Determining whether a system is consistent or inconsistent from the linear dependence of its columns
Clash Royale CLAN TAG#URR8PPP
$begingroup$
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 Feb 20 at 19:01
s0ulr3aper07
578111
asked Feb 20 at 18:39
Elena TorreElena Torre
596
$endgroup$
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 Feb 20 at 19:01
s0ulr3aper07
578111
asked Feb 20 at 18:39
Elena TorreElena Torre
596
$endgroup$
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 Feb 20 at 19:01
s0ulr3aper07
578111
asked Feb 20 at 18:39
Elena TorreElena Torre
596
$endgroup$
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 Feb 20 at 19:01
s0ulr3aper07
578111
asked Feb 20 at 18:39
Elena TorreElena Torre
596
edited Feb 20 at 19:01
s0ulr3aper07
578111
asked Feb 20 at 18:39
Elena TorreElena Torre
596
edited Feb 20 at 19:01
s0ulr3aper07
578111
edited Feb 20 at 19:01
s0ulr3aper07
578111
edited Feb 20 at 19:01
s0ulr3aper07
578111
578111
asked Feb 20 at 18:39
Elena TorreElena Torre
596
asked Feb 20 at 18:39
Elena TorreElena Torre
596
asked Feb 20 at 18:39
Elena TorreElena Torre
596
596
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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For $x=beginpmatrixx_1\x_2\x_3endpmatrix$ 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 Feb 20 at 18:45
David HillDavid Hill
9,2381619
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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 Feb 20 at 18:56
amdamd
31k21051
$endgroup$
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 Feb 20 at 18:58
s0ulr3aper07s0ulr3aper07
578111
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For $x=beginpmatrixx_1\x_2\x_3endpmatrix$ 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 Feb 20 at 18:45
David HillDavid Hill
9,2381619
$endgroup$
add a comment |
$begingroup$
For $x=beginpmatrixx_1\x_2\x_3endpmatrix$ 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 Feb 20 at 18:45
David HillDavid Hill
9,2381619
$endgroup$
add a comment |
$begingroup$
For $x=beginpmatrixx_1\x_2\x_3endpmatrix$ 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 Feb 20 at 18:45
David HillDavid Hill
9,2381619
$endgroup$
For $x=beginpmatrixx_1\x_2\x_3endpmatrix$ 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 Feb 20 at 18:45
David HillDavid Hill
9,2381619
answered Feb 20 at 18:45
David HillDavid Hill
9,2381619
answered Feb 20 at 18:45
David HillDavid Hill
9,2381619
answered Feb 20 at 18:45
David HillDavid Hill
9,2381619
9,2381619
add a comment |
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 Feb 20 at 18:56
amdamd
31k21051
$endgroup$
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 Feb 20 at 18:56
amdamd
31k21051
$endgroup$
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 Feb 20 at 18:56
amdamd
31k21051
$endgroup$
Hint: For the equation $Ax=b$ to have any solution at all, $b$ must be an element of the column space of $A$.
answered Feb 20 at 18:56
amdamd
31k21051
answered Feb 20 at 18:56
amdamd
31k21051
answered Feb 20 at 18:56
amdamd
31k21051
answered Feb 20 at 18:56
amdamd
31k21051
31k21051
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 Feb 20 at 18:58
s0ulr3aper07s0ulr3aper07
578111
$endgroup$
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 Feb 20 at 18:58
s0ulr3aper07s0ulr3aper07
578111
$endgroup$
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 Feb 20 at 18:58
s0ulr3aper07s0ulr3aper07
578111
$endgroup$
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 Feb 20 at 18:58
s0ulr3aper07s0ulr3aper07
578111
answered Feb 20 at 18:58
s0ulr3aper07s0ulr3aper07
578111
answered Feb 20 at 18:58
s0ulr3aper07s0ulr3aper07
578111
answered Feb 20 at 18:58
s0ulr3aper07s0ulr3aper07
578111
578111
add a comment |
add a comment |
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