Find trivial and non-trivial solutions to a system with a parameter











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begin{cases} x + (2p−1)y−pz = 0 \
(2−p)x + y−z = 0 \
x + py−z = 0 end{cases}




  • For which values of $p$, system has trivial solutions?

  • For which values of $p$, system has nontrivial solutions? Write them


Now I have come to $p_1=1, p_2=1, p_3=1/2$ with Sarrus rule. Now how do I get trivial and nontrivial solutions ? I am just looking for instructions






linear-algebra systems-of-equations







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  • Well if $x=y=z = 0$ then any value of $p$ will have trivial solutions. It the equations are linear dependent then those are the only solutions. But if they aren't linearly independent they can have non-trivial solutions. Which can only occur if the determinate is $0$. So this question boils down to nothing more or less than: For what values of $p$ will the determinate be $0$. (Note: there is nothing in the question about finding the solutions).
    – fleablood
    Nov 22 at 18:03















up vote
1
down vote

favorite












begin{cases} x + (2p−1)y−pz = 0 \
(2−p)x + y−z = 0 \
x + py−z = 0 end{cases}




  • For which values of $p$, system has trivial solutions?

  • For which values of $p$, system has nontrivial solutions? Write them


Now I have come to $p_1=1, p_2=1, p_3=1/2$ with Sarrus rule. Now how do I get trivial and nontrivial solutions ? I am just looking for instructions






linear-algebra systems-of-equations







share|cite|improve this question
























  • Well if $x=y=z = 0$ then any value of $p$ will have trivial solutions. It the equations are linear dependent then those are the only solutions. But if they aren't linearly independent they can have non-trivial solutions. Which can only occur if the determinate is $0$. So this question boils down to nothing more or less than: For what values of $p$ will the determinate be $0$. (Note: there is nothing in the question about finding the solutions).
    – fleablood
    Nov 22 at 18:03













up vote
1
down vote

favorite









up vote
1
down vote

favorite











begin{cases} x + (2p−1)y−pz = 0 \
(2−p)x + y−z = 0 \
x + py−z = 0 end{cases}




  • For which values of $p$, system has trivial solutions?

  • For which values of $p$, system has nontrivial solutions? Write them


Now I have come to $p_1=1, p_2=1, p_3=1/2$ with Sarrus rule. Now how do I get trivial and nontrivial solutions ? I am just looking for instructions






linear-algebra systems-of-equations







share|cite|improve this question















begin{cases} x + (2p−1)y−pz = 0 \
(2−p)x + y−z = 0 \
x + py−z = 0 end{cases}




  • For which values of $p$, system has trivial solutions?

  • For which values of $p$, system has nontrivial solutions? Write them


Now I have come to $p_1=1, p_2=1, p_3=1/2$ with Sarrus rule. Now how do I get trivial and nontrivial solutions ? I am just looking for instructions






linear-algebra systems-of-equations




linear-algebra systems-of-equations






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edited Nov 22 at 17:46









Lorenzo B.

1,8322520




1,8322520










asked Nov 22 at 17:23









Miko

103




103












  • Well if $x=y=z = 0$ then any value of $p$ will have trivial solutions. It the equations are linear dependent then those are the only solutions. But if they aren't linearly independent they can have non-trivial solutions. Which can only occur if the determinate is $0$. So this question boils down to nothing more or less than: For what values of $p$ will the determinate be $0$. (Note: there is nothing in the question about finding the solutions).
    – fleablood
    Nov 22 at 18:03


















  • Well if $x=y=z = 0$ then any value of $p$ will have trivial solutions. It the equations are linear dependent then those are the only solutions. But if they aren't linearly independent they can have non-trivial solutions. Which can only occur if the determinate is $0$. So this question boils down to nothing more or less than: For what values of $p$ will the determinate be $0$. (Note: there is nothing in the question about finding the solutions).
    – fleablood
    Nov 22 at 18:03
















Well if $x=y=z = 0$ then any value of $p$ will have trivial solutions. It the equations are linear dependent then those are the only solutions. But if they aren't linearly independent they can have non-trivial solutions. Which can only occur if the determinate is $0$. So this question boils down to nothing more or less than: For what values of $p$ will the determinate be $0$. (Note: there is nothing in the question about finding the solutions).
– fleablood
Nov 22 at 18:03




Well if $x=y=z = 0$ then any value of $p$ will have trivial solutions. It the equations are linear dependent then those are the only solutions. But if they aren't linearly independent they can have non-trivial solutions. Which can only occur if the determinate is $0$. So this question boils down to nothing more or less than: For what values of $p$ will the determinate be $0$. (Note: there is nothing in the question about finding the solutions).
– fleablood
Nov 22 at 18:03










3 Answers
3






active

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up vote
0
down vote



accepted










Well if $x=y=z = 0$ then any value of $p$ will have trivial solutions.



So the answer to the first is: For all values of $p$ there will be trivial solutions.



It the equations are linear dependent then those are the only solutions. But if they aren't linearly independent they can have non-trivial solutions. Which can only occur if the determinate is $0$. So this question boils down to nothing more or less than: For what values of $p$ will the determinate be $0$. (Note: there is nothing in the question about finding the solutions).



Using rule of sarrus the determinate is:



$-1 - (2p-1)-p^2(2-p) +p + p +(2-p)(2p-1) =$



$ p^3 - 4p^2 +5p-2= (p-1)^2(p-2)$



Which if we set to $0$ and solve... we get



$p =1$ or $p =2$.



And $p = 1$ we get the system:



$x +y - z = 0; x+y-z = 0; x+y-z =0$ or $z= x+y$ which has infinitely many solutions.



And if $p=2$ we get the system:



$x+3y -2z =0; y-z = 0; x+2y - z=0$ or $y=z=-x$ which has infinitely many solutions.



Bear in mind $begin{cases} a_1x + a_2y+a_3z = 0 \
b_1x + b_2y+b_3z = 0 \
c_1x + c_2y+c_3z = 0 end{cases}$



Will either have exactly one trivial solution $x=y=z =0$ if the equation or independent. But if the equations are not independent there will always be an infinite number of solutions.






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    up vote
    0
    down vote













    Hint :



    Apply Gauss-Elimination to :



    $$left(begin{array}{ccc|c} 1 & 2p-1 & -p &0 \ 2-p & 1 & -1 & 0 \ 1 & p & -1 & 0end{array}right)$$






    share|cite|improve this answer





















    • Is that necessary? I used Sarrus rule to avoid Gauss-Elimination as it can get complicated with those p
      – Miko
      Nov 22 at 17:45










    • @Miko By applying Gauss-Elimination, you will yield an expression involving $p$ that you will have to investigate in order to answer to your exercise. This is the most straight-forward and classic way. I don't understand what those $p_i$s are ? The Rule of Sarrus is a memorisation technique involving determinants, certainly not the math tool to use here.
      – Rebellos
      Nov 22 at 17:51


















    up vote
    0
    down vote













    Hint:



    This system has non-trivial solutions if and only if its determinant is zero . This determinant can be calculated by row and column operations:
    begin{align}
    begin{vmatrix}
    1&2p-1&-p \ 2-p&1&-1 \ 1&p&-1
    end{vmatrix} &=
    begin{vmatrix}
    1&2p-1&-p \ 0&1+(p-2)(2p-1)&-1-p(p-2) \ 0&1-p&p-1
    end{vmatrix}
    =begin{vmatrix}
    1&p-1&-p \ 0&(p-2)(p-1)&-1-p(p-2) \ 0&0&p-1
    end{vmatrix}\[1ex]& = (p-1)begin{vmatrix}
    1& 1&-p \ 0&p-2 &-1-p(p-2) \0&0&p-1
    end{vmatrix} = (p-1)begin{vmatrix}
    p-2 &-1-p(p-2) \ 0&p-1
    end{vmatrix} \[1ex]
    &=(p-1)^2(p-2).
    end{align}






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

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      active

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      up vote
      0
      down vote



      accepted










      Well if $x=y=z = 0$ then any value of $p$ will have trivial solutions.



      So the answer to the first is: For all values of $p$ there will be trivial solutions.



      It the equations are linear dependent then those are the only solutions. But if they aren't linearly independent they can have non-trivial solutions. Which can only occur if the determinate is $0$. So this question boils down to nothing more or less than: For what values of $p$ will the determinate be $0$. (Note: there is nothing in the question about finding the solutions).



      Using rule of sarrus the determinate is:



      $-1 - (2p-1)-p^2(2-p) +p + p +(2-p)(2p-1) =$



      $ p^3 - 4p^2 +5p-2= (p-1)^2(p-2)$



      Which if we set to $0$ and solve... we get



      $p =1$ or $p =2$.



      And $p = 1$ we get the system:



      $x +y - z = 0; x+y-z = 0; x+y-z =0$ or $z= x+y$ which has infinitely many solutions.



      And if $p=2$ we get the system:



      $x+3y -2z =0; y-z = 0; x+2y - z=0$ or $y=z=-x$ which has infinitely many solutions.



      Bear in mind $begin{cases} a_1x + a_2y+a_3z = 0 \
      b_1x + b_2y+b_3z = 0 \
      c_1x + c_2y+c_3z = 0 end{cases}$



      Will either have exactly one trivial solution $x=y=z =0$ if the equation or independent. But if the equations are not independent there will always be an infinite number of solutions.






      share|cite|improve this answer



























        up vote
        0
        down vote



        accepted










        Well if $x=y=z = 0$ then any value of $p$ will have trivial solutions.



        So the answer to the first is: For all values of $p$ there will be trivial solutions.



        It the equations are linear dependent then those are the only solutions. But if they aren't linearly independent they can have non-trivial solutions. Which can only occur if the determinate is $0$. So this question boils down to nothing more or less than: For what values of $p$ will the determinate be $0$. (Note: there is nothing in the question about finding the solutions).



        Using rule of sarrus the determinate is:



        $-1 - (2p-1)-p^2(2-p) +p + p +(2-p)(2p-1) =$



        $ p^3 - 4p^2 +5p-2= (p-1)^2(p-2)$



        Which if we set to $0$ and solve... we get



        $p =1$ or $p =2$.



        And $p = 1$ we get the system:



        $x +y - z = 0; x+y-z = 0; x+y-z =0$ or $z= x+y$ which has infinitely many solutions.



        And if $p=2$ we get the system:



        $x+3y -2z =0; y-z = 0; x+2y - z=0$ or $y=z=-x$ which has infinitely many solutions.



        Bear in mind $begin{cases} a_1x + a_2y+a_3z = 0 \
        b_1x + b_2y+b_3z = 0 \
        c_1x + c_2y+c_3z = 0 end{cases}$



        Will either have exactly one trivial solution $x=y=z =0$ if the equation or independent. But if the equations are not independent there will always be an infinite number of solutions.






        share|cite|improve this answer

























          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          Well if $x=y=z = 0$ then any value of $p$ will have trivial solutions.



          So the answer to the first is: For all values of $p$ there will be trivial solutions.



          It the equations are linear dependent then those are the only solutions. But if they aren't linearly independent they can have non-trivial solutions. Which can only occur if the determinate is $0$. So this question boils down to nothing more or less than: For what values of $p$ will the determinate be $0$. (Note: there is nothing in the question about finding the solutions).



          Using rule of sarrus the determinate is:



          $-1 - (2p-1)-p^2(2-p) +p + p +(2-p)(2p-1) =$



          $ p^3 - 4p^2 +5p-2= (p-1)^2(p-2)$



          Which if we set to $0$ and solve... we get



          $p =1$ or $p =2$.



          And $p = 1$ we get the system:



          $x +y - z = 0; x+y-z = 0; x+y-z =0$ or $z= x+y$ which has infinitely many solutions.



          And if $p=2$ we get the system:



          $x+3y -2z =0; y-z = 0; x+2y - z=0$ or $y=z=-x$ which has infinitely many solutions.



          Bear in mind $begin{cases} a_1x + a_2y+a_3z = 0 \
          b_1x + b_2y+b_3z = 0 \
          c_1x + c_2y+c_3z = 0 end{cases}$



          Will either have exactly one trivial solution $x=y=z =0$ if the equation or independent. But if the equations are not independent there will always be an infinite number of solutions.






          share|cite|improve this answer














          Well if $x=y=z = 0$ then any value of $p$ will have trivial solutions.



          So the answer to the first is: For all values of $p$ there will be trivial solutions.



          It the equations are linear dependent then those are the only solutions. But if they aren't linearly independent they can have non-trivial solutions. Which can only occur if the determinate is $0$. So this question boils down to nothing more or less than: For what values of $p$ will the determinate be $0$. (Note: there is nothing in the question about finding the solutions).



          Using rule of sarrus the determinate is:



          $-1 - (2p-1)-p^2(2-p) +p + p +(2-p)(2p-1) =$



          $ p^3 - 4p^2 +5p-2= (p-1)^2(p-2)$



          Which if we set to $0$ and solve... we get



          $p =1$ or $p =2$.



          And $p = 1$ we get the system:



          $x +y - z = 0; x+y-z = 0; x+y-z =0$ or $z= x+y$ which has infinitely many solutions.



          And if $p=2$ we get the system:



          $x+3y -2z =0; y-z = 0; x+2y - z=0$ or $y=z=-x$ which has infinitely many solutions.



          Bear in mind $begin{cases} a_1x + a_2y+a_3z = 0 \
          b_1x + b_2y+b_3z = 0 \
          c_1x + c_2y+c_3z = 0 end{cases}$



          Will either have exactly one trivial solution $x=y=z =0$ if the equation or independent. But if the equations are not independent there will always be an infinite number of solutions.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 22 at 19:11

























          answered Nov 22 at 18:50









          fleablood

          67.6k22684




          67.6k22684






















              up vote
              0
              down vote













              Hint :



              Apply Gauss-Elimination to :



              $$left(begin{array}{ccc|c} 1 & 2p-1 & -p &0 \ 2-p & 1 & -1 & 0 \ 1 & p & -1 & 0end{array}right)$$






              share|cite|improve this answer





















              • Is that necessary? I used Sarrus rule to avoid Gauss-Elimination as it can get complicated with those p
                – Miko
                Nov 22 at 17:45










              • @Miko By applying Gauss-Elimination, you will yield an expression involving $p$ that you will have to investigate in order to answer to your exercise. This is the most straight-forward and classic way. I don't understand what those $p_i$s are ? The Rule of Sarrus is a memorisation technique involving determinants, certainly not the math tool to use here.
                – Rebellos
                Nov 22 at 17:51















              up vote
              0
              down vote













              Hint :



              Apply Gauss-Elimination to :



              $$left(begin{array}{ccc|c} 1 & 2p-1 & -p &0 \ 2-p & 1 & -1 & 0 \ 1 & p & -1 & 0end{array}right)$$






              share|cite|improve this answer





















              • Is that necessary? I used Sarrus rule to avoid Gauss-Elimination as it can get complicated with those p
                – Miko
                Nov 22 at 17:45










              • @Miko By applying Gauss-Elimination, you will yield an expression involving $p$ that you will have to investigate in order to answer to your exercise. This is the most straight-forward and classic way. I don't understand what those $p_i$s are ? The Rule of Sarrus is a memorisation technique involving determinants, certainly not the math tool to use here.
                – Rebellos
                Nov 22 at 17:51













              up vote
              0
              down vote










              up vote
              0
              down vote









              Hint :



              Apply Gauss-Elimination to :



              $$left(begin{array}{ccc|c} 1 & 2p-1 & -p &0 \ 2-p & 1 & -1 & 0 \ 1 & p & -1 & 0end{array}right)$$






              share|cite|improve this answer












              Hint :



              Apply Gauss-Elimination to :



              $$left(begin{array}{ccc|c} 1 & 2p-1 & -p &0 \ 2-p & 1 & -1 & 0 \ 1 & p & -1 & 0end{array}right)$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 22 at 17:39









              Rebellos

              13.9k31243




              13.9k31243












              • Is that necessary? I used Sarrus rule to avoid Gauss-Elimination as it can get complicated with those p
                – Miko
                Nov 22 at 17:45










              • @Miko By applying Gauss-Elimination, you will yield an expression involving $p$ that you will have to investigate in order to answer to your exercise. This is the most straight-forward and classic way. I don't understand what those $p_i$s are ? The Rule of Sarrus is a memorisation technique involving determinants, certainly not the math tool to use here.
                – Rebellos
                Nov 22 at 17:51


















              • Is that necessary? I used Sarrus rule to avoid Gauss-Elimination as it can get complicated with those p
                – Miko
                Nov 22 at 17:45










              • @Miko By applying Gauss-Elimination, you will yield an expression involving $p$ that you will have to investigate in order to answer to your exercise. This is the most straight-forward and classic way. I don't understand what those $p_i$s are ? The Rule of Sarrus is a memorisation technique involving determinants, certainly not the math tool to use here.
                – Rebellos
                Nov 22 at 17:51
















              Is that necessary? I used Sarrus rule to avoid Gauss-Elimination as it can get complicated with those p
              – Miko
              Nov 22 at 17:45




              Is that necessary? I used Sarrus rule to avoid Gauss-Elimination as it can get complicated with those p
              – Miko
              Nov 22 at 17:45












              @Miko By applying Gauss-Elimination, you will yield an expression involving $p$ that you will have to investigate in order to answer to your exercise. This is the most straight-forward and classic way. I don't understand what those $p_i$s are ? The Rule of Sarrus is a memorisation technique involving determinants, certainly not the math tool to use here.
              – Rebellos
              Nov 22 at 17:51




              @Miko By applying Gauss-Elimination, you will yield an expression involving $p$ that you will have to investigate in order to answer to your exercise. This is the most straight-forward and classic way. I don't understand what those $p_i$s are ? The Rule of Sarrus is a memorisation technique involving determinants, certainly not the math tool to use here.
              – Rebellos
              Nov 22 at 17:51










              up vote
              0
              down vote













              Hint:



              This system has non-trivial solutions if and only if its determinant is zero . This determinant can be calculated by row and column operations:
              begin{align}
              begin{vmatrix}
              1&2p-1&-p \ 2-p&1&-1 \ 1&p&-1
              end{vmatrix} &=
              begin{vmatrix}
              1&2p-1&-p \ 0&1+(p-2)(2p-1)&-1-p(p-2) \ 0&1-p&p-1
              end{vmatrix}
              =begin{vmatrix}
              1&p-1&-p \ 0&(p-2)(p-1)&-1-p(p-2) \ 0&0&p-1
              end{vmatrix}\[1ex]& = (p-1)begin{vmatrix}
              1& 1&-p \ 0&p-2 &-1-p(p-2) \0&0&p-1
              end{vmatrix} = (p-1)begin{vmatrix}
              p-2 &-1-p(p-2) \ 0&p-1
              end{vmatrix} \[1ex]
              &=(p-1)^2(p-2).
              end{align}






              share|cite|improve this answer

























                up vote
                0
                down vote













                Hint:



                This system has non-trivial solutions if and only if its determinant is zero . This determinant can be calculated by row and column operations:
                begin{align}
                begin{vmatrix}
                1&2p-1&-p \ 2-p&1&-1 \ 1&p&-1
                end{vmatrix} &=
                begin{vmatrix}
                1&2p-1&-p \ 0&1+(p-2)(2p-1)&-1-p(p-2) \ 0&1-p&p-1
                end{vmatrix}
                =begin{vmatrix}
                1&p-1&-p \ 0&(p-2)(p-1)&-1-p(p-2) \ 0&0&p-1
                end{vmatrix}\[1ex]& = (p-1)begin{vmatrix}
                1& 1&-p \ 0&p-2 &-1-p(p-2) \0&0&p-1
                end{vmatrix} = (p-1)begin{vmatrix}
                p-2 &-1-p(p-2) \ 0&p-1
                end{vmatrix} \[1ex]
                &=(p-1)^2(p-2).
                end{align}






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Hint:



                  This system has non-trivial solutions if and only if its determinant is zero . This determinant can be calculated by row and column operations:
                  begin{align}
                  begin{vmatrix}
                  1&2p-1&-p \ 2-p&1&-1 \ 1&p&-1
                  end{vmatrix} &=
                  begin{vmatrix}
                  1&2p-1&-p \ 0&1+(p-2)(2p-1)&-1-p(p-2) \ 0&1-p&p-1
                  end{vmatrix}
                  =begin{vmatrix}
                  1&p-1&-p \ 0&(p-2)(p-1)&-1-p(p-2) \ 0&0&p-1
                  end{vmatrix}\[1ex]& = (p-1)begin{vmatrix}
                  1& 1&-p \ 0&p-2 &-1-p(p-2) \0&0&p-1
                  end{vmatrix} = (p-1)begin{vmatrix}
                  p-2 &-1-p(p-2) \ 0&p-1
                  end{vmatrix} \[1ex]
                  &=(p-1)^2(p-2).
                  end{align}






                  share|cite|improve this answer












                  Hint:



                  This system has non-trivial solutions if and only if its determinant is zero . This determinant can be calculated by row and column operations:
                  begin{align}
                  begin{vmatrix}
                  1&2p-1&-p \ 2-p&1&-1 \ 1&p&-1
                  end{vmatrix} &=
                  begin{vmatrix}
                  1&2p-1&-p \ 0&1+(p-2)(2p-1)&-1-p(p-2) \ 0&1-p&p-1
                  end{vmatrix}
                  =begin{vmatrix}
                  1&p-1&-p \ 0&(p-2)(p-1)&-1-p(p-2) \ 0&0&p-1
                  end{vmatrix}\[1ex]& = (p-1)begin{vmatrix}
                  1& 1&-p \ 0&p-2 &-1-p(p-2) \0&0&p-1
                  end{vmatrix} = (p-1)begin{vmatrix}
                  p-2 &-1-p(p-2) \ 0&p-1
                  end{vmatrix} \[1ex]
                  &=(p-1)^2(p-2).
                  end{align}







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                  answered Nov 22 at 20:53









                  Bernard

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