How could I determine the solution of this system of equations?












1












$begingroup$


The problem I'm facing is this one (with the answer of the question underlined):



Problem



I reduced the matrix from this:



begin{bmatrix}1&2&-1&4\3&-1&5&2\4&1&(a^2-14)&a+2end{bmatrix}



to this:
begin{bmatrix}1&2&-1&4\0&1&-8/7&10/7\0&0&(a^2-18)&a-4end{bmatrix}



But after that I don't understand why does the system of equations has a solution when $a != 4$ or $a != -4$.



What I learned, applied to this problem, is that:




  • If $a^2 - 18 = 0 $, the system of equations could have infinite solutions or none solution.


Please, could somebody give me a hand on this?
Thanks.






linear-algebra systems-of-equations







share|cite|improve this question









$endgroup$












  • $begingroup$
    As a note: you can do neq to get $neq$.
    $endgroup$
    – Dave
    Dec 31 '18 at 0:16










  • $begingroup$
    If $a^2=18$, then the system is inconsistent. For it to have an infinite number of solutions, the last row of the reduced matrix must consist entirely of zeros, which can never happen.
    $endgroup$
    – amd
    Dec 31 '18 at 0:38
















1












$begingroup$


The problem I'm facing is this one (with the answer of the question underlined):



Problem



I reduced the matrix from this:



begin{bmatrix}1&2&-1&4\3&-1&5&2\4&1&(a^2-14)&a+2end{bmatrix}



to this:
begin{bmatrix}1&2&-1&4\0&1&-8/7&10/7\0&0&(a^2-18)&a-4end{bmatrix}



But after that I don't understand why does the system of equations has a solution when $a != 4$ or $a != -4$.



What I learned, applied to this problem, is that:




  • If $a^2 - 18 = 0 $, the system of equations could have infinite solutions or none solution.


Please, could somebody give me a hand on this?
Thanks.






linear-algebra systems-of-equations







share|cite|improve this question









$endgroup$












  • $begingroup$
    As a note: you can do neq to get $neq$.
    $endgroup$
    – Dave
    Dec 31 '18 at 0:16










  • $begingroup$
    If $a^2=18$, then the system is inconsistent. For it to have an infinite number of solutions, the last row of the reduced matrix must consist entirely of zeros, which can never happen.
    $endgroup$
    – amd
    Dec 31 '18 at 0:38














1












1








1





$begingroup$


The problem I'm facing is this one (with the answer of the question underlined):



Problem



I reduced the matrix from this:



begin{bmatrix}1&2&-1&4\3&-1&5&2\4&1&(a^2-14)&a+2end{bmatrix}



to this:
begin{bmatrix}1&2&-1&4\0&1&-8/7&10/7\0&0&(a^2-18)&a-4end{bmatrix}



But after that I don't understand why does the system of equations has a solution when $a != 4$ or $a != -4$.



What I learned, applied to this problem, is that:




  • If $a^2 - 18 = 0 $, the system of equations could have infinite solutions or none solution.


Please, could somebody give me a hand on this?
Thanks.






linear-algebra systems-of-equations







share|cite|improve this question









$endgroup$




The problem I'm facing is this one (with the answer of the question underlined):



Problem



I reduced the matrix from this:



begin{bmatrix}1&2&-1&4\3&-1&5&2\4&1&(a^2-14)&a+2end{bmatrix}



to this:
begin{bmatrix}1&2&-1&4\0&1&-8/7&10/7\0&0&(a^2-18)&a-4end{bmatrix}



But after that I don't understand why does the system of equations has a solution when $a != 4$ or $a != -4$.



What I learned, applied to this problem, is that:




  • If $a^2 - 18 = 0 $, the system of equations could have infinite solutions or none solution.


Please, could somebody give me a hand on this?
Thanks.






linear-algebra systems-of-equations




linear-algebra systems-of-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 31 '18 at 0:15









Carlos Córdova S.Carlos Córdova S.

134




134












  • $begingroup$
    As a note: you can do neq to get $neq$.
    $endgroup$
    – Dave
    Dec 31 '18 at 0:16










  • $begingroup$
    If $a^2=18$, then the system is inconsistent. For it to have an infinite number of solutions, the last row of the reduced matrix must consist entirely of zeros, which can never happen.
    $endgroup$
    – amd
    Dec 31 '18 at 0:38


















  • $begingroup$
    As a note: you can do neq to get $neq$.
    $endgroup$
    – Dave
    Dec 31 '18 at 0:16










  • $begingroup$
    If $a^2=18$, then the system is inconsistent. For it to have an infinite number of solutions, the last row of the reduced matrix must consist entirely of zeros, which can never happen.
    $endgroup$
    – amd
    Dec 31 '18 at 0:38
















$begingroup$
As a note: you can do neq to get $neq$.
$endgroup$
– Dave
Dec 31 '18 at 0:16




$begingroup$
As a note: you can do neq to get $neq$.
$endgroup$
– Dave
Dec 31 '18 at 0:16












$begingroup$
If $a^2=18$, then the system is inconsistent. For it to have an infinite number of solutions, the last row of the reduced matrix must consist entirely of zeros, which can never happen.
$endgroup$
– amd
Dec 31 '18 at 0:38




$begingroup$
If $a^2=18$, then the system is inconsistent. For it to have an infinite number of solutions, the last row of the reduced matrix must consist entirely of zeros, which can never happen.
$endgroup$
– amd
Dec 31 '18 at 0:38










2 Answers
2






active

oldest

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0












$begingroup$

Let us do it without matrix calculations. We have the linear equations
$$x+2 y-z=4 tag 1$$
$$3 x-y+5 z=2 tag 2$$
$$4 x+y+left(a^2-14right) z=2+a tag 3$$
Using $(1)$ and $(2)$ eliminate $x$ and $y$ as functions of $z$
$$x=frac{8}{7}-frac{9 z}{7} qquad text{and} qquad y=frac{8 z}{7}+frac{10}{7}$$ Plug these results in $(3)$ to get
$$left(a^2-18right) z=a-4$$
S0, the only problem is $a^2=18$ which makes $z$ undefined. If $a^2neq 18$, whatever could be $a$ there are solutions for $x,y,z$.



This is what you did show. Well done and $to +1$ for your post.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    I checked your matrix reduction and it looks good. Your approach looks good too. From here, you would have no solutions if and only if $a^2=18$ since your second matrix would become $$begin{bmatrix}1&2&-1&|&4 \ 0&1&-8/7&|&10/7 \ 0&0&0&|&pmsqrt{18}-4end{bmatrix}$$ which has no solutions (it could not have infinitely many solutions in this case). If $a=pm 4$ the system does have solutions.



    Perhaps there is an error in the given solution, because your calculations and method look fine to me (also, I'm not familiar with the language in which the problem is written, so I am not sure exactly what the problem is asking for).






    share|cite|improve this answer











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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Let us do it without matrix calculations. We have the linear equations
      $$x+2 y-z=4 tag 1$$
      $$3 x-y+5 z=2 tag 2$$
      $$4 x+y+left(a^2-14right) z=2+a tag 3$$
      Using $(1)$ and $(2)$ eliminate $x$ and $y$ as functions of $z$
      $$x=frac{8}{7}-frac{9 z}{7} qquad text{and} qquad y=frac{8 z}{7}+frac{10}{7}$$ Plug these results in $(3)$ to get
      $$left(a^2-18right) z=a-4$$
      S0, the only problem is $a^2=18$ which makes $z$ undefined. If $a^2neq 18$, whatever could be $a$ there are solutions for $x,y,z$.



      This is what you did show. Well done and $to +1$ for your post.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Let us do it without matrix calculations. We have the linear equations
        $$x+2 y-z=4 tag 1$$
        $$3 x-y+5 z=2 tag 2$$
        $$4 x+y+left(a^2-14right) z=2+a tag 3$$
        Using $(1)$ and $(2)$ eliminate $x$ and $y$ as functions of $z$
        $$x=frac{8}{7}-frac{9 z}{7} qquad text{and} qquad y=frac{8 z}{7}+frac{10}{7}$$ Plug these results in $(3)$ to get
        $$left(a^2-18right) z=a-4$$
        S0, the only problem is $a^2=18$ which makes $z$ undefined. If $a^2neq 18$, whatever could be $a$ there are solutions for $x,y,z$.



        This is what you did show. Well done and $to +1$ for your post.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Let us do it without matrix calculations. We have the linear equations
          $$x+2 y-z=4 tag 1$$
          $$3 x-y+5 z=2 tag 2$$
          $$4 x+y+left(a^2-14right) z=2+a tag 3$$
          Using $(1)$ and $(2)$ eliminate $x$ and $y$ as functions of $z$
          $$x=frac{8}{7}-frac{9 z}{7} qquad text{and} qquad y=frac{8 z}{7}+frac{10}{7}$$ Plug these results in $(3)$ to get
          $$left(a^2-18right) z=a-4$$
          S0, the only problem is $a^2=18$ which makes $z$ undefined. If $a^2neq 18$, whatever could be $a$ there are solutions for $x,y,z$.



          This is what you did show. Well done and $to +1$ for your post.






          share|cite|improve this answer









          $endgroup$



          Let us do it without matrix calculations. We have the linear equations
          $$x+2 y-z=4 tag 1$$
          $$3 x-y+5 z=2 tag 2$$
          $$4 x+y+left(a^2-14right) z=2+a tag 3$$
          Using $(1)$ and $(2)$ eliminate $x$ and $y$ as functions of $z$
          $$x=frac{8}{7}-frac{9 z}{7} qquad text{and} qquad y=frac{8 z}{7}+frac{10}{7}$$ Plug these results in $(3)$ to get
          $$left(a^2-18right) z=a-4$$
          S0, the only problem is $a^2=18$ which makes $z$ undefined. If $a^2neq 18$, whatever could be $a$ there are solutions for $x,y,z$.



          This is what you did show. Well done and $to +1$ for your post.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 31 '18 at 2:21









          Claude LeiboviciClaude Leibovici

          125k1158135




          125k1158135























              0












              $begingroup$

              I checked your matrix reduction and it looks good. Your approach looks good too. From here, you would have no solutions if and only if $a^2=18$ since your second matrix would become $$begin{bmatrix}1&2&-1&|&4 \ 0&1&-8/7&|&10/7 \ 0&0&0&|&pmsqrt{18}-4end{bmatrix}$$ which has no solutions (it could not have infinitely many solutions in this case). If $a=pm 4$ the system does have solutions.



              Perhaps there is an error in the given solution, because your calculations and method look fine to me (also, I'm not familiar with the language in which the problem is written, so I am not sure exactly what the problem is asking for).






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                I checked your matrix reduction and it looks good. Your approach looks good too. From here, you would have no solutions if and only if $a^2=18$ since your second matrix would become $$begin{bmatrix}1&2&-1&|&4 \ 0&1&-8/7&|&10/7 \ 0&0&0&|&pmsqrt{18}-4end{bmatrix}$$ which has no solutions (it could not have infinitely many solutions in this case). If $a=pm 4$ the system does have solutions.



                Perhaps there is an error in the given solution, because your calculations and method look fine to me (also, I'm not familiar with the language in which the problem is written, so I am not sure exactly what the problem is asking for).






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I checked your matrix reduction and it looks good. Your approach looks good too. From here, you would have no solutions if and only if $a^2=18$ since your second matrix would become $$begin{bmatrix}1&2&-1&|&4 \ 0&1&-8/7&|&10/7 \ 0&0&0&|&pmsqrt{18}-4end{bmatrix}$$ which has no solutions (it could not have infinitely many solutions in this case). If $a=pm 4$ the system does have solutions.



                  Perhaps there is an error in the given solution, because your calculations and method look fine to me (also, I'm not familiar with the language in which the problem is written, so I am not sure exactly what the problem is asking for).






                  share|cite|improve this answer











                  $endgroup$



                  I checked your matrix reduction and it looks good. Your approach looks good too. From here, you would have no solutions if and only if $a^2=18$ since your second matrix would become $$begin{bmatrix}1&2&-1&|&4 \ 0&1&-8/7&|&10/7 \ 0&0&0&|&pmsqrt{18}-4end{bmatrix}$$ which has no solutions (it could not have infinitely many solutions in this case). If $a=pm 4$ the system does have solutions.



                  Perhaps there is an error in the given solution, because your calculations and method look fine to me (also, I'm not familiar with the language in which the problem is written, so I am not sure exactly what the problem is asking for).







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 31 '18 at 0:42

























                  answered Dec 31 '18 at 0:26









                  DaveDave

                  9,05311033




                  9,05311033






























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