingroup$
$0+b$? Because those are the solutions to the given system?
$endgroup$
– Future Math person
Dec 17 '18 at 7:14
$begingroup$
$0+tcdot b$. Yes according to the if statement in the proposition $E$.
$endgroup$
– Yadati Kiran
Dec 17 '18 at 7:16
|
show 5 more comments
$begingroup$
($D$) follows from ($E$) considering $b=0$. $(A)$ is false because otherwise $A$ were invertible.
$endgroup$
– Yadati Kiran
Dec 17 '18 at 6:55
$begingroup$
But why is $(E)$ true? How are you able to express the solution as a linear combination like that?
$endgroup$
– Future Math person
Dec 17 '18 at 7:06
$begingroup$
Knowing the fact that matrices are always linear, $A(x_1+tx_2)=Ax_1+tcdot Ax_2=?$
$endgroup$
– Yadati Kiran
Dec 17 '18 at 7:10
$begingroup$
$0+b$? Because those are the solutions to the given system?
$endgroup$
– Future Math person
Dec 17 '18 at 7:14
$begingroup$
$0+tcdot b$. Yes according to the if statement in the proposition $E$.
$endgroup$
– Yadati Kiran
Dec 17 '18 at 7:16
$begingroup$
($D$) follows from ($E$) considering $b=0$. $(A)$ is false because otherwise $A$ were invertible.
$endgroup$
– Yadati Kiran
Dec 17 '18 at 6:55
$begingroup$
($D$) follows from ($E$) considering $b=0$. $(A)$ is false because otherwise $A$ were invertible.
$endgroup$
– Yadati Kiran
Dec 17 '18 at 6:55
$begingroup$
But why is $(E)$ true? How are you able to express the solution as a linear combination like that?
$endgroup$
– Future Math person
Dec 17 '18 at 7:06
$begingroup$
But why is $(E)$ true? How are you able to express the solution as a linear combination like that?
$endgroup$
– Future Math person
Dec 17 '18 at 7:06
$begingroup$
Knowing the fact that matrices are always linear, $A(x_1+tx_2)=Ax_1+tcdot Ax_2=?$
$endgroup$
– Yadati Kiran
Dec 17 '18 at 7:10
$begingroup$
Knowing the fact that matrices are always linear, $A(x_1+tx_2)=Ax_1+tcdot Ax_2=?$
$endgroup$
– Yadati Kiran
Dec 17 '18 at 7:10
$begingroup$
$0+b$? Because those are the solutions to the given system?
$endgroup$
– Future Math person
Dec 17 '18 at 7:14
$begingroup$
$0+b$? Because those are the solutions to the given system?
$endgroup$
– Future Math person
Dec 17 '18 at 7:14
$begingroup$
$0+tcdot b$. Yes according to the if statement in the proposition $E$.
$endgroup$
– Yadati Kiran
Dec 17 '18 at 7:16
$begingroup$
$0+tcdot b$. Yes according to the if statement in the proposition $E$.
$endgroup$
– Yadati Kiran
Dec 17 '18 at 7:16
|
show 5 more comments
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$begingroup$
($D$) follows from ($E$) considering $b=0$. $(A)$ is false because otherwise $A$ were invertible.
$endgroup$
– Yadati Kiran
Dec 17 '18 at 6:55
$begingroup$
But why is $(E)$ true? How are you able to express the solution as a linear combination like that?
$endgroup$
– Future Math person
Dec 17 '18 at 7:06
$begingroup$
Knowing the fact that matrices are always linear, $A(x_1+tx_2)=Ax_1+tcdot Ax_2=?$
$endgroup$
– Yadati Kiran
Dec 17 '18 at 7:10
$begingroup$
$0+b$? Because those are the solutions to the given system?
$endgroup$
– Future Math person
Dec 17 '18 at 7:14
$begingroup$
$0+tcdot b$. Yes according to the if statement in the proposition $E$.
$endgroup$
– Yadati Kiran
Dec 17 '18 at 7:16