Determining whether a transformation is linear based on input and output coordinates












1












$begingroup$


I was asked to check if there is a linear transformation that lives up to the requirements. If so, I should come up with T(x,y,x). Otherwise, I should explain why not.



a) $T:R^3 to R^2$, so that $T(0,2,-1)=(-1,3)$, $T(2,-2,-1)=(1,4)$ and $T(1,-4,1)=(2,1)$.



b) $T:R^3 to R^2$, so that $T(1,1,-1)=(-1,3)$, $T(1,1,1)=(1,4)$ and $T(1,0,1)=(2,1)$.



The way I resolved it was as follows:
a) I created the following matrices:
$$begin{matrix}a&b&c\d&e&fend{matrix}$$



multiplied by



$$begin{matrix}0&2&1\2&-2&4\-1&-1&1end{matrix}$$



equals



$$begin{matrix}-1&1&2\3&4&1end{matrix}$$



Then I wrote down the resulting two sets of equations. First:



2b-c=-1



2a-2b-c=1



a-4b+c=2



and resolved this through a matrix.



Next:



2e-f=3



2d-2e-f=4



d-4e+1f=1



and tried to resolve this via a matrix, but there was a contradiction, so no solutions. I assumed that meant that the transformation is not linear.



I did exactly the same for b), but this time the resulting two systems of equations both came out with one solution each. I assumed that this means this is indeed a linear transformation.



Is the reasoning correct?



Thank you!






linear-algebra linear-transformations







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I was asked to check if there is a linear transformation that lives up to the requirements. If so, I should come up with T(x,y,x). Otherwise, I should explain why not.



    a) $T:R^3 to R^2$, so that $T(0,2,-1)=(-1,3)$, $T(2,-2,-1)=(1,4)$ and $T(1,-4,1)=(2,1)$.



    b) $T:R^3 to R^2$, so that $T(1,1,-1)=(-1,3)$, $T(1,1,1)=(1,4)$ and $T(1,0,1)=(2,1)$.



    The way I resolved it was as follows:
    a) I created the following matrices:
    $$begin{matrix}a&b&c\d&e&fend{matrix}$$



    multiplied by



    $$begin{matrix}0&2&1\2&-2&4\-1&-1&1end{matrix}$$



    equals



    $$begin{matrix}-1&1&2\3&4&1end{matrix}$$



    Then I wrote down the resulting two sets of equations. First:



    2b-c=-1



    2a-2b-c=1



    a-4b+c=2



    and resolved this through a matrix.



    Next:



    2e-f=3



    2d-2e-f=4



    d-4e+1f=1



    and tried to resolve this via a matrix, but there was a contradiction, so no solutions. I assumed that meant that the transformation is not linear.



    I did exactly the same for b), but this time the resulting two systems of equations both came out with one solution each. I assumed that this means this is indeed a linear transformation.



    Is the reasoning correct?



    Thank you!






    linear-algebra linear-transformations







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I was asked to check if there is a linear transformation that lives up to the requirements. If so, I should come up with T(x,y,x). Otherwise, I should explain why not.



      a) $T:R^3 to R^2$, so that $T(0,2,-1)=(-1,3)$, $T(2,-2,-1)=(1,4)$ and $T(1,-4,1)=(2,1)$.



      b) $T:R^3 to R^2$, so that $T(1,1,-1)=(-1,3)$, $T(1,1,1)=(1,4)$ and $T(1,0,1)=(2,1)$.



      The way I resolved it was as follows:
      a) I created the following matrices:
      $$begin{matrix}a&b&c\d&e&fend{matrix}$$



      multiplied by



      $$begin{matrix}0&2&1\2&-2&4\-1&-1&1end{matrix}$$



      equals



      $$begin{matrix}-1&1&2\3&4&1end{matrix}$$



      Then I wrote down the resulting two sets of equations. First:



      2b-c=-1



      2a-2b-c=1



      a-4b+c=2



      and resolved this through a matrix.



      Next:



      2e-f=3



      2d-2e-f=4



      d-4e+1f=1



      and tried to resolve this via a matrix, but there was a contradiction, so no solutions. I assumed that meant that the transformation is not linear.



      I did exactly the same for b), but this time the resulting two systems of equations both came out with one solution each. I assumed that this means this is indeed a linear transformation.



      Is the reasoning correct?



      Thank you!






      linear-algebra linear-transformations







      share|cite|improve this question











      $endgroup$




      I was asked to check if there is a linear transformation that lives up to the requirements. If so, I should come up with T(x,y,x). Otherwise, I should explain why not.



      a) $T:R^3 to R^2$, so that $T(0,2,-1)=(-1,3)$, $T(2,-2,-1)=(1,4)$ and $T(1,-4,1)=(2,1)$.



      b) $T:R^3 to R^2$, so that $T(1,1,-1)=(-1,3)$, $T(1,1,1)=(1,4)$ and $T(1,0,1)=(2,1)$.



      The way I resolved it was as follows:
      a) I created the following matrices:
      $$begin{matrix}a&b&c\d&e&fend{matrix}$$



      multiplied by



      $$begin{matrix}0&2&1\2&-2&4\-1&-1&1end{matrix}$$



      equals



      $$begin{matrix}-1&1&2\3&4&1end{matrix}$$



      Then I wrote down the resulting two sets of equations. First:



      2b-c=-1



      2a-2b-c=1



      a-4b+c=2



      and resolved this through a matrix.



      Next:



      2e-f=3



      2d-2e-f=4



      d-4e+1f=1



      and tried to resolve this via a matrix, but there was a contradiction, so no solutions. I assumed that meant that the transformation is not linear.



      I did exactly the same for b), but this time the resulting two systems of equations both came out with one solution each. I assumed that this means this is indeed a linear transformation.



      Is the reasoning correct?



      Thank you!






      linear-algebra linear-transformations




      linear-algebra linear-transformations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Jan 1 at 20:55









      A.Γ.

      22.9k32656




      22.9k32656










      asked Jan 1 at 20:47









      daltadalta

      1508




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          $begingroup$

          Actually, the transformation in the first case is also linear. To find the matrix we just have to set up the linear system of equations
          $$
          begin{bmatrix}a&b&c\d&e&fend{bmatrix}
          begin{bmatrix}0&2&1\2&-2&4\-1&-1&1end{bmatrix}=
          begin{bmatrix}-1&1&2\3&4&1end{bmatrix}.
          $$
          The square matrix of the system is invertible, hence, the solution is
          $$
          begin{bmatrix}a&b&c\d&e&fend{bmatrix}=
          begin{bmatrix}-1&1&2\3&4&1end{bmatrix}
          begin{bmatrix}0&2&1\2&-2&4\-1&-1&1end{bmatrix}^{-1}
          $$
          Similar system can be solved in the second case.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks! But then how can the second transformation be linear? And the matrices definitely come out coherent in that case...
            $endgroup$
            – dalta
            Jan 1 at 21:19










          • $begingroup$
            @dalta Sorry for misleading you, I have edited the answer.
            $endgroup$
            – A.Γ.
            Jan 1 at 21:32












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          $begingroup$

          Actually, the transformation in the first case is also linear. To find the matrix we just have to set up the linear system of equations
          $$
          begin{bmatrix}a&b&c\d&e&fend{bmatrix}
          begin{bmatrix}0&2&1\2&-2&4\-1&-1&1end{bmatrix}=
          begin{bmatrix}-1&1&2\3&4&1end{bmatrix}.
          $$
          The square matrix of the system is invertible, hence, the solution is
          $$
          begin{bmatrix}a&b&c\d&e&fend{bmatrix}=
          begin{bmatrix}-1&1&2\3&4&1end{bmatrix}
          begin{bmatrix}0&2&1\2&-2&4\-1&-1&1end{bmatrix}^{-1}
          $$
          Similar system can be solved in the second case.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks! But then how can the second transformation be linear? And the matrices definitely come out coherent in that case...
            $endgroup$
            – dalta
            Jan 1 at 21:19










          • $begingroup$
            @dalta Sorry for misleading you, I have edited the answer.
            $endgroup$
            – A.Γ.
            Jan 1 at 21:32
















          0












          $begingroup$

          Actually, the transformation in the first case is also linear. To find the matrix we just have to set up the linear system of equations
          $$
          begin{bmatrix}a&b&c\d&e&fend{bmatrix}
          begin{bmatrix}0&2&1\2&-2&4\-1&-1&1end{bmatrix}=
          begin{bmatrix}-1&1&2\3&4&1end{bmatrix}.
          $$
          The square matrix of the system is invertible, hence, the solution is
          $$
          begin{bmatrix}a&b&c\d&e&fend{bmatrix}=
          begin{bmatrix}-1&1&2\3&4&1end{bmatrix}
          begin{bmatrix}0&2&1\2&-2&4\-1&-1&1end{bmatrix}^{-1}
          $$
          Similar system can be solved in the second case.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks! But then how can the second transformation be linear? And the matrices definitely come out coherent in that case...
            $endgroup$
            – dalta
            Jan 1 at 21:19










          • $begingroup$
            @dalta Sorry for misleading you, I have edited the answer.
            $endgroup$
            – A.Γ.
            Jan 1 at 21:32














          0












          0








          0





          $begingroup$

          Actually, the transformation in the first case is also linear. To find the matrix we just have to set up the linear system of equations
          $$
          begin{bmatrix}a&b&c\d&e&fend{bmatrix}
          begin{bmatrix}0&2&1\2&-2&4\-1&-1&1end{bmatrix}=
          begin{bmatrix}-1&1&2\3&4&1end{bmatrix}.
          $$
          The square matrix of the system is invertible, hence, the solution is
          $$
          begin{bmatrix}a&b&c\d&e&fend{bmatrix}=
          begin{bmatrix}-1&1&2\3&4&1end{bmatrix}
          begin{bmatrix}0&2&1\2&-2&4\-1&-1&1end{bmatrix}^{-1}
          $$
          Similar system can be solved in the second case.






          share|cite|improve this answer











          $endgroup$



          Actually, the transformation in the first case is also linear. To find the matrix we just have to set up the linear system of equations
          $$
          begin{bmatrix}a&b&c\d&e&fend{bmatrix}
          begin{bmatrix}0&2&1\2&-2&4\-1&-1&1end{bmatrix}=
          begin{bmatrix}-1&1&2\3&4&1end{bmatrix}.
          $$
          The square matrix of the system is invertible, hence, the solution is
          $$
          begin{bmatrix}a&b&c\d&e&fend{bmatrix}=
          begin{bmatrix}-1&1&2\3&4&1end{bmatrix}
          begin{bmatrix}0&2&1\2&-2&4\-1&-1&1end{bmatrix}^{-1}
          $$
          Similar system can be solved in the second case.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 1 at 21:28

























          answered Jan 1 at 21:03









          A.Γ.A.Γ.

          22.9k32656




          22.9k32656












          • $begingroup$
            Thanks! But then how can the second transformation be linear? And the matrices definitely come out coherent in that case...
            $endgroup$
            – dalta
            Jan 1 at 21:19










          • $begingroup$
            @dalta Sorry for misleading you, I have edited the answer.
            $endgroup$
            – A.Γ.
            Jan 1 at 21:32


















          • $begingroup$
            Thanks! But then how can the second transformation be linear? And the matrices definitely come out coherent in that case...
            $endgroup$
            – dalta
            Jan 1 at 21:19










          • $begingroup$
            @dalta Sorry for misleading you, I have edited the answer.
            $endgroup$
            – A.Γ.
            Jan 1 at 21:32
















          $begingroup$
          Thanks! But then how can the second transformation be linear? And the matrices definitely come out coherent in that case...
          $endgroup$
          – dalta
          Jan 1 at 21:19




          $begingroup$
          Thanks! But then how can the second transformation be linear? And the matrices definitely come out coherent in that case...
          $endgroup$
          – dalta
          Jan 1 at 21:19












          $begingroup$
          @dalta Sorry for misleading you, I have edited the answer.
          $endgroup$
          – A.Γ.
          Jan 1 at 21:32




          $begingroup$
          @dalta Sorry for misleading you, I have edited the answer.
          $endgroup$
          – A.Γ.
          Jan 1 at 21:32


















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