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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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
















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











$endgroup$



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














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











$endgroup$




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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 14:41









Shubham Johri

4,920717




4,920717










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


















  • $begingroup$
    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










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









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    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









    $endgroup$


















      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









      $endgroup$
















        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









        $endgroup$



        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












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 5 at 14:58









        Shubham JohriShubham Johri

        4,920717




        4,920717















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