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Basis for nullspace - Free variables and basis for N(A)

2

$begingroup$

I'm trying to understand the solution for a textbook question where I am asked to find a basis for N(A). I took a screenshot of the page, and I circled the portion I don't quite understand. I don't understand where the values for s and t came from? Like how did they get s = 1, and t = -2 for the first part in that tuple? I tried finding a correlation between these values and the RREF matrix but I just couldn't see the connection. Any help or explanation would be appreciated.

linear-algebra matrices

share|cite|improve this question

asked 2 hours ago

GilmoreGirlingGilmoreGirling

425

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know about free variables?
  $endgroup$
  – Tojrah
  2 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

I'm trying to understand the solution for a textbook question where I am asked to find a basis for N(A). I took a screenshot of the page, and I circled the portion I don't quite understand. I don't understand where the values for s and t came from? Like how did they get s = 1, and t = -2 for the first part in that tuple? I tried finding a correlation between these values and the RREF matrix but I just couldn't see the connection. Any help or explanation would be appreciated.

linear-algebra matrices

share|cite|improve this question

asked 2 hours ago

GilmoreGirlingGilmoreGirling

425

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know about free variables?
  $endgroup$
  – Tojrah
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I'm trying to understand the solution for a textbook question where I am asked to find a basis for N(A). I took a screenshot of the page, and I circled the portion I don't quite understand. I don't understand where the values for s and t came from? Like how did they get s = 1, and t = -2 for the first part in that tuple? I tried finding a correlation between these values and the RREF matrix but I just couldn't see the connection. Any help or explanation would be appreciated.

linear-algebra matrices

share|cite|improve this question

asked 2 hours ago

GilmoreGirlingGilmoreGirling

425

$endgroup$

share|cite|improve this question

asked 2 hours ago

GilmoreGirlingGilmoreGirling

425

share|cite|improve this question

asked 2 hours ago

GilmoreGirlingGilmoreGirling

425

asked 2 hours ago

GilmoreGirlingGilmoreGirling

425

asked 2 hours ago

GilmoreGirlingGilmoreGirling

425

2 Answers
2

active

oldest

votes

3

$begingroup$

We already transform $A$ into $$begin{pmatrix} 1&-1&0&2\0&0&1&-1end{pmatrix}$$
Since first and third column contains the pivot, so $x_1$ and $x_3$ are pivot variables. That is, $x_2$ and $x_4$ are free.
The task now is to solve $$x_1-x_2+2x_4=0\x_3-x_4=0$$
Set $x_2=1$ and $x_4=0$ to see $$begin{pmatrix} 1\1\0\0 end{pmatrix}$$ is a special solution. Similarly set $x_2=0$ and $x_4=1$ to see $$begin{pmatrix} -2\0\1\1 end{pmatrix}$$ is a special solution. Now the combination $$sbegin{pmatrix} 1\1\0\0 end{pmatrix}+tbegin{pmatrix} -2\0\1\1 end{pmatrix}$$ gives all solution s to $Ax=0$

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Chinnapparaj RChinnapparaj R

6,61721029

$endgroup$

add a comment | 

3

$begingroup$

Columns 1 and 3 have the pivots. So the other two columns (2 and 4) correspond to the free variables.

Then call $x_4=t$. Then the last equation says $$x_3 - x_4 =0 leftrightarrow x_3 = x_4= t$$

Call the other free variable $x_2=s$.
Then the first equation becomes, after substiting what we know so far:

$$x_1 - s + 2t = 0$$ from which

$$x_1 = s -2t$$ follows.

share|cite|improve this answer

answered 2 hours ago

Henno BrandsmaHenno Brandsma

117k349127

$endgroup$

add a comment | 

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3

$begingroup$

We already transform $A$ into $$begin{pmatrix} 1&-1&0&2\0&0&1&-1end{pmatrix}$$
Since first and third column contains the pivot, so $x_1$ and $x_3$ are pivot variables. That is, $x_2$ and $x_4$ are free.
The task now is to solve $$x_1-x_2+2x_4=0\x_3-x_4=0$$
Set $x_2=1$ and $x_4=0$ to see $$begin{pmatrix} 1\1\0\0 end{pmatrix}$$ is a special solution. Similarly set $x_2=0$ and $x_4=1$ to see $$begin{pmatrix} -2\0\1\1 end{pmatrix}$$ is a special solution. Now the combination $$sbegin{pmatrix} 1\1\0\0 end{pmatrix}+tbegin{pmatrix} -2\0\1\1 end{pmatrix}$$ gives all solution s to $Ax=0$

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Chinnapparaj RChinnapparaj R

6,61721029

$endgroup$

add a comment | 

3

$begingroup$

We already transform $A$ into $$begin{pmatrix} 1&-1&0&2\0&0&1&-1end{pmatrix}$$
Since first and third column contains the pivot, so $x_1$ and $x_3$ are pivot variables. That is, $x_2$ and $x_4$ are free.
The task now is to solve $$x_1-x_2+2x_4=0\x_3-x_4=0$$
Set $x_2=1$ and $x_4=0$ to see $$begin{pmatrix} 1\1\0\0 end{pmatrix}$$ is a special solution. Similarly set $x_2=0$ and $x_4=1$ to see $$begin{pmatrix} -2\0\1\1 end{pmatrix}$$ is a special solution. Now the combination $$sbegin{pmatrix} 1\1\0\0 end{pmatrix}+tbegin{pmatrix} -2\0\1\1 end{pmatrix}$$ gives all solution s to $Ax=0$

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Chinnapparaj RChinnapparaj R

6,61721029

$endgroup$

add a comment | 

$begingroup$

We already transform $A$ into $$begin{pmatrix} 1&-1&0&2\0&0&1&-1end{pmatrix}$$
Since first and third column contains the pivot, so $x_1$ and $x_3$ are pivot variables. That is, $x_2$ and $x_4$ are free.
The task now is to solve $$x_1-x_2+2x_4=0\x_3-x_4=0$$
Set $x_2=1$ and $x_4=0$ to see $$begin{pmatrix} 1\1\0\0 end{pmatrix}$$ is a special solution. Similarly set $x_2=0$ and $x_4=1$ to see $$begin{pmatrix} -2\0\1\1 end{pmatrix}$$ is a special solution. Now the combination $$sbegin{pmatrix} 1\1\0\0 end{pmatrix}+tbegin{pmatrix} -2\0\1\1 end{pmatrix}$$ gives all solution s to $Ax=0$

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Chinnapparaj RChinnapparaj R

6,61721029

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

Chinnapparaj RChinnapparaj R

6,61721029

answered 2 hours ago

Chinnapparaj RChinnapparaj R

6,61721029

answered 2 hours ago

Chinnapparaj RChinnapparaj R

6,61721029

3

$begingroup$

Columns 1 and 3 have the pivots. So the other two columns (2 and 4) correspond to the free variables.

Then call $x_4=t$. Then the last equation says $$x_3 - x_4 =0 leftrightarrow x_3 = x_4= t$$

Call the other free variable $x_2=s$.
Then the first equation becomes, after substiting what we know so far:

$$x_1 - s + 2t = 0$$ from which

$$x_1 = s -2t$$ follows.

share|cite|improve this answer

answered 2 hours ago

Henno BrandsmaHenno Brandsma

117k349127

$endgroup$

add a comment | 

3

$begingroup$

Columns 1 and 3 have the pivots. So the other two columns (2 and 4) correspond to the free variables.

Then call $x_4=t$. Then the last equation says $$x_3 - x_4 =0 leftrightarrow x_3 = x_4= t$$

Call the other free variable $x_2=s$.
Then the first equation becomes, after substiting what we know so far:

$$x_1 - s + 2t = 0$$ from which

$$x_1 = s -2t$$ follows.

share|cite|improve this answer

answered 2 hours ago

Henno BrandsmaHenno Brandsma

117k349127

$endgroup$

add a comment | 

$begingroup$

Columns 1 and 3 have the pivots. So the other two columns (2 and 4) correspond to the free variables.

Then call $x_4=t$. Then the last equation says $$x_3 - x_4 =0 leftrightarrow x_3 = x_4= t$$

Call the other free variable $x_2=s$.
Then the first equation becomes, after substiting what we know so far:

$$x_1 - s + 2t = 0$$ from which

$$x_1 = s -2t$$ follows.

share|cite|improve this answer

answered 2 hours ago

Henno BrandsmaHenno Brandsma

117k349127

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Henno BrandsmaHenno Brandsma

117k349127

answered 2 hours ago

Henno BrandsmaHenno Brandsma

117k349127

answered 2 hours ago

Henno BrandsmaHenno Brandsma

117k349127