Need help solving for a linear system $x_1$ + $2x_2$ = $λx_1$ & $2x_1$ + $2x_2$ = $λx_2$












0












$begingroup$


I'm working on an exercise problem that is as follows. Find all real values of λ for which the system has a non trivial solution



$x_1$ + $2x_2$ = $λx_1$



$2x_1$ + $x_2$ = $λx_2$



I'm not sure if I did it correctly but I rewrote the equation as:



$$
begin{matrix}
1 & 2\
2 & 1\
end{matrix}
$$



multiplied by



$$
begin{matrix}
x_1\
x_2\
end{matrix}
$$



which equals



$$
begin{matrix}
λx_1\
λx_2\
end{matrix}
$$



I then multiplied the both sides by the inverse of the the 2x2 matrix giving me just
$$
begin{matrix}
x_1\
x_2\
end{matrix}
$$



on the left hand side and I solved for $x_1$ and $x_2$ in terms of $λx_1$ and $λx_2$. After this I plugged the values I got for $x_1$ and $x_2$ back into the first equation in the linear system. I feel like I'm on the wrong track, so any help on how to solve this would be appreciated.






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Sorry I corrected it. I wrote down the intial linear system wrong.
    $endgroup$
    – PCR
    Feb 17 '16 at 20:32










  • $begingroup$
    Move everything to one side of the equal sign, correct like terms, turn it into a matrix. Solve for when the determinant is zero.
    $endgroup$
    – Kaynex
    Feb 17 '16 at 20:51
















0












$begingroup$


I'm working on an exercise problem that is as follows. Find all real values of λ for which the system has a non trivial solution



$x_1$ + $2x_2$ = $λx_1$



$2x_1$ + $x_2$ = $λx_2$



I'm not sure if I did it correctly but I rewrote the equation as:



$$
begin{matrix}
1 & 2\
2 & 1\
end{matrix}
$$



multiplied by



$$
begin{matrix}
x_1\
x_2\
end{matrix}
$$



which equals



$$
begin{matrix}
λx_1\
λx_2\
end{matrix}
$$



I then multiplied the both sides by the inverse of the the 2x2 matrix giving me just
$$
begin{matrix}
x_1\
x_2\
end{matrix}
$$



on the left hand side and I solved for $x_1$ and $x_2$ in terms of $λx_1$ and $λx_2$. After this I plugged the values I got for $x_1$ and $x_2$ back into the first equation in the linear system. I feel like I'm on the wrong track, so any help on how to solve this would be appreciated.






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Sorry I corrected it. I wrote down the intial linear system wrong.
    $endgroup$
    – PCR
    Feb 17 '16 at 20:32










  • $begingroup$
    Move everything to one side of the equal sign, correct like terms, turn it into a matrix. Solve for when the determinant is zero.
    $endgroup$
    – Kaynex
    Feb 17 '16 at 20:51














0












0








0


0



$begingroup$


I'm working on an exercise problem that is as follows. Find all real values of λ for which the system has a non trivial solution



$x_1$ + $2x_2$ = $λx_1$



$2x_1$ + $x_2$ = $λx_2$



I'm not sure if I did it correctly but I rewrote the equation as:



$$
begin{matrix}
1 & 2\
2 & 1\
end{matrix}
$$



multiplied by



$$
begin{matrix}
x_1\
x_2\
end{matrix}
$$



which equals



$$
begin{matrix}
λx_1\
λx_2\
end{matrix}
$$



I then multiplied the both sides by the inverse of the the 2x2 matrix giving me just
$$
begin{matrix}
x_1\
x_2\
end{matrix}
$$



on the left hand side and I solved for $x_1$ and $x_2$ in terms of $λx_1$ and $λx_2$. After this I plugged the values I got for $x_1$ and $x_2$ back into the first equation in the linear system. I feel like I'm on the wrong track, so any help on how to solve this would be appreciated.






linear-algebra







share|cite|improve this question











$endgroup$




I'm working on an exercise problem that is as follows. Find all real values of λ for which the system has a non trivial solution



$x_1$ + $2x_2$ = $λx_1$



$2x_1$ + $x_2$ = $λx_2$



I'm not sure if I did it correctly but I rewrote the equation as:



$$
begin{matrix}
1 & 2\
2 & 1\
end{matrix}
$$



multiplied by



$$
begin{matrix}
x_1\
x_2\
end{matrix}
$$



which equals



$$
begin{matrix}
λx_1\
λx_2\
end{matrix}
$$



I then multiplied the both sides by the inverse of the the 2x2 matrix giving me just
$$
begin{matrix}
x_1\
x_2\
end{matrix}
$$



on the left hand side and I solved for $x_1$ and $x_2$ in terms of $λx_1$ and $λx_2$. After this I plugged the values I got for $x_1$ and $x_2$ back into the first equation in the linear system. I feel like I'm on the wrong track, so any help on how to solve this would be appreciated.






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 8:31









Math Girl

627318




627318










asked Feb 17 '16 at 20:23









PCRPCR

216




216












  • $begingroup$
    Sorry I corrected it. I wrote down the intial linear system wrong.
    $endgroup$
    – PCR
    Feb 17 '16 at 20:32










  • $begingroup$
    Move everything to one side of the equal sign, correct like terms, turn it into a matrix. Solve for when the determinant is zero.
    $endgroup$
    – Kaynex
    Feb 17 '16 at 20:51


















  • $begingroup$
    Sorry I corrected it. I wrote down the intial linear system wrong.
    $endgroup$
    – PCR
    Feb 17 '16 at 20:32










  • $begingroup$
    Move everything to one side of the equal sign, correct like terms, turn it into a matrix. Solve for when the determinant is zero.
    $endgroup$
    – Kaynex
    Feb 17 '16 at 20:51
















$begingroup$
Sorry I corrected it. I wrote down the intial linear system wrong.
$endgroup$
– PCR
Feb 17 '16 at 20:32




$begingroup$
Sorry I corrected it. I wrote down the intial linear system wrong.
$endgroup$
– PCR
Feb 17 '16 at 20:32












$begingroup$
Move everything to one side of the equal sign, correct like terms, turn it into a matrix. Solve for when the determinant is zero.
$endgroup$
– Kaynex
Feb 17 '16 at 20:51




$begingroup$
Move everything to one side of the equal sign, correct like terms, turn it into a matrix. Solve for when the determinant is zero.
$endgroup$
– Kaynex
Feb 17 '16 at 20:51










3 Answers
3






active

oldest

votes


















1












$begingroup$

If you do not know linear algebra, then rewrite the first equation to $x_2=frac{x_1(lambda-1)}{2}$ and substitute this into the second equation. Then we obtain
$$
(lambda+1)(lambda-3)x_1=0.
$$
Now argue that we have a non-trivial solution only for $(lambda+1)(lambda-3)=0$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    You have



    $$begin{bmatrix}
    1 & 2 \
    2 & 2
    end{bmatrix} begin{bmatrix}
    x_1 \
    x_2
    end{bmatrix}=
    lambda begin{bmatrix}
    x_1 \
    x_2
    end{bmatrix},$$
    so $lambda$ is eigenvalue of matrix $begin{bmatrix}
    1 & 2 \
    2 & 2
    end{bmatrix}$. Can you find all eigenvalues of that matrix using, for example, characteristic polynomial?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I don't know what an eigenvalue is. Looking ahead in the textbook, it's a topic in the next section, so I'm assuming the textbook wants me to solve it without knowing what an eigenvalue is.
      $endgroup$
      – PCR
      Feb 17 '16 at 20:34










    • $begingroup$
      @PCR what if your textbook wants to lure you to the idea of eigenvctors and eigenvalues? In my time, I had much, much more trouble with those teachers who did not want me to look ahead.
      $endgroup$
      – Gyro Gearloose
      Feb 17 '16 at 20:42



















    0












    $begingroup$

    The mistake you made: You have chosen $A=begin{bmatrix}1 & 2\2 & 1end{bmatrix}$ and $x=begin{bmatrix}x_1\x_2end{bmatrix}$. So you have $Ax=lambda x$ in the matrix equation form. When you multiply $A^{-1}$, you will get $x=Ix=A^{-1}Ax=lambda A^{-1}x$. Observe the right hand side involves another matrix $A^{-1}$. Hence you are not getting the solution. Consider the technique suggested by 'Dietrich Burde' and try to generalize. Your text book is probably designed in such a way that you can understand the generalizations in the next sections.






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      If you do not know linear algebra, then rewrite the first equation to $x_2=frac{x_1(lambda-1)}{2}$ and substitute this into the second equation. Then we obtain
      $$
      (lambda+1)(lambda-3)x_1=0.
      $$
      Now argue that we have a non-trivial solution only for $(lambda+1)(lambda-3)=0$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        If you do not know linear algebra, then rewrite the first equation to $x_2=frac{x_1(lambda-1)}{2}$ and substitute this into the second equation. Then we obtain
        $$
        (lambda+1)(lambda-3)x_1=0.
        $$
        Now argue that we have a non-trivial solution only for $(lambda+1)(lambda-3)=0$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          If you do not know linear algebra, then rewrite the first equation to $x_2=frac{x_1(lambda-1)}{2}$ and substitute this into the second equation. Then we obtain
          $$
          (lambda+1)(lambda-3)x_1=0.
          $$
          Now argue that we have a non-trivial solution only for $(lambda+1)(lambda-3)=0$.






          share|cite|improve this answer









          $endgroup$



          If you do not know linear algebra, then rewrite the first equation to $x_2=frac{x_1(lambda-1)}{2}$ and substitute this into the second equation. Then we obtain
          $$
          (lambda+1)(lambda-3)x_1=0.
          $$
          Now argue that we have a non-trivial solution only for $(lambda+1)(lambda-3)=0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 17 '16 at 20:42









          Dietrich BurdeDietrich Burde

          80.2k647104




          80.2k647104























              0












              $begingroup$

              You have



              $$begin{bmatrix}
              1 & 2 \
              2 & 2
              end{bmatrix} begin{bmatrix}
              x_1 \
              x_2
              end{bmatrix}=
              lambda begin{bmatrix}
              x_1 \
              x_2
              end{bmatrix},$$
              so $lambda$ is eigenvalue of matrix $begin{bmatrix}
              1 & 2 \
              2 & 2
              end{bmatrix}$. Can you find all eigenvalues of that matrix using, for example, characteristic polynomial?






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                I don't know what an eigenvalue is. Looking ahead in the textbook, it's a topic in the next section, so I'm assuming the textbook wants me to solve it without knowing what an eigenvalue is.
                $endgroup$
                – PCR
                Feb 17 '16 at 20:34










              • $begingroup$
                @PCR what if your textbook wants to lure you to the idea of eigenvctors and eigenvalues? In my time, I had much, much more trouble with those teachers who did not want me to look ahead.
                $endgroup$
                – Gyro Gearloose
                Feb 17 '16 at 20:42
















              0












              $begingroup$

              You have



              $$begin{bmatrix}
              1 & 2 \
              2 & 2
              end{bmatrix} begin{bmatrix}
              x_1 \
              x_2
              end{bmatrix}=
              lambda begin{bmatrix}
              x_1 \
              x_2
              end{bmatrix},$$
              so $lambda$ is eigenvalue of matrix $begin{bmatrix}
              1 & 2 \
              2 & 2
              end{bmatrix}$. Can you find all eigenvalues of that matrix using, for example, characteristic polynomial?






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                I don't know what an eigenvalue is. Looking ahead in the textbook, it's a topic in the next section, so I'm assuming the textbook wants me to solve it without knowing what an eigenvalue is.
                $endgroup$
                – PCR
                Feb 17 '16 at 20:34










              • $begingroup$
                @PCR what if your textbook wants to lure you to the idea of eigenvctors and eigenvalues? In my time, I had much, much more trouble with those teachers who did not want me to look ahead.
                $endgroup$
                – Gyro Gearloose
                Feb 17 '16 at 20:42














              0












              0








              0





              $begingroup$

              You have



              $$begin{bmatrix}
              1 & 2 \
              2 & 2
              end{bmatrix} begin{bmatrix}
              x_1 \
              x_2
              end{bmatrix}=
              lambda begin{bmatrix}
              x_1 \
              x_2
              end{bmatrix},$$
              so $lambda$ is eigenvalue of matrix $begin{bmatrix}
              1 & 2 \
              2 & 2
              end{bmatrix}$. Can you find all eigenvalues of that matrix using, for example, characteristic polynomial?






              share|cite|improve this answer









              $endgroup$



              You have



              $$begin{bmatrix}
              1 & 2 \
              2 & 2
              end{bmatrix} begin{bmatrix}
              x_1 \
              x_2
              end{bmatrix}=
              lambda begin{bmatrix}
              x_1 \
              x_2
              end{bmatrix},$$
              so $lambda$ is eigenvalue of matrix $begin{bmatrix}
              1 & 2 \
              2 & 2
              end{bmatrix}$. Can you find all eigenvalues of that matrix using, for example, characteristic polynomial?







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Feb 17 '16 at 20:30









              aghaagha

              9,03641533




              9,03641533












              • $begingroup$
                I don't know what an eigenvalue is. Looking ahead in the textbook, it's a topic in the next section, so I'm assuming the textbook wants me to solve it without knowing what an eigenvalue is.
                $endgroup$
                – PCR
                Feb 17 '16 at 20:34










              • $begingroup$
                @PCR what if your textbook wants to lure you to the idea of eigenvctors and eigenvalues? In my time, I had much, much more trouble with those teachers who did not want me to look ahead.
                $endgroup$
                – Gyro Gearloose
                Feb 17 '16 at 20:42


















              • $begingroup$
                I don't know what an eigenvalue is. Looking ahead in the textbook, it's a topic in the next section, so I'm assuming the textbook wants me to solve it without knowing what an eigenvalue is.
                $endgroup$
                – PCR
                Feb 17 '16 at 20:34










              • $begingroup$
                @PCR what if your textbook wants to lure you to the idea of eigenvctors and eigenvalues? In my time, I had much, much more trouble with those teachers who did not want me to look ahead.
                $endgroup$
                – Gyro Gearloose
                Feb 17 '16 at 20:42
















              $begingroup$
              I don't know what an eigenvalue is. Looking ahead in the textbook, it's a topic in the next section, so I'm assuming the textbook wants me to solve it without knowing what an eigenvalue is.
              $endgroup$
              – PCR
              Feb 17 '16 at 20:34




              $begingroup$
              I don't know what an eigenvalue is. Looking ahead in the textbook, it's a topic in the next section, so I'm assuming the textbook wants me to solve it without knowing what an eigenvalue is.
              $endgroup$
              – PCR
              Feb 17 '16 at 20:34












              $begingroup$
              @PCR what if your textbook wants to lure you to the idea of eigenvctors and eigenvalues? In my time, I had much, much more trouble with those teachers who did not want me to look ahead.
              $endgroup$
              – Gyro Gearloose
              Feb 17 '16 at 20:42




              $begingroup$
              @PCR what if your textbook wants to lure you to the idea of eigenvctors and eigenvalues? In my time, I had much, much more trouble with those teachers who did not want me to look ahead.
              $endgroup$
              – Gyro Gearloose
              Feb 17 '16 at 20:42











              0












              $begingroup$

              The mistake you made: You have chosen $A=begin{bmatrix}1 & 2\2 & 1end{bmatrix}$ and $x=begin{bmatrix}x_1\x_2end{bmatrix}$. So you have $Ax=lambda x$ in the matrix equation form. When you multiply $A^{-1}$, you will get $x=Ix=A^{-1}Ax=lambda A^{-1}x$. Observe the right hand side involves another matrix $A^{-1}$. Hence you are not getting the solution. Consider the technique suggested by 'Dietrich Burde' and try to generalize. Your text book is probably designed in such a way that you can understand the generalizations in the next sections.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                The mistake you made: You have chosen $A=begin{bmatrix}1 & 2\2 & 1end{bmatrix}$ and $x=begin{bmatrix}x_1\x_2end{bmatrix}$. So you have $Ax=lambda x$ in the matrix equation form. When you multiply $A^{-1}$, you will get $x=Ix=A^{-1}Ax=lambda A^{-1}x$. Observe the right hand side involves another matrix $A^{-1}$. Hence you are not getting the solution. Consider the technique suggested by 'Dietrich Burde' and try to generalize. Your text book is probably designed in such a way that you can understand the generalizations in the next sections.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The mistake you made: You have chosen $A=begin{bmatrix}1 & 2\2 & 1end{bmatrix}$ and $x=begin{bmatrix}x_1\x_2end{bmatrix}$. So you have $Ax=lambda x$ in the matrix equation form. When you multiply $A^{-1}$, you will get $x=Ix=A^{-1}Ax=lambda A^{-1}x$. Observe the right hand side involves another matrix $A^{-1}$. Hence you are not getting the solution. Consider the technique suggested by 'Dietrich Burde' and try to generalize. Your text book is probably designed in such a way that you can understand the generalizations in the next sections.






                  share|cite|improve this answer









                  $endgroup$



                  The mistake you made: You have chosen $A=begin{bmatrix}1 & 2\2 & 1end{bmatrix}$ and $x=begin{bmatrix}x_1\x_2end{bmatrix}$. So you have $Ax=lambda x$ in the matrix equation form. When you multiply $A^{-1}$, you will get $x=Ix=A^{-1}Ax=lambda A^{-1}x$. Observe the right hand side involves another matrix $A^{-1}$. Hence you are not getting the solution. Consider the technique suggested by 'Dietrich Burde' and try to generalize. Your text book is probably designed in such a way that you can understand the generalizations in the next sections.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 18 '16 at 15:20









                  G_0_pi_i_eG_0_pi_i_e

                  603515




                  603515






























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