Question about solution to matrix equation
0
$begingroup$
In the question above, I am unsure how the very first step of the solution came about. How does fact that R is in reduced row echelon form help deduce the value of d?
linear-algebra matrix-equations
edited Dec 8 '18 at 16:33
gimusi
92.8k84494
asked Dec 8 '18 at 16:03
Javier LimJavier Lim
82
$endgroup$
add a comment |
0
$begingroup$
In the question above, I am unsure how the very first step of the solution came about. How does fact that R is in reduced row echelon form help deduce the value of d?
linear-algebra matrix-equations
edited Dec 8 '18 at 16:33
gimusi
92.8k84494
asked Dec 8 '18 at 16:03
Javier LimJavier Lim
82
$endgroup$
add a comment |
0
0
0
$begingroup$
In the question above, I am unsure how the very first step of the solution came about. How does fact that R is in reduced row echelon form help deduce the value of d?
linear-algebra matrix-equations
edited Dec 8 '18 at 16:33
gimusi
92.8k84494
asked Dec 8 '18 at 16:03
Javier LimJavier Lim
82
$endgroup$
In the question above, I am unsure how the very first step of the solution came about. How does fact that R is in reduced row echelon form help deduce the value of d?
linear-algebra matrix-equations
linear-algebra matrix-equations
edited Dec 8 '18 at 16:33
gimusi
92.8k84494
asked Dec 8 '18 at 16:03
Javier LimJavier Lim
82
edited Dec 8 '18 at 16:33
gimusi
92.8k84494
asked Dec 8 '18 at 16:03
Javier LimJavier Lim
82
edited Dec 8 '18 at 16:33
gimusi
92.8k84494
edited Dec 8 '18 at 16:33
gimusi
92.8k84494
edited Dec 8 '18 at 16:33
gimusi
92.8k84494
92.8k84494
asked Dec 8 '18 at 16:03
Javier LimJavier Lim
82
asked Dec 8 '18 at 16:03
Javier LimJavier Lim
82
asked Dec 8 '18 at 16:03
Javier LimJavier Lim
82
82
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
0
$begingroup$
Recall that for the general solution of a linear system is given by
$$x=x_p+x_h$$
with
$x_p$ particular solution such that $$Ax_p=b$$
$x_h$ homogeneous solution such that $$Ax_h=0$$
ans since R has rank $1$ it is equal to
$$R=begin{bmatrix}1&0&0\0&0&0\0&0&0end{bmatrix}$$
What can we conclude now?
answered Dec 8 '18 at 16:10
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
I'm really sorry this might be a trivial question, but I still can't seem to deduce anything.
$endgroup$
– Javier Lim
Dec 8 '18 at 16:22
$begingroup$
@JavierLim Firstly you need to have clear how the complete solution for a linear system work. Have you clear the mening of the terms for $$x=x_p+c_1x_{h1}+c_2x_{h2}$$
$endgroup$
– gimusi
Dec 8 '18 at 16:36
$begingroup$
$x_p$ stands for the particular solution, so $Ax_p = b$ and $c_1x_{h1} + c_2x_{h2}$ stand for vectors in the nullspace of $A$, correct?
$endgroup$
– Javier Lim
Dec 8 '18 at 16:44
$begingroup$
@JavierLim Exactly! Then also for $R$ we have that $Rx_p=d$ and $Rx_h=0$ . Now you need to figure out what $R$ is.
$endgroup$
– gimusi
Dec 8 '18 at 16:47
$begingroup$
Ahh, thank you! I was thinking too hard. Thanks for the help!
$endgroup$
– Javier Lim
Dec 8 '18 at 16:52
|
show 1 more comment
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
$begingroup$
Recall that for the general solution of a linear system is given by
$$x=x_p+x_h$$
with
$x_p$ particular solution such that $$Ax_p=b$$
$x_h$ homogeneous solution such that $$Ax_h=0$$
ans since R has rank $1$ it is equal to
$$R=begin{bmatrix}1&0&0\0&0&0\0&0&0end{bmatrix}$$
What can we conclude now?
answered Dec 8 '18 at 16:10
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
I'm really sorry this might be a trivial question, but I still can't seem to deduce anything.
$endgroup$
– Javier Lim
Dec 8 '18 at 16:22
$begingroup$
@JavierLim Firstly you need to have clear how the complete solution for a linear system work. Have you clear the mening of the terms for $$x=x_p+c_1x_{h1}+c_2x_{h2}$$
$endgroup$
– gimusi
Dec 8 '18 at 16:36
$begingroup$
$x_p$ stands for the particular solution, so $Ax_p = b$ and $c_1x_{h1} + c_2x_{h2}$ stand for vectors in the nullspace of $A$, correct?
$endgroup$
– Javier Lim
Dec 8 '18 at 16:44
$begingroup$
@JavierLim Exactly! Then also for $R$ we have that $Rx_p=d$ and $Rx_h=0$ . Now you need to figure out what $R$ is.
$endgroup$
– gimusi
Dec 8 '18 at 16:47
$begingroup$
Ahh, thank you! I was thinking too hard. Thanks for the help!
$endgroup$
– Javier Lim
Dec 8 '18 at 16:52
|
show 1 more comment
0
$begingroup$
Recall that for the general solution of a linear system is given by
$$x=x_p+x_h$$
with
$x_p$ particular solution such that $$Ax_p=b$$
$x_h$ homogeneous solution such that $$Ax_h=0$$
ans since R has rank $1$ it is equal to
$$R=begin{bmatrix}1&0&0\0&0&0\0&0&0end{bmatrix}$$
What can we conclude now?
answered Dec 8 '18 at 16:10
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
I'm really sorry this might be a trivial question, but I still can't seem to deduce anything.
$endgroup$
– Javier Lim
Dec 8 '18 at 16:22
$begingroup$
@JavierLim Firstly you need to have clear how the complete solution for a linear system work. Have you clear the mening of the terms for $$x=x_p+c_1x_{h1}+c_2x_{h2}$$
$endgroup$
– gimusi
Dec 8 '18 at 16:36
$begingroup$
$x_p$ stands for the particular solution, so $Ax_p = b$ and $c_1x_{h1} + c_2x_{h2}$ stand for vectors in the nullspace of $A$, correct?
$endgroup$
– Javier Lim
Dec 8 '18 at 16:44
$begingroup$
@JavierLim Exactly! Then also for $R$ we have that $Rx_p=d$ and $Rx_h=0$ . Now you need to figure out what $R$ is.
$endgroup$
– gimusi
Dec 8 '18 at 16:47
$begingroup$
Ahh, thank you! I was thinking too hard. Thanks for the help!
$endgroup$
– Javier Lim
Dec 8 '18 at 16:52
|
show 1 more comment
0
0
0
$begingroup$
Recall that for the general solution of a linear system is given by
$$x=x_p+x_h$$
with
$x_p$ particular solution such that $$Ax_p=b$$
$x_h$ homogeneous solution such that $$Ax_h=0$$
ans since R has rank $1$ it is equal to
$$R=begin{bmatrix}1&0&0\0&0&0\0&0&0end{bmatrix}$$
What can we conclude now?
answered Dec 8 '18 at 16:10
gimusigimusi
92.8k84494
$endgroup$
Recall that for the general solution of a linear system is given by
$$x=x_p+x_h$$
with
$x_p$ particular solution such that $$Ax_p=b$$
$x_h$ homogeneous solution such that $$Ax_h=0$$
ans since R has rank $1$ it is equal to
$$R=begin{bmatrix}1&0&0\0&0&0\0&0&0end{bmatrix}$$
What can we conclude now?
answered Dec 8 '18 at 16:10
gimusigimusi
92.8k84494
edited Dec 8 '18 at 16:15
answered Dec 8 '18 at 16:10
gimusigimusi
92.8k84494
answered Dec 8 '18 at 16:10
gimusigimusi
92.8k84494
answered Dec 8 '18 at 16:10
gimusigimusi
92.8k84494
92.8k84494
$begingroup$
I'm really sorry this might be a trivial question, but I still can't seem to deduce anything.
$endgroup$
– Javier Lim
Dec 8 '18 at 16:22
$begingroup$
@JavierLim Firstly you need to have clear how the complete solution for a linear system work. Have you clear the mening of the terms for $$x=x_p+c_1x_{h1}+c_2x_{h2}$$
$endgroup$
– gimusi
Dec 8 '18 at 16:36
$begingroup$
$x_p$ stands for the particular solution, so $Ax_p = b$ and $c_1x_{h1} + c_2x_{h2}$ stand for vectors in the nullspace of $A$, correct?
$endgroup$
– Javier Lim
Dec 8 '18 at 16:44
$begingroup$
@JavierLim Exactly! Then also for $R$ we have that $Rx_p=d$ and $Rx_h=0$ . Now you need to figure out what $R$ is.
$endgroup$
– gimusi
Dec 8 '18 at 16:47
$begingroup$
Ahh, thank you! I was thinking too hard. Thanks for the help!
$endgroup$
– Javier Lim
Dec 8 '18 at 16:52
|
show 1 more comment
$begingroup$
I'm really sorry this might be a trivial question, but I still can't seem to deduce anything.
$endgroup$
– Javier Lim
Dec 8 '18 at 16:22
$begingroup$
@JavierLim Firstly you need to have clear how the complete solution for a linear system work. Have you clear the mening of the terms for $$x=x_p+c_1x_{h1}+c_2x_{h2}$$
$endgroup$
– gimusi
Dec 8 '18 at 16:36
$begingroup$
$x_p$ stands for the particular solution, so $Ax_p = b$ and $c_1x_{h1} + c_2x_{h2}$ stand for vectors in the nullspace of $A$, correct?
$endgroup$
– Javier Lim
Dec 8 '18 at 16:44
$begingroup$
@JavierLim Exactly! Then also for $R$ we have that $Rx_p=d$ and $Rx_h=0$ . Now you need to figure out what $R$ is.
$endgroup$
– gimusi
Dec 8 '18 at 16:47
$begingroup$
Ahh, thank you! I was thinking too hard. Thanks for the help!
$endgroup$
– Javier Lim
Dec 8 '18 at 16:52
$begingroup$
I'm really sorry this might be a trivial question, but I still can't seem to deduce anything.
$endgroup$
– Javier Lim
Dec 8 '18 at 16:22
$begingroup$
I'm really sorry this might be a trivial question, but I still can't seem to deduce anything.
$endgroup$
– Javier Lim
Dec 8 '18 at 16:22
$begingroup$
@JavierLim Firstly you need to have clear how the complete solution for a linear system work. Have you clear the mening of the terms for $$x=x_p+c_1x_{h1}+c_2x_{h2}$$
$endgroup$
– gimusi
Dec 8 '18 at 16:36
$begingroup$
@JavierLim Firstly you need to have clear how the complete solution for a linear system work. Have you clear the mening of the terms for $$x=x_p+c_1x_{h1}+c_2x_{h2}$$
$endgroup$
– gimusi
Dec 8 '18 at 16:36
$begingroup$
$x_p$ stands for the particular solution, so $Ax_p = b$ and $c_1x_{h1} + c_2x_{h2}$ stand for vectors in the nullspace of $A$, correct?
$endgroup$
– Javier Lim
Dec 8 '18 at 16:44
$begingroup$
$x_p$ stands for the particular solution, so $Ax_p = b$ and $c_1x_{h1} + c_2x_{h2}$ stand for vectors in the nullspace of $A$, correct?
$endgroup$
– Javier Lim
Dec 8 '18 at 16:44
$begingroup$
@JavierLim Exactly! Then also for $R$ we have that $Rx_p=d$ and $Rx_h=0$ . Now you need to figure out what $R$ is.
$endgroup$
– gimusi
Dec 8 '18 at 16:47
$begingroup$
@JavierLim Exactly! Then also for $R$ we have that $Rx_p=d$ and $Rx_h=0$ . Now you need to figure out what $R$ is.
$endgroup$
– gimusi
Dec 8 '18 at 16:47
$begingroup$
Ahh, thank you! I was thinking too hard. Thanks for the help!
$endgroup$
– Javier Lim
Dec 8 '18 at 16:52
$begingroup$
Ahh, thank you! I was thinking too hard. Thanks for the help!
$endgroup$
– Javier Lim
Dec 8 '18 at 16:52
|
show 1 more comment
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