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Find the Image $2x-3y=0$ given matrix transformation


Find matrix of linear transformation relative to new basesFind the image under $T$ Linear Transformation - Linear Algebrafind the standard matrix for the composition of reflection and rotationFind the matrix of a linear transformationHow do I determine the transformation matrix T of the coordinate transformation from the base E to the base B?Effect of simple linear transformationMatrix associated to a linear transformation given eigenvaluesFinding Matrix representation of linear transformation from rectangular matrix/imageFinding the standard matrix of the transformation, is it unique?Why do you use the inverse matrix to find the image of a curve in the plane?













2












$begingroup$


Be $M$ the linear transformation represented by matrix $T$:



$$T =
left[ beginmatrix
5 & 2 \
4 & 1 \
endmatrixright ]
$$



Mark the correct answer which show the image by $M$ of the line $2x-3y = 0$



(A) $19x - 8y = 0$



(B) $19x - 14y = 0$ (answer)



(C) $19x -15y = 0$



(D) $15x-14y = 0$



(E) $4x - 5y = 0$



Any hints?



My first attempt was related the tranformation the each coordanaties from line r: 2x-3y = 0 with matrix T.



r as a vector: $vecr = left[beginmatrix 2\-3endmatrixright]$



$vecz = T . vecr = left[beginmatrix 4\-5endmatrixright]$ (wrong)










share|cite|improve this question









$endgroup$
















    2












    $begingroup$


    Be $M$ the linear transformation represented by matrix $T$:



    $$T =
    left[ beginmatrix
    5 & 2 \
    4 & 1 \
    endmatrixright ]
    $$



    Mark the correct answer which show the image by $M$ of the line $2x-3y = 0$



    (A) $19x - 8y = 0$



    (B) $19x - 14y = 0$ (answer)



    (C) $19x -15y = 0$



    (D) $15x-14y = 0$



    (E) $4x - 5y = 0$



    Any hints?



    My first attempt was related the tranformation the each coordanaties from line r: 2x-3y = 0 with matrix T.



    r as a vector: $vecr = left[beginmatrix 2\-3endmatrixright]$



    $vecz = T . vecr = left[beginmatrix 4\-5endmatrixright]$ (wrong)










    share|cite|improve this question









    $endgroup$














      2












      2








      2





      $begingroup$


      Be $M$ the linear transformation represented by matrix $T$:



      $$T =
      left[ beginmatrix
      5 & 2 \
      4 & 1 \
      endmatrixright ]
      $$



      Mark the correct answer which show the image by $M$ of the line $2x-3y = 0$



      (A) $19x - 8y = 0$



      (B) $19x - 14y = 0$ (answer)



      (C) $19x -15y = 0$



      (D) $15x-14y = 0$



      (E) $4x - 5y = 0$



      Any hints?



      My first attempt was related the tranformation the each coordanaties from line r: 2x-3y = 0 with matrix T.



      r as a vector: $vecr = left[beginmatrix 2\-3endmatrixright]$



      $vecz = T . vecr = left[beginmatrix 4\-5endmatrixright]$ (wrong)










      share|cite|improve this question









      $endgroup$




      Be $M$ the linear transformation represented by matrix $T$:



      $$T =
      left[ beginmatrix
      5 & 2 \
      4 & 1 \
      endmatrixright ]
      $$



      Mark the correct answer which show the image by $M$ of the line $2x-3y = 0$



      (A) $19x - 8y = 0$



      (B) $19x - 14y = 0$ (answer)



      (C) $19x -15y = 0$



      (D) $15x-14y = 0$



      (E) $4x - 5y = 0$



      Any hints?



      My first attempt was related the tranformation the each coordanaties from line r: 2x-3y = 0 with matrix T.



      r as a vector: $vecr = left[beginmatrix 2\-3endmatrixright]$



      $vecz = T . vecr = left[beginmatrix 4\-5endmatrixright]$ (wrong)







      linear-transformations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 15 at 17:40









      miguel747miguel747

      13711




      13711




















          4 Answers
          4






          active

          oldest

          votes


















          1












          $begingroup$

          The line $2x-3y=0$ defines a subspace. The vector $r$ that you used is orthogonal to this line.



          The correct point of view is to look at the line as set of all $(x,y)inmathbb R^2$ such that $(x,y)=(x, frac23x)$ since $y=frac23x$. So it suffices to check where $T$ sends, say $(3,2)$ (which is on the line). Thus, computing
          $beginpmatrix 5& 2\ 4& 1endpmatrix beginbmatrix 3\ 2endbmatrix$
          should give the answer.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            misunderstand with some concepts but got that point. Thank you anyway
            $endgroup$
            – miguel747
            Mar 15 at 17:57










          • $begingroup$
            You should at least accept this as an answer then.
            $endgroup$
            – chhro
            Mar 15 at 18:08


















          3












          $begingroup$

          The simplest way to obtain the image of the given line



          $$2x-3y = 0$$



          is to obtain images of its $2$ points - why don't select the natural points $A=[0,0], B=[3,2]$?



          Computing their images we obtain



          $$left( beginmatrix
          5 & 2 \
          4 & 1 \
          endmatrixright )cdot
          left( beginmatrix
          0 \
          0 \
          endmatrixright ) = colorred
          left( beginmatrix
          0\
          0 \
          endmatrixright ),quadleft( beginmatrix
          5 & 2 \
          4 & 1 \
          endmatrixright )cdot
          left( beginmatrix
          3 \
          2 \
          endmatrixright ) = colorred
          left( beginmatrix
          19\
          14 \
          endmatrixright ),$$



          Now we write the equation of the line passing through resulting points $colorred(0, 0)$ and $colorred(19, 14)$, obtaining the correct result



          $$colorred14x-19y=0,$$



          which is similar to your incorrect result (B).



          enter image description here






          share|cite|improve this answer











          $endgroup$




















            1












            $begingroup$

            It might help to think of the problem this way: If you’ve been given some equations that describe the relationship between the pairs of variables $(x,y)$ and $(x',y')$, how might you express the equation $2x-3y=0$ in terms of $x'$ and $y'$? One way is to solve those first equations for $x$ and $y$, then substitute into $2x-3y$ and simplify. In this case, you’re given that $$beginbmatrixx'\y'endbmatrix = T beginbmatrixx\yendbmatrix,tag1$$ so solving for $x$ and $y$ is a simple matter of computing $T^-1$.



            You could also work directly with vectors and matrices. The equation of the line can be written as $$beginbmatrix2&-3endbmatrix beginbmatrixx\yendbmatrix = 0tag2$$ (verify this). Substituting from (1) for $[x,y]^T$, this becomes $$beginbmatrix2&-3endbmatrixT^-1beginbmatrixx'\y'endbmatrix = 0tag3,$$ so the coefficients of the transformed equation are the elements of the row vector $$beginbmatrix2&-3endbmatrix beginbmatrix5&2\4&1endbmatrix^-1.$$ I trust that you’re able to compute the value of this expression at this stage of your studies.



            Incidentally, the vector $[2,-3]^T$ is orthogonal (normal) to the line. Equation (3) shows that such normal vectors transform differently than do vectors that represent points: If the point transformation is given by $mathbf v' = Mmathbf v$, then a normal vector $mathbf n$ transforms as $mathbf n' = M^-Tmathbf n$. (The $-T$ superscript means “inverse transpose.”)






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              But $T$ might not be invertible everytime.
              $endgroup$
              – chhro
              Mar 15 at 22:05










            • $begingroup$
              @chhro That’s no different from not being able to solve for $x$ and $y$ in the first paragraph. In that case, you have to use another method. Happily, $T^-1$ exists in the specific instance. It wasn’t my intent to provide a comprehensive answer replete with obfuscating details. One could quibble that the given line might not pass through the origin, in which case the method in your answer doesn’t quite apply.
              $endgroup$
              – amd
              Mar 15 at 23:15











            • $begingroup$
              Very interesting approach @amd . Ty for take your time to add more details and rich information. :)
              $endgroup$
              – miguel747
              Mar 16 at 1:55










            • $begingroup$
              I'm sure my argument works even if the given line is not a subspace since it is independent of the given linear transformation $T$. It's essentially the idea in MarianD's answer.
              $endgroup$
              – chhro
              Mar 19 at 20:43










            • $begingroup$
              @chhro You’re tacitly taking advantage of the fact that $T(0)=0$. When the line isn’t a subspace, it does not suffice to check where $T$ sends a single point on the line.
              $endgroup$
              – amd
              Mar 19 at 20:47



















            0












            $begingroup$

            STEP 1. Realize that $《x, y》: 2x + 3y = 0$ is a subspace.



            STEP 2. find the basis for the subspace. It is $(1, 2/3)$.



            STEP 3. Do matrix multiplication.



            $$left[ beginmatrix
            5 & 2 \
            4 & 1 \
            endmatrixright ]
            left[ beginmatrix
            1 \
            2/3 \
            endmatrixright ] =
            left[ beginmatrix
            19/3\
            14/3 \
            endmatrixright ]$$



            STEP 4. Translate back into equation form.



            $$P_1 = alpha left[ beginmatrix
            19/3 \
            4/3 \
            endmatrixright ]$$
            and $$P_2 = beta left[ beginmatrix
            19/3 \
            4/3 \
            endmatrixright ]$$
            . Plug these two points (interpreting the top entries as x coordinate and bottom entries as y-coordinate). Use the point-slope equation $$y = fracy_2-y_1x_2 - x_1x$$.






            share|cite|improve this answer











            $endgroup$








            • 1




              $begingroup$
              y2 is the y-coordinate of P2. x2 is the x-coordinate of P2. y1 is the y-coordinate of P1. x1 is the x-coordinate of P1.
              $endgroup$
              – Tomislav Ostojich
              Mar 16 at 2:01










            Your Answer





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            4 Answers
            4






            active

            oldest

            votes








            4 Answers
            4






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            The line $2x-3y=0$ defines a subspace. The vector $r$ that you used is orthogonal to this line.



            The correct point of view is to look at the line as set of all $(x,y)inmathbb R^2$ such that $(x,y)=(x, frac23x)$ since $y=frac23x$. So it suffices to check where $T$ sends, say $(3,2)$ (which is on the line). Thus, computing
            $beginpmatrix 5& 2\ 4& 1endpmatrix beginbmatrix 3\ 2endbmatrix$
            should give the answer.






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              misunderstand with some concepts but got that point. Thank you anyway
              $endgroup$
              – miguel747
              Mar 15 at 17:57










            • $begingroup$
              You should at least accept this as an answer then.
              $endgroup$
              – chhro
              Mar 15 at 18:08















            1












            $begingroup$

            The line $2x-3y=0$ defines a subspace. The vector $r$ that you used is orthogonal to this line.



            The correct point of view is to look at the line as set of all $(x,y)inmathbb R^2$ such that $(x,y)=(x, frac23x)$ since $y=frac23x$. So it suffices to check where $T$ sends, say $(3,2)$ (which is on the line). Thus, computing
            $beginpmatrix 5& 2\ 4& 1endpmatrix beginbmatrix 3\ 2endbmatrix$
            should give the answer.






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              misunderstand with some concepts but got that point. Thank you anyway
              $endgroup$
              – miguel747
              Mar 15 at 17:57










            • $begingroup$
              You should at least accept this as an answer then.
              $endgroup$
              – chhro
              Mar 15 at 18:08













            1












            1








            1





            $begingroup$

            The line $2x-3y=0$ defines a subspace. The vector $r$ that you used is orthogonal to this line.



            The correct point of view is to look at the line as set of all $(x,y)inmathbb R^2$ such that $(x,y)=(x, frac23x)$ since $y=frac23x$. So it suffices to check where $T$ sends, say $(3,2)$ (which is on the line). Thus, computing
            $beginpmatrix 5& 2\ 4& 1endpmatrix beginbmatrix 3\ 2endbmatrix$
            should give the answer.






            share|cite|improve this answer









            $endgroup$



            The line $2x-3y=0$ defines a subspace. The vector $r$ that you used is orthogonal to this line.



            The correct point of view is to look at the line as set of all $(x,y)inmathbb R^2$ such that $(x,y)=(x, frac23x)$ since $y=frac23x$. So it suffices to check where $T$ sends, say $(3,2)$ (which is on the line). Thus, computing
            $beginpmatrix 5& 2\ 4& 1endpmatrix beginbmatrix 3\ 2endbmatrix$
            should give the answer.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 15 at 17:48









            chhrochhro

            1,434311




            1,434311











            • $begingroup$
              misunderstand with some concepts but got that point. Thank you anyway
              $endgroup$
              – miguel747
              Mar 15 at 17:57










            • $begingroup$
              You should at least accept this as an answer then.
              $endgroup$
              – chhro
              Mar 15 at 18:08
















            • $begingroup$
              misunderstand with some concepts but got that point. Thank you anyway
              $endgroup$
              – miguel747
              Mar 15 at 17:57










            • $begingroup$
              You should at least accept this as an answer then.
              $endgroup$
              – chhro
              Mar 15 at 18:08















            $begingroup$
            misunderstand with some concepts but got that point. Thank you anyway
            $endgroup$
            – miguel747
            Mar 15 at 17:57




            $begingroup$
            misunderstand with some concepts but got that point. Thank you anyway
            $endgroup$
            – miguel747
            Mar 15 at 17:57












            $begingroup$
            You should at least accept this as an answer then.
            $endgroup$
            – chhro
            Mar 15 at 18:08




            $begingroup$
            You should at least accept this as an answer then.
            $endgroup$
            – chhro
            Mar 15 at 18:08











            3












            $begingroup$

            The simplest way to obtain the image of the given line



            $$2x-3y = 0$$



            is to obtain images of its $2$ points - why don't select the natural points $A=[0,0], B=[3,2]$?



            Computing their images we obtain



            $$left( beginmatrix
            5 & 2 \
            4 & 1 \
            endmatrixright )cdot
            left( beginmatrix
            0 \
            0 \
            endmatrixright ) = colorred
            left( beginmatrix
            0\
            0 \
            endmatrixright ),quadleft( beginmatrix
            5 & 2 \
            4 & 1 \
            endmatrixright )cdot
            left( beginmatrix
            3 \
            2 \
            endmatrixright ) = colorred
            left( beginmatrix
            19\
            14 \
            endmatrixright ),$$



            Now we write the equation of the line passing through resulting points $colorred(0, 0)$ and $colorred(19, 14)$, obtaining the correct result



            $$colorred14x-19y=0,$$



            which is similar to your incorrect result (B).



            enter image description here






            share|cite|improve this answer











            $endgroup$

















              3












              $begingroup$

              The simplest way to obtain the image of the given line



              $$2x-3y = 0$$



              is to obtain images of its $2$ points - why don't select the natural points $A=[0,0], B=[3,2]$?



              Computing their images we obtain



              $$left( beginmatrix
              5 & 2 \
              4 & 1 \
              endmatrixright )cdot
              left( beginmatrix
              0 \
              0 \
              endmatrixright ) = colorred
              left( beginmatrix
              0\
              0 \
              endmatrixright ),quadleft( beginmatrix
              5 & 2 \
              4 & 1 \
              endmatrixright )cdot
              left( beginmatrix
              3 \
              2 \
              endmatrixright ) = colorred
              left( beginmatrix
              19\
              14 \
              endmatrixright ),$$



              Now we write the equation of the line passing through resulting points $colorred(0, 0)$ and $colorred(19, 14)$, obtaining the correct result



              $$colorred14x-19y=0,$$



              which is similar to your incorrect result (B).



              enter image description here






              share|cite|improve this answer











              $endgroup$















                3












                3








                3





                $begingroup$

                The simplest way to obtain the image of the given line



                $$2x-3y = 0$$



                is to obtain images of its $2$ points - why don't select the natural points $A=[0,0], B=[3,2]$?



                Computing their images we obtain



                $$left( beginmatrix
                5 & 2 \
                4 & 1 \
                endmatrixright )cdot
                left( beginmatrix
                0 \
                0 \
                endmatrixright ) = colorred
                left( beginmatrix
                0\
                0 \
                endmatrixright ),quadleft( beginmatrix
                5 & 2 \
                4 & 1 \
                endmatrixright )cdot
                left( beginmatrix
                3 \
                2 \
                endmatrixright ) = colorred
                left( beginmatrix
                19\
                14 \
                endmatrixright ),$$



                Now we write the equation of the line passing through resulting points $colorred(0, 0)$ and $colorred(19, 14)$, obtaining the correct result



                $$colorred14x-19y=0,$$



                which is similar to your incorrect result (B).



                enter image description here






                share|cite|improve this answer











                $endgroup$



                The simplest way to obtain the image of the given line



                $$2x-3y = 0$$



                is to obtain images of its $2$ points - why don't select the natural points $A=[0,0], B=[3,2]$?



                Computing their images we obtain



                $$left( beginmatrix
                5 & 2 \
                4 & 1 \
                endmatrixright )cdot
                left( beginmatrix
                0 \
                0 \
                endmatrixright ) = colorred
                left( beginmatrix
                0\
                0 \
                endmatrixright ),quadleft( beginmatrix
                5 & 2 \
                4 & 1 \
                endmatrixright )cdot
                left( beginmatrix
                3 \
                2 \
                endmatrixright ) = colorred
                left( beginmatrix
                19\
                14 \
                endmatrixright ),$$



                Now we write the equation of the line passing through resulting points $colorred(0, 0)$ and $colorred(19, 14)$, obtaining the correct result



                $$colorred14x-19y=0,$$



                which is similar to your incorrect result (B).



                enter image description here







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Mar 16 at 2:44

























                answered Mar 16 at 2:04









                MarianDMarianD

                1,5311616




                1,5311616





















                    1












                    $begingroup$

                    It might help to think of the problem this way: If you’ve been given some equations that describe the relationship between the pairs of variables $(x,y)$ and $(x',y')$, how might you express the equation $2x-3y=0$ in terms of $x'$ and $y'$? One way is to solve those first equations for $x$ and $y$, then substitute into $2x-3y$ and simplify. In this case, you’re given that $$beginbmatrixx'\y'endbmatrix = T beginbmatrixx\yendbmatrix,tag1$$ so solving for $x$ and $y$ is a simple matter of computing $T^-1$.



                    You could also work directly with vectors and matrices. The equation of the line can be written as $$beginbmatrix2&-3endbmatrix beginbmatrixx\yendbmatrix = 0tag2$$ (verify this). Substituting from (1) for $[x,y]^T$, this becomes $$beginbmatrix2&-3endbmatrixT^-1beginbmatrixx'\y'endbmatrix = 0tag3,$$ so the coefficients of the transformed equation are the elements of the row vector $$beginbmatrix2&-3endbmatrix beginbmatrix5&2\4&1endbmatrix^-1.$$ I trust that you’re able to compute the value of this expression at this stage of your studies.



                    Incidentally, the vector $[2,-3]^T$ is orthogonal (normal) to the line. Equation (3) shows that such normal vectors transform differently than do vectors that represent points: If the point transformation is given by $mathbf v' = Mmathbf v$, then a normal vector $mathbf n$ transforms as $mathbf n' = M^-Tmathbf n$. (The $-T$ superscript means “inverse transpose.”)






                    share|cite|improve this answer









                    $endgroup$












                    • $begingroup$
                      But $T$ might not be invertible everytime.
                      $endgroup$
                      – chhro
                      Mar 15 at 22:05










                    • $begingroup$
                      @chhro That’s no different from not being able to solve for $x$ and $y$ in the first paragraph. In that case, you have to use another method. Happily, $T^-1$ exists in the specific instance. It wasn’t my intent to provide a comprehensive answer replete with obfuscating details. One could quibble that the given line might not pass through the origin, in which case the method in your answer doesn’t quite apply.
                      $endgroup$
                      – amd
                      Mar 15 at 23:15











                    • $begingroup$
                      Very interesting approach @amd . Ty for take your time to add more details and rich information. :)
                      $endgroup$
                      – miguel747
                      Mar 16 at 1:55










                    • $begingroup$
                      I'm sure my argument works even if the given line is not a subspace since it is independent of the given linear transformation $T$. It's essentially the idea in MarianD's answer.
                      $endgroup$
                      – chhro
                      Mar 19 at 20:43










                    • $begingroup$
                      @chhro You’re tacitly taking advantage of the fact that $T(0)=0$. When the line isn’t a subspace, it does not suffice to check where $T$ sends a single point on the line.
                      $endgroup$
                      – amd
                      Mar 19 at 20:47
















                    1












                    $begingroup$

                    It might help to think of the problem this way: If you’ve been given some equations that describe the relationship between the pairs of variables $(x,y)$ and $(x',y')$, how might you express the equation $2x-3y=0$ in terms of $x'$ and $y'$? One way is to solve those first equations for $x$ and $y$, then substitute into $2x-3y$ and simplify. In this case, you’re given that $$beginbmatrixx'\y'endbmatrix = T beginbmatrixx\yendbmatrix,tag1$$ so solving for $x$ and $y$ is a simple matter of computing $T^-1$.



                    You could also work directly with vectors and matrices. The equation of the line can be written as $$beginbmatrix2&-3endbmatrix beginbmatrixx\yendbmatrix = 0tag2$$ (verify this). Substituting from (1) for $[x,y]^T$, this becomes $$beginbmatrix2&-3endbmatrixT^-1beginbmatrixx'\y'endbmatrix = 0tag3,$$ so the coefficients of the transformed equation are the elements of the row vector $$beginbmatrix2&-3endbmatrix beginbmatrix5&2\4&1endbmatrix^-1.$$ I trust that you’re able to compute the value of this expression at this stage of your studies.



                    Incidentally, the vector $[2,-3]^T$ is orthogonal (normal) to the line. Equation (3) shows that such normal vectors transform differently than do vectors that represent points: If the point transformation is given by $mathbf v' = Mmathbf v$, then a normal vector $mathbf n$ transforms as $mathbf n' = M^-Tmathbf n$. (The $-T$ superscript means “inverse transpose.”)






                    share|cite|improve this answer









                    $endgroup$












                    • $begingroup$
                      But $T$ might not be invertible everytime.
                      $endgroup$
                      – chhro
                      Mar 15 at 22:05










                    • $begingroup$
                      @chhro That’s no different from not being able to solve for $x$ and $y$ in the first paragraph. In that case, you have to use another method. Happily, $T^-1$ exists in the specific instance. It wasn’t my intent to provide a comprehensive answer replete with obfuscating details. One could quibble that the given line might not pass through the origin, in which case the method in your answer doesn’t quite apply.
                      $endgroup$
                      – amd
                      Mar 15 at 23:15











                    • $begingroup$
                      Very interesting approach @amd . Ty for take your time to add more details and rich information. :)
                      $endgroup$
                      – miguel747
                      Mar 16 at 1:55










                    • $begingroup$
                      I'm sure my argument works even if the given line is not a subspace since it is independent of the given linear transformation $T$. It's essentially the idea in MarianD's answer.
                      $endgroup$
                      – chhro
                      Mar 19 at 20:43










                    • $begingroup$
                      @chhro You’re tacitly taking advantage of the fact that $T(0)=0$. When the line isn’t a subspace, it does not suffice to check where $T$ sends a single point on the line.
                      $endgroup$
                      – amd
                      Mar 19 at 20:47














                    1












                    1








                    1





                    $begingroup$

                    It might help to think of the problem this way: If you’ve been given some equations that describe the relationship between the pairs of variables $(x,y)$ and $(x',y')$, how might you express the equation $2x-3y=0$ in terms of $x'$ and $y'$? One way is to solve those first equations for $x$ and $y$, then substitute into $2x-3y$ and simplify. In this case, you’re given that $$beginbmatrixx'\y'endbmatrix = T beginbmatrixx\yendbmatrix,tag1$$ so solving for $x$ and $y$ is a simple matter of computing $T^-1$.



                    You could also work directly with vectors and matrices. The equation of the line can be written as $$beginbmatrix2&-3endbmatrix beginbmatrixx\yendbmatrix = 0tag2$$ (verify this). Substituting from (1) for $[x,y]^T$, this becomes $$beginbmatrix2&-3endbmatrixT^-1beginbmatrixx'\y'endbmatrix = 0tag3,$$ so the coefficients of the transformed equation are the elements of the row vector $$beginbmatrix2&-3endbmatrix beginbmatrix5&2\4&1endbmatrix^-1.$$ I trust that you’re able to compute the value of this expression at this stage of your studies.



                    Incidentally, the vector $[2,-3]^T$ is orthogonal (normal) to the line. Equation (3) shows that such normal vectors transform differently than do vectors that represent points: If the point transformation is given by $mathbf v' = Mmathbf v$, then a normal vector $mathbf n$ transforms as $mathbf n' = M^-Tmathbf n$. (The $-T$ superscript means “inverse transpose.”)






                    share|cite|improve this answer









                    $endgroup$



                    It might help to think of the problem this way: If you’ve been given some equations that describe the relationship between the pairs of variables $(x,y)$ and $(x',y')$, how might you express the equation $2x-3y=0$ in terms of $x'$ and $y'$? One way is to solve those first equations for $x$ and $y$, then substitute into $2x-3y$ and simplify. In this case, you’re given that $$beginbmatrixx'\y'endbmatrix = T beginbmatrixx\yendbmatrix,tag1$$ so solving for $x$ and $y$ is a simple matter of computing $T^-1$.



                    You could also work directly with vectors and matrices. The equation of the line can be written as $$beginbmatrix2&-3endbmatrix beginbmatrixx\yendbmatrix = 0tag2$$ (verify this). Substituting from (1) for $[x,y]^T$, this becomes $$beginbmatrix2&-3endbmatrixT^-1beginbmatrixx'\y'endbmatrix = 0tag3,$$ so the coefficients of the transformed equation are the elements of the row vector $$beginbmatrix2&-3endbmatrix beginbmatrix5&2\4&1endbmatrix^-1.$$ I trust that you’re able to compute the value of this expression at this stage of your studies.



                    Incidentally, the vector $[2,-3]^T$ is orthogonal (normal) to the line. Equation (3) shows that such normal vectors transform differently than do vectors that represent points: If the point transformation is given by $mathbf v' = Mmathbf v$, then a normal vector $mathbf n$ transforms as $mathbf n' = M^-Tmathbf n$. (The $-T$ superscript means “inverse transpose.”)







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 15 at 19:53









                    amdamd

                    31.1k21051




                    31.1k21051











                    • $begingroup$
                      But $T$ might not be invertible everytime.
                      $endgroup$
                      – chhro
                      Mar 15 at 22:05










                    • $begingroup$
                      @chhro That’s no different from not being able to solve for $x$ and $y$ in the first paragraph. In that case, you have to use another method. Happily, $T^-1$ exists in the specific instance. It wasn’t my intent to provide a comprehensive answer replete with obfuscating details. One could quibble that the given line might not pass through the origin, in which case the method in your answer doesn’t quite apply.
                      $endgroup$
                      – amd
                      Mar 15 at 23:15











                    • $begingroup$
                      Very interesting approach @amd . Ty for take your time to add more details and rich information. :)
                      $endgroup$
                      – miguel747
                      Mar 16 at 1:55










                    • $begingroup$
                      I'm sure my argument works even if the given line is not a subspace since it is independent of the given linear transformation $T$. It's essentially the idea in MarianD's answer.
                      $endgroup$
                      – chhro
                      Mar 19 at 20:43










                    • $begingroup$
                      @chhro You’re tacitly taking advantage of the fact that $T(0)=0$. When the line isn’t a subspace, it does not suffice to check where $T$ sends a single point on the line.
                      $endgroup$
                      – amd
                      Mar 19 at 20:47

















                    • $begingroup$
                      But $T$ might not be invertible everytime.
                      $endgroup$
                      – chhro
                      Mar 15 at 22:05










                    • $begingroup$
                      @chhro That’s no different from not being able to solve for $x$ and $y$ in the first paragraph. In that case, you have to use another method. Happily, $T^-1$ exists in the specific instance. It wasn’t my intent to provide a comprehensive answer replete with obfuscating details. One could quibble that the given line might not pass through the origin, in which case the method in your answer doesn’t quite apply.
                      $endgroup$
                      – amd
                      Mar 15 at 23:15











                    • $begingroup$
                      Very interesting approach @amd . Ty for take your time to add more details and rich information. :)
                      $endgroup$
                      – miguel747
                      Mar 16 at 1:55










                    • $begingroup$
                      I'm sure my argument works even if the given line is not a subspace since it is independent of the given linear transformation $T$. It's essentially the idea in MarianD's answer.
                      $endgroup$
                      – chhro
                      Mar 19 at 20:43










                    • $begingroup$
                      @chhro You’re tacitly taking advantage of the fact that $T(0)=0$. When the line isn’t a subspace, it does not suffice to check where $T$ sends a single point on the line.
                      $endgroup$
                      – amd
                      Mar 19 at 20:47
















                    $begingroup$
                    But $T$ might not be invertible everytime.
                    $endgroup$
                    – chhro
                    Mar 15 at 22:05




                    $begingroup$
                    But $T$ might not be invertible everytime.
                    $endgroup$
                    – chhro
                    Mar 15 at 22:05












                    $begingroup$
                    @chhro That’s no different from not being able to solve for $x$ and $y$ in the first paragraph. In that case, you have to use another method. Happily, $T^-1$ exists in the specific instance. It wasn’t my intent to provide a comprehensive answer replete with obfuscating details. One could quibble that the given line might not pass through the origin, in which case the method in your answer doesn’t quite apply.
                    $endgroup$
                    – amd
                    Mar 15 at 23:15





                    $begingroup$
                    @chhro That’s no different from not being able to solve for $x$ and $y$ in the first paragraph. In that case, you have to use another method. Happily, $T^-1$ exists in the specific instance. It wasn’t my intent to provide a comprehensive answer replete with obfuscating details. One could quibble that the given line might not pass through the origin, in which case the method in your answer doesn’t quite apply.
                    $endgroup$
                    – amd
                    Mar 15 at 23:15













                    $begingroup$
                    Very interesting approach @amd . Ty for take your time to add more details and rich information. :)
                    $endgroup$
                    – miguel747
                    Mar 16 at 1:55




                    $begingroup$
                    Very interesting approach @amd . Ty for take your time to add more details and rich information. :)
                    $endgroup$
                    – miguel747
                    Mar 16 at 1:55












                    $begingroup$
                    I'm sure my argument works even if the given line is not a subspace since it is independent of the given linear transformation $T$. It's essentially the idea in MarianD's answer.
                    $endgroup$
                    – chhro
                    Mar 19 at 20:43




                    $begingroup$
                    I'm sure my argument works even if the given line is not a subspace since it is independent of the given linear transformation $T$. It's essentially the idea in MarianD's answer.
                    $endgroup$
                    – chhro
                    Mar 19 at 20:43












                    $begingroup$
                    @chhro You’re tacitly taking advantage of the fact that $T(0)=0$. When the line isn’t a subspace, it does not suffice to check where $T$ sends a single point on the line.
                    $endgroup$
                    – amd
                    Mar 19 at 20:47





                    $begingroup$
                    @chhro You’re tacitly taking advantage of the fact that $T(0)=0$. When the line isn’t a subspace, it does not suffice to check where $T$ sends a single point on the line.
                    $endgroup$
                    – amd
                    Mar 19 at 20:47












                    0












                    $begingroup$

                    STEP 1. Realize that $《x, y》: 2x + 3y = 0$ is a subspace.



                    STEP 2. find the basis for the subspace. It is $(1, 2/3)$.



                    STEP 3. Do matrix multiplication.



                    $$left[ beginmatrix
                    5 & 2 \
                    4 & 1 \
                    endmatrixright ]
                    left[ beginmatrix
                    1 \
                    2/3 \
                    endmatrixright ] =
                    left[ beginmatrix
                    19/3\
                    14/3 \
                    endmatrixright ]$$



                    STEP 4. Translate back into equation form.



                    $$P_1 = alpha left[ beginmatrix
                    19/3 \
                    4/3 \
                    endmatrixright ]$$
                    and $$P_2 = beta left[ beginmatrix
                    19/3 \
                    4/3 \
                    endmatrixright ]$$
                    . Plug these two points (interpreting the top entries as x coordinate and bottom entries as y-coordinate). Use the point-slope equation $$y = fracy_2-y_1x_2 - x_1x$$.






                    share|cite|improve this answer











                    $endgroup$








                    • 1




                      $begingroup$
                      y2 is the y-coordinate of P2. x2 is the x-coordinate of P2. y1 is the y-coordinate of P1. x1 is the x-coordinate of P1.
                      $endgroup$
                      – Tomislav Ostojich
                      Mar 16 at 2:01















                    0












                    $begingroup$

                    STEP 1. Realize that $《x, y》: 2x + 3y = 0$ is a subspace.



                    STEP 2. find the basis for the subspace. It is $(1, 2/3)$.



                    STEP 3. Do matrix multiplication.



                    $$left[ beginmatrix
                    5 & 2 \
                    4 & 1 \
                    endmatrixright ]
                    left[ beginmatrix
                    1 \
                    2/3 \
                    endmatrixright ] =
                    left[ beginmatrix
                    19/3\
                    14/3 \
                    endmatrixright ]$$



                    STEP 4. Translate back into equation form.



                    $$P_1 = alpha left[ beginmatrix
                    19/3 \
                    4/3 \
                    endmatrixright ]$$
                    and $$P_2 = beta left[ beginmatrix
                    19/3 \
                    4/3 \
                    endmatrixright ]$$
                    . Plug these two points (interpreting the top entries as x coordinate and bottom entries as y-coordinate). Use the point-slope equation $$y = fracy_2-y_1x_2 - x_1x$$.






                    share|cite|improve this answer











                    $endgroup$








                    • 1




                      $begingroup$
                      y2 is the y-coordinate of P2. x2 is the x-coordinate of P2. y1 is the y-coordinate of P1. x1 is the x-coordinate of P1.
                      $endgroup$
                      – Tomislav Ostojich
                      Mar 16 at 2:01













                    0












                    0








                    0





                    $begingroup$

                    STEP 1. Realize that $《x, y》: 2x + 3y = 0$ is a subspace.



                    STEP 2. find the basis for the subspace. It is $(1, 2/3)$.



                    STEP 3. Do matrix multiplication.



                    $$left[ beginmatrix
                    5 & 2 \
                    4 & 1 \
                    endmatrixright ]
                    left[ beginmatrix
                    1 \
                    2/3 \
                    endmatrixright ] =
                    left[ beginmatrix
                    19/3\
                    14/3 \
                    endmatrixright ]$$



                    STEP 4. Translate back into equation form.



                    $$P_1 = alpha left[ beginmatrix
                    19/3 \
                    4/3 \
                    endmatrixright ]$$
                    and $$P_2 = beta left[ beginmatrix
                    19/3 \
                    4/3 \
                    endmatrixright ]$$
                    . Plug these two points (interpreting the top entries as x coordinate and bottom entries as y-coordinate). Use the point-slope equation $$y = fracy_2-y_1x_2 - x_1x$$.






                    share|cite|improve this answer











                    $endgroup$



                    STEP 1. Realize that $《x, y》: 2x + 3y = 0$ is a subspace.



                    STEP 2. find the basis for the subspace. It is $(1, 2/3)$.



                    STEP 3. Do matrix multiplication.



                    $$left[ beginmatrix
                    5 & 2 \
                    4 & 1 \
                    endmatrixright ]
                    left[ beginmatrix
                    1 \
                    2/3 \
                    endmatrixright ] =
                    left[ beginmatrix
                    19/3\
                    14/3 \
                    endmatrixright ]$$



                    STEP 4. Translate back into equation form.



                    $$P_1 = alpha left[ beginmatrix
                    19/3 \
                    4/3 \
                    endmatrixright ]$$
                    and $$P_2 = beta left[ beginmatrix
                    19/3 \
                    4/3 \
                    endmatrixright ]$$
                    . Plug these two points (interpreting the top entries as x coordinate and bottom entries as y-coordinate). Use the point-slope equation $$y = fracy_2-y_1x_2 - x_1x$$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Mar 16 at 2:01

























                    answered Mar 16 at 1:00









                    Tomislav OstojichTomislav Ostojich

                    761718




                    761718







                    • 1




                      $begingroup$
                      y2 is the y-coordinate of P2. x2 is the x-coordinate of P2. y1 is the y-coordinate of P1. x1 is the x-coordinate of P1.
                      $endgroup$
                      – Tomislav Ostojich
                      Mar 16 at 2:01












                    • 1




                      $begingroup$
                      y2 is the y-coordinate of P2. x2 is the x-coordinate of P2. y1 is the y-coordinate of P1. x1 is the x-coordinate of P1.
                      $endgroup$
                      – Tomislav Ostojich
                      Mar 16 at 2:01







                    1




                    1




                    $begingroup$
                    y2 is the y-coordinate of P2. x2 is the x-coordinate of P2. y1 is the y-coordinate of P1. x1 is the x-coordinate of P1.
                    $endgroup$
                    – Tomislav Ostojich
                    Mar 16 at 2:01




                    $begingroup$
                    y2 is the y-coordinate of P2. x2 is the x-coordinate of P2. y1 is the y-coordinate of P1. x1 is the x-coordinate of P1.
                    $endgroup$
                    – Tomislav Ostojich
                    Mar 16 at 2:01

















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