Give the linear systems: $(i) x_1+2x_2=2,3x_1+7x_2=8\(ii) x_1+2x_2=1,3x_1+7x_2=7$ [closed]

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Give the linear systems:

(i) $x_1+2x_2=2\3x_1+7x_2=8$

(ii) $x_1+2x_2=1\3x_1+7x_2=7$

Solve both systems by incorporating the right-hand sides into a $2times2$ matrix $B$ and computing the reduced row echelon form of

$$[A|B]=begin{bmatrix}1&2&Big|&2&1\3&7&Big|&8&7end{bmatrix}$$

Both two sides of this form is $2times2$, so I have no idea how to reduce the row echelon form of this one.

linear-algebra matrix-equations

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edited Jan 5 at 14:41

Shubham Johri

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asked Jan 5 at 13:01

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closed as off-topic by Henrik, KM101, darij grinberg, max_zorn, Ali Caglayan Jan 5 at 19:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, KM101, darij grinberg, max_zorn, Ali Caglayan

If this question can be reworded to fit the rules in the help center, please edit the question.

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  Welcome to the website. I have done it for you this time, but in future, please typeset your equations using Mathjax for better presentation and readability.
  $endgroup$
  – Shubham Johri
  Jan 5 at 14:43
  

  
  
  
  

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0

$begingroup$

Give the linear systems:

(i) $x_1+2x_2=2\3x_1+7x_2=8$

(ii) $x_1+2x_2=1\3x_1+7x_2=7$

Solve both systems by incorporating the right-hand sides into a $2times2$ matrix $B$ and computing the reduced row echelon form of

$$[A|B]=begin{bmatrix}1&2&Big|&2&1\3&7&Big|&8&7end{bmatrix}$$

Both two sides of this form is $2times2$, so I have no idea how to reduce the row echelon form of this one.

linear-algebra matrix-equations

share|cite|improve this question

edited Jan 5 at 14:41

Shubham Johri

4,920717

asked Jan 5 at 13:01

Shadow ZShadow Z

396

$endgroup$

closed as off-topic by Henrik, KM101, darij grinberg, max_zorn, Ali Caglayan Jan 5 at 19:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, KM101, darij grinberg, max_zorn, Ali Caglayan

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to the website. I have done it for you this time, but in future, please typeset your equations using Mathjax for better presentation and readability.
  $endgroup$
  – Shubham Johri
  Jan 5 at 14:43
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

Give the linear systems:

(i) $x_1+2x_2=2\3x_1+7x_2=8$

(ii) $x_1+2x_2=1\3x_1+7x_2=7$

Solve both systems by incorporating the right-hand sides into a $2times2$ matrix $B$ and computing the reduced row echelon form of

$$[A|B]=begin{bmatrix}1&2&Big|&2&1\3&7&Big|&8&7end{bmatrix}$$

Both two sides of this form is $2times2$, so I have no idea how to reduce the row echelon form of this one.

linear-algebra matrix-equations

share|cite|improve this question

edited Jan 5 at 14:41

Shubham Johri

4,920717

asked Jan 5 at 13:01

Shadow ZShadow Z

396

$endgroup$

Give the linear systems:

(i) $x_1+2x_2=2\3x_1+7x_2=8$

(ii) $x_1+2x_2=1\3x_1+7x_2=7$

Solve both systems by incorporating the right-hand sides into a $2times2$ matrix $B$ and computing the reduced row echelon form of

$$[A|B]=begin{bmatrix}1&2&Big|&2&1\3&7&Big|&8&7end{bmatrix}$$

Both two sides of this form is $2times2$, so I have no idea how to reduce the row echelon form of this one.

linear-algebra matrix-equations

linear-algebra matrix-equations

share|cite|improve this question

edited Jan 5 at 14:41

Shubham Johri

4,920717

asked Jan 5 at 13:01

Shadow ZShadow Z

396

share|cite|improve this question

edited Jan 5 at 14:41

Shubham Johri

4,920717

asked Jan 5 at 13:01

Shadow ZShadow Z

396

share|cite|improve this question

share|cite|improve this question

edited Jan 5 at 14:41

Shubham Johri

4,920717

edited Jan 5 at 14:41

Shubham Johri

4,920717

edited Jan 5 at 14:41

Shubham Johri

4,920717

4,920717

asked Jan 5 at 13:01

Shadow ZShadow Z

396

asked Jan 5 at 13:01

Shadow ZShadow Z

396

asked Jan 5 at 13:01

Shadow ZShadow Z

396

396

closed as off-topic by Henrik, KM101, darij grinberg, max_zorn, Ali Caglayan Jan 5 at 19:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, KM101, darij grinberg, max_zorn, Ali Caglayan

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Henrik, KM101, darij grinberg, max_zorn, Ali Caglayan Jan 5 at 19:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, KM101, darij grinberg, max_zorn, Ali Caglayan

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to the website. I have done it for you this time, but in future, please typeset your equations using Mathjax for better presentation and readability.
  $endgroup$
  – Shubham Johri
  Jan 5 at 14:43
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to the website. I have done it for you this time, but in future, please typeset your equations using Mathjax for better presentation and readability.
  $endgroup$
  – Shubham Johri
  Jan 5 at 14:43
  

  
  
  
  

$begingroup$
Welcome to the website. I have done it for you this time, but in future, please typeset your equations using Mathjax for better presentation and readability.
$endgroup$
– Shubham Johri
Jan 5 at 14:43

$begingroup$
Welcome to the website. I have done it for you this time, but in future, please typeset your equations using Mathjax for better presentation and readability.
$endgroup$
– Shubham Johri
Jan 5 at 14:43

add a comment | 

1 Answer
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Since the coefficient matrix for both systems is $A=begin{bmatrix}1&2\3&7end{bmatrix}$, we can solve them simultaneously. This is because the row transformations required to be done on $A$ to convert it into its row echelon form will be the same for both systems.

$begin{bmatrix}1&2&Big|&2&1\3&7&Big|&8&7end{bmatrix}xrightarrow{R_2to R_2-3R_1}begin{bmatrix}1&2&Big|&2&1\0&1&Big|&2&4end{bmatrix}xrightarrow{R_1to R_1-2R_2}begin{bmatrix}1&0&Big|&-2&-7\0&1&Big|&2&4end{bmatrix}$

Now the solution for the first system is given $$begin{bmatrix}1&0&Big|&-2\0&1&Big|&2end{bmatrix}$$ and the solution of the second system is given by $$begin{bmatrix}1&0&Big|&-7\0&1&Big|&4end{bmatrix}$$

share|cite|improve this answer

answered Jan 5 at 14:58

Shubham JohriShubham Johri

4,920717

$endgroup$

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

Since the coefficient matrix for both systems is $A=begin{bmatrix}1&2\3&7end{bmatrix}$, we can solve them simultaneously. This is because the row transformations required to be done on $A$ to convert it into its row echelon form will be the same for both systems.

$begin{bmatrix}1&2&Big|&2&1\3&7&Big|&8&7end{bmatrix}xrightarrow{R_2to R_2-3R_1}begin{bmatrix}1&2&Big|&2&1\0&1&Big|&2&4end{bmatrix}xrightarrow{R_1to R_1-2R_2}begin{bmatrix}1&0&Big|&-2&-7\0&1&Big|&2&4end{bmatrix}$

Now the solution for the first system is given $$begin{bmatrix}1&0&Big|&-2\0&1&Big|&2end{bmatrix}$$ and the solution of the second system is given by $$begin{bmatrix}1&0&Big|&-7\0&1&Big|&4end{bmatrix}$$

share|cite|improve this answer

answered Jan 5 at 14:58

Shubham JohriShubham Johri

4,920717

$endgroup$

add a comment | 

1

$begingroup$

Since the coefficient matrix for both systems is $A=begin{bmatrix}1&2\3&7end{bmatrix}$, we can solve them simultaneously. This is because the row transformations required to be done on $A$ to convert it into its row echelon form will be the same for both systems.

$begin{bmatrix}1&2&Big|&2&1\3&7&Big|&8&7end{bmatrix}xrightarrow{R_2to R_2-3R_1}begin{bmatrix}1&2&Big|&2&1\0&1&Big|&2&4end{bmatrix}xrightarrow{R_1to R_1-2R_2}begin{bmatrix}1&0&Big|&-2&-7\0&1&Big|&2&4end{bmatrix}$

Now the solution for the first system is given $$begin{bmatrix}1&0&Big|&-2\0&1&Big|&2end{bmatrix}$$ and the solution of the second system is given by $$begin{bmatrix}1&0&Big|&-7\0&1&Big|&4end{bmatrix}$$

share|cite|improve this answer

answered Jan 5 at 14:58

Shubham JohriShubham Johri

4,920717

$endgroup$

add a comment | 

1

1

1

$begingroup$

Since the coefficient matrix for both systems is $A=begin{bmatrix}1&2\3&7end{bmatrix}$, we can solve them simultaneously. This is because the row transformations required to be done on $A$ to convert it into its row echelon form will be the same for both systems.

$begin{bmatrix}1&2&Big|&2&1\3&7&Big|&8&7end{bmatrix}xrightarrow{R_2to R_2-3R_1}begin{bmatrix}1&2&Big|&2&1\0&1&Big|&2&4end{bmatrix}xrightarrow{R_1to R_1-2R_2}begin{bmatrix}1&0&Big|&-2&-7\0&1&Big|&2&4end{bmatrix}$

Now the solution for the first system is given $$begin{bmatrix}1&0&Big|&-2\0&1&Big|&2end{bmatrix}$$ and the solution of the second system is given by $$begin{bmatrix}1&0&Big|&-7\0&1&Big|&4end{bmatrix}$$

share|cite|improve this answer

answered Jan 5 at 14:58

Shubham JohriShubham Johri

4,920717

$endgroup$

Since the coefficient matrix for both systems is $A=begin{bmatrix}1&2\3&7end{bmatrix}$, we can solve them simultaneously. This is because the row transformations required to be done on $A$ to convert it into its row echelon form will be the same for both systems.

$begin{bmatrix}1&2&Big|&2&1\3&7&Big|&8&7end{bmatrix}xrightarrow{R_2to R_2-3R_1}begin{bmatrix}1&2&Big|&2&1\0&1&Big|&2&4end{bmatrix}xrightarrow{R_1to R_1-2R_2}begin{bmatrix}1&0&Big|&-2&-7\0&1&Big|&2&4end{bmatrix}$

Now the solution for the first system is given $$begin{bmatrix}1&0&Big|&-2\0&1&Big|&2end{bmatrix}$$ and the solution of the second system is given by $$begin{bmatrix}1&0&Big|&-7\0&1&Big|&4end{bmatrix}$$

share|cite|improve this answer

answered Jan 5 at 14:58

Shubham JohriShubham Johri

4,920717

share|cite|improve this answer

share|cite|improve this answer

answered Jan 5 at 14:58

Shubham JohriShubham Johri

4,920717

answered Jan 5 at 14:58

Shubham JohriShubham Johri

4,920717

answered Jan 5 at 14:58

Shubham JohriShubham Johri

4,920717

4,920717

add a comment | 

add a comment | 

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