Determining whether a system is consistent or inconsistent from the linear dependence of its columns
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A is a $5 times 3$ matrix that has column vectors {$v_1, v_2, v_3$}, the only solution to $Ax = 0$ is $x= 0$, and the set of vectors {$v_1, v_2, v_3, b $} is linearly dependent. Is the system $Ax = b$ consistent, inconsistent or could be either one?
So far, I understand that if a $Ax = 0 $ has only the trivial solution ($x = 0$), then its columns are linearly independent. This means that the column vectors of A are linearly independent. Since the set of vectors {$v_1, v_2, v_3, b $} is linearly dependent, this leads me to believe that b is a combination of the columns of A, but I'm not sure how that affects the whether or not the system is consistent.
linear-algebra
edited yesterday
s0ulr3aper07
456111
asked yesterday
Elena TorreElena Torre
404
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$begingroup$
A is a $5 times 3$ matrix that has column vectors {$v_1, v_2, v_3$}, the only solution to $Ax = 0$ is $x= 0$, and the set of vectors {$v_1, v_2, v_3, b $} is linearly dependent. Is the system $Ax = b$ consistent, inconsistent or could be either one?
So far, I understand that if a $Ax = 0 $ has only the trivial solution ($x = 0$), then its columns are linearly independent. This means that the column vectors of A are linearly independent. Since the set of vectors {$v_1, v_2, v_3, b $} is linearly dependent, this leads me to believe that b is a combination of the columns of A, but I'm not sure how that affects the whether or not the system is consistent.
linear-algebra
edited yesterday
s0ulr3aper07
456111
asked yesterday
Elena TorreElena Torre
404
$endgroup$
add a comment |
$begingroup$
A is a $5 times 3$ matrix that has column vectors {$v_1, v_2, v_3$}, the only solution to $Ax = 0$ is $x= 0$, and the set of vectors {$v_1, v_2, v_3, b $} is linearly dependent. Is the system $Ax = b$ consistent, inconsistent or could be either one?
So far, I understand that if a $Ax = 0 $ has only the trivial solution ($x = 0$), then its columns are linearly independent. This means that the column vectors of A are linearly independent. Since the set of vectors {$v_1, v_2, v_3, b $} is linearly dependent, this leads me to believe that b is a combination of the columns of A, but I'm not sure how that affects the whether or not the system is consistent.
linear-algebra
edited yesterday
s0ulr3aper07
456111
asked yesterday
Elena TorreElena Torre
404
$endgroup$
A is a $5 times 3$ matrix that has column vectors {$v_1, v_2, v_3$}, the only solution to $Ax = 0$ is $x= 0$, and the set of vectors {$v_1, v_2, v_3, b $} is linearly dependent. Is the system $Ax = b$ consistent, inconsistent or could be either one?
So far, I understand that if a $Ax = 0 $ has only the trivial solution ($x = 0$), then its columns are linearly independent. This means that the column vectors of A are linearly independent. Since the set of vectors {$v_1, v_2, v_3, b $} is linearly dependent, this leads me to believe that b is a combination of the columns of A, but I'm not sure how that affects the whether or not the system is consistent.
linear-algebra
linear-algebra
edited yesterday
s0ulr3aper07
456111
asked yesterday
Elena TorreElena Torre
404
edited yesterday
s0ulr3aper07
456111
asked yesterday
Elena TorreElena Torre
404
edited yesterday
s0ulr3aper07
456111
edited yesterday
s0ulr3aper07
456111
edited yesterday
s0ulr3aper07
456111
456111
asked yesterday
Elena TorreElena Torre
404
asked yesterday
Elena TorreElena Torre
404
asked yesterday
Elena TorreElena Torre
404
404
add a comment |
add a comment |
3 Answers
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For $x=begin{pmatrix}x_1\x_2\x_3end{pmatrix}$ you can compute
$$Ax=x_1v_1+x_2v_3+x_3v_3.$$
so, if $b=alpha v_1+beta v_2+gamma v_3$, can you find $x$ so that $Ax=b$?
answered yesterday
David HillDavid Hill
9,1111619
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Hint: For the equation $Ax=b$ to have any solution at all, $b$ must be an element of the column space of $A$.
answered yesterday
amdamd
30.5k21050
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Your reasoning about $b$ being a Linear Combination of the columns of $A$ is correct. Just realize now that $b=(alpha_1cdot v_1)+(alpha_2cdot v_2)+(alpha_3cdot v_3)Rightarrow Ax=b where x={ alpha_1,alpha_2,alpha_3}$
Thus, the system has at least one solution and must be consistent.
answered yesterday
s0ulr3aper07s0ulr3aper07
456111
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add a comment |
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3 Answers
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3 Answers
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$begingroup$
For $x=begin{pmatrix}x_1\x_2\x_3end{pmatrix}$ you can compute
$$Ax=x_1v_1+x_2v_3+x_3v_3.$$
so, if $b=alpha v_1+beta v_2+gamma v_3$, can you find $x$ so that $Ax=b$?
answered yesterday
David HillDavid Hill
9,1111619
$endgroup$
add a comment |
$begingroup$
For $x=begin{pmatrix}x_1\x_2\x_3end{pmatrix}$ you can compute
$$Ax=x_1v_1+x_2v_3+x_3v_3.$$
so, if $b=alpha v_1+beta v_2+gamma v_3$, can you find $x$ so that $Ax=b$?
answered yesterday
David HillDavid Hill
9,1111619
$endgroup$
add a comment |
$begingroup$
For $x=begin{pmatrix}x_1\x_2\x_3end{pmatrix}$ you can compute
$$Ax=x_1v_1+x_2v_3+x_3v_3.$$
so, if $b=alpha v_1+beta v_2+gamma v_3$, can you find $x$ so that $Ax=b$?
answered yesterday
David HillDavid Hill
9,1111619
$endgroup$
For $x=begin{pmatrix}x_1\x_2\x_3end{pmatrix}$ you can compute
$$Ax=x_1v_1+x_2v_3+x_3v_3.$$
so, if $b=alpha v_1+beta v_2+gamma v_3$, can you find $x$ so that $Ax=b$?
answered yesterday
David HillDavid Hill
9,1111619
answered yesterday
David HillDavid Hill
9,1111619
answered yesterday
David HillDavid Hill
9,1111619
answered yesterday
David HillDavid Hill
9,1111619
9,1111619
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$begingroup$
Hint: For the equation $Ax=b$ to have any solution at all, $b$ must be an element of the column space of $A$.
answered yesterday
amdamd
30.5k21050
$endgroup$
add a comment |
$begingroup$
Hint: For the equation $Ax=b$ to have any solution at all, $b$ must be an element of the column space of $A$.
answered yesterday
amdamd
30.5k21050
$endgroup$
add a comment |
$begingroup$
Hint: For the equation $Ax=b$ to have any solution at all, $b$ must be an element of the column space of $A$.
answered yesterday
amdamd
30.5k21050
$endgroup$
Hint: For the equation $Ax=b$ to have any solution at all, $b$ must be an element of the column space of $A$.
answered yesterday
amdamd
30.5k21050
answered yesterday
amdamd
30.5k21050
answered yesterday
amdamd
30.5k21050
answered yesterday
amdamd
30.5k21050
30.5k21050
add a comment |
add a comment |
$begingroup$
Your reasoning about $b$ being a Linear Combination of the columns of $A$ is correct. Just realize now that $b=(alpha_1cdot v_1)+(alpha_2cdot v_2)+(alpha_3cdot v_3)Rightarrow Ax=b where x={ alpha_1,alpha_2,alpha_3}$
Thus, the system has at least one solution and must be consistent.
answered yesterday
s0ulr3aper07s0ulr3aper07
456111
$endgroup$
add a comment |
$begingroup$
Your reasoning about $b$ being a Linear Combination of the columns of $A$ is correct. Just realize now that $b=(alpha_1cdot v_1)+(alpha_2cdot v_2)+(alpha_3cdot v_3)Rightarrow Ax=b where x={ alpha_1,alpha_2,alpha_3}$
Thus, the system has at least one solution and must be consistent.
answered yesterday
s0ulr3aper07s0ulr3aper07
456111
$endgroup$
add a comment |
$begingroup$
Your reasoning about $b$ being a Linear Combination of the columns of $A$ is correct. Just realize now that $b=(alpha_1cdot v_1)+(alpha_2cdot v_2)+(alpha_3cdot v_3)Rightarrow Ax=b where x={ alpha_1,alpha_2,alpha_3}$
Thus, the system has at least one solution and must be consistent.
answered yesterday
s0ulr3aper07s0ulr3aper07
456111
$endgroup$
Your reasoning about $b$ being a Linear Combination of the columns of $A$ is correct. Just realize now that $b=(alpha_1cdot v_1)+(alpha_2cdot v_2)+(alpha_3cdot v_3)Rightarrow Ax=b where x={ alpha_1,alpha_2,alpha_3}$
Thus, the system has at least one solution and must be consistent.
answered yesterday
s0ulr3aper07s0ulr3aper07
456111
answered yesterday
s0ulr3aper07s0ulr3aper07
456111
answered yesterday
s0ulr3aper07s0ulr3aper07
456111
answered yesterday
s0ulr3aper07s0ulr3aper07
456111
456111
add a comment |
add a comment |
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