State whether the following simultaneous equations have a unique solution












2














I have searched online and through my textbooks but I'm completely stuck, I'm asked to verify if the following simultaneous equations have unique solutions;



$$y+2z=1\
-x+y=3\
2x+3z=2$$



I have the determinant of the matrix, but unsure how to proceed to find what is asked.






linear-algebra







share|cite|improve this question
























  • Gaussian elimination? en.wikipedia.org/wiki/Gaussian_elimination
    – Lord Shark the Unknown
    1 hour ago










  • The determinant is non-zero, so there exists a unique solution given by $$begin{bmatrix}x\y\zend{bmatrix}=begin{bmatrix}0&1&2\-1&1&0\2&0&3 end{bmatrix}^{-1}begin{bmatrix}1\3\2end{bmatrix}$$
    – Shubham Johri
    1 hour ago












  • ... and what is the value of the determinant?
    – A.Γ.
    1 hour ago
















2














I have searched online and through my textbooks but I'm completely stuck, I'm asked to verify if the following simultaneous equations have unique solutions;



$$y+2z=1\
-x+y=3\
2x+3z=2$$



I have the determinant of the matrix, but unsure how to proceed to find what is asked.






linear-algebra







share|cite|improve this question
























  • Gaussian elimination? en.wikipedia.org/wiki/Gaussian_elimination
    – Lord Shark the Unknown
    1 hour ago










  • The determinant is non-zero, so there exists a unique solution given by $$begin{bmatrix}x\y\zend{bmatrix}=begin{bmatrix}0&1&2\-1&1&0\2&0&3 end{bmatrix}^{-1}begin{bmatrix}1\3\2end{bmatrix}$$
    – Shubham Johri
    1 hour ago












  • ... and what is the value of the determinant?
    – A.Γ.
    1 hour ago














2












2








2







I have searched online and through my textbooks but I'm completely stuck, I'm asked to verify if the following simultaneous equations have unique solutions;



$$y+2z=1\
-x+y=3\
2x+3z=2$$



I have the determinant of the matrix, but unsure how to proceed to find what is asked.






linear-algebra







share|cite|improve this question















I have searched online and through my textbooks but I'm completely stuck, I'm asked to verify if the following simultaneous equations have unique solutions;



$$y+2z=1\
-x+y=3\
2x+3z=2$$



I have the determinant of the matrix, but unsure how to proceed to find what is asked.






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 1 hour ago









A.Γ.

22k22455




22k22455










asked 1 hour ago









RocketKangaroo

284




284












  • Gaussian elimination? en.wikipedia.org/wiki/Gaussian_elimination
    – Lord Shark the Unknown
    1 hour ago










  • The determinant is non-zero, so there exists a unique solution given by $$begin{bmatrix}x\y\zend{bmatrix}=begin{bmatrix}0&1&2\-1&1&0\2&0&3 end{bmatrix}^{-1}begin{bmatrix}1\3\2end{bmatrix}$$
    – Shubham Johri
    1 hour ago












  • ... and what is the value of the determinant?
    – A.Γ.
    1 hour ago


















  • Gaussian elimination? en.wikipedia.org/wiki/Gaussian_elimination
    – Lord Shark the Unknown
    1 hour ago










  • The determinant is non-zero, so there exists a unique solution given by $$begin{bmatrix}x\y\zend{bmatrix}=begin{bmatrix}0&1&2\-1&1&0\2&0&3 end{bmatrix}^{-1}begin{bmatrix}1\3\2end{bmatrix}$$
    – Shubham Johri
    1 hour ago












  • ... and what is the value of the determinant?
    – A.Γ.
    1 hour ago
















Gaussian elimination? en.wikipedia.org/wiki/Gaussian_elimination
– Lord Shark the Unknown
1 hour ago




Gaussian elimination? en.wikipedia.org/wiki/Gaussian_elimination
– Lord Shark the Unknown
1 hour ago












The determinant is non-zero, so there exists a unique solution given by $$begin{bmatrix}x\y\zend{bmatrix}=begin{bmatrix}0&1&2\-1&1&0\2&0&3 end{bmatrix}^{-1}begin{bmatrix}1\3\2end{bmatrix}$$
– Shubham Johri
1 hour ago






The determinant is non-zero, so there exists a unique solution given by $$begin{bmatrix}x\y\zend{bmatrix}=begin{bmatrix}0&1&2\-1&1&0\2&0&3 end{bmatrix}^{-1}begin{bmatrix}1\3\2end{bmatrix}$$
– Shubham Johri
1 hour ago














... and what is the value of the determinant?
– A.Γ.
1 hour ago




... and what is the value of the determinant?
– A.Γ.
1 hour ago










3 Answers
3






active

oldest

votes


















4














Well, once you have this system of equations, its easy to write it in matrix form as you seem to have done:



$$
underbrace{begin{bmatrix}
0&1&2\
-1&1&0\
2&0&3
end{bmatrix}}_{A}
begin{bmatrix}
x\
y\
z
end{bmatrix}=
begin{bmatrix}
1\
3\
2
end{bmatrix}.$$
We have $det(A)=0cdot(1cdot 3-0cdot 0)+1cdot(1cdot 3)+2(1cdot 0-2cdot 1)=3-4=-1.$ So, the determinant is nonsingular. This implies that the matrix represents an isomorphism $mathbb{R}^3to mathbb{R}^3$. So, for any element of $mathbb{R}^3$, there exists a unique solution. In particular, the matrix equation above admits a unique solution.






share|cite|improve this answer





























    2














    Multiplying the second by $2$ and adding to the third we get
    $$2y+3z=8$$
    $$y+2z=1$$ From here we get $$z=-6,x=10,y=13$$






    share|cite|improve this answer





























      0














      The general criterion for the existence of a solution of a non-homogeneous linear system is the matrix of the homogeneous part and the augmented matrix have the same rank, i.e. here
      $$operatorname{rank}begin{bmatrix}0&1&2\-1&1&0\2&0&3end{bmatrix}=operatorname{rank}begin{bmatrix}0&1&2&1\-1&1&0&3\2&0&3&2end{bmatrix}.$$
      Furthermore, this rank is the codimension of the (affine) subspace of solutions.



      The homogeneous linear system has maximal rank ($3$) as its determinant is non-zero. Hence the augmented matrix also has rank $3$, and there exists a unique solution.






      share|cite|improve this answer





















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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4














        Well, once you have this system of equations, its easy to write it in matrix form as you seem to have done:



        $$
        underbrace{begin{bmatrix}
        0&1&2\
        -1&1&0\
        2&0&3
        end{bmatrix}}_{A}
        begin{bmatrix}
        x\
        y\
        z
        end{bmatrix}=
        begin{bmatrix}
        1\
        3\
        2
        end{bmatrix}.$$
        We have $det(A)=0cdot(1cdot 3-0cdot 0)+1cdot(1cdot 3)+2(1cdot 0-2cdot 1)=3-4=-1.$ So, the determinant is nonsingular. This implies that the matrix represents an isomorphism $mathbb{R}^3to mathbb{R}^3$. So, for any element of $mathbb{R}^3$, there exists a unique solution. In particular, the matrix equation above admits a unique solution.






        share|cite|improve this answer


























          4














          Well, once you have this system of equations, its easy to write it in matrix form as you seem to have done:



          $$
          underbrace{begin{bmatrix}
          0&1&2\
          -1&1&0\
          2&0&3
          end{bmatrix}}_{A}
          begin{bmatrix}
          x\
          y\
          z
          end{bmatrix}=
          begin{bmatrix}
          1\
          3\
          2
          end{bmatrix}.$$
          We have $det(A)=0cdot(1cdot 3-0cdot 0)+1cdot(1cdot 3)+2(1cdot 0-2cdot 1)=3-4=-1.$ So, the determinant is nonsingular. This implies that the matrix represents an isomorphism $mathbb{R}^3to mathbb{R}^3$. So, for any element of $mathbb{R}^3$, there exists a unique solution. In particular, the matrix equation above admits a unique solution.






          share|cite|improve this answer
























            4












            4








            4






            Well, once you have this system of equations, its easy to write it in matrix form as you seem to have done:



            $$
            underbrace{begin{bmatrix}
            0&1&2\
            -1&1&0\
            2&0&3
            end{bmatrix}}_{A}
            begin{bmatrix}
            x\
            y\
            z
            end{bmatrix}=
            begin{bmatrix}
            1\
            3\
            2
            end{bmatrix}.$$
            We have $det(A)=0cdot(1cdot 3-0cdot 0)+1cdot(1cdot 3)+2(1cdot 0-2cdot 1)=3-4=-1.$ So, the determinant is nonsingular. This implies that the matrix represents an isomorphism $mathbb{R}^3to mathbb{R}^3$. So, for any element of $mathbb{R}^3$, there exists a unique solution. In particular, the matrix equation above admits a unique solution.






            share|cite|improve this answer












            Well, once you have this system of equations, its easy to write it in matrix form as you seem to have done:



            $$
            underbrace{begin{bmatrix}
            0&1&2\
            -1&1&0\
            2&0&3
            end{bmatrix}}_{A}
            begin{bmatrix}
            x\
            y\
            z
            end{bmatrix}=
            begin{bmatrix}
            1\
            3\
            2
            end{bmatrix}.$$
            We have $det(A)=0cdot(1cdot 3-0cdot 0)+1cdot(1cdot 3)+2(1cdot 0-2cdot 1)=3-4=-1.$ So, the determinant is nonsingular. This implies that the matrix represents an isomorphism $mathbb{R}^3to mathbb{R}^3$. So, for any element of $mathbb{R}^3$, there exists a unique solution. In particular, the matrix equation above admits a unique solution.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            Antonios-Alexandros Robotis

            9,19741640




            9,19741640























                2














                Multiplying the second by $2$ and adding to the third we get
                $$2y+3z=8$$
                $$y+2z=1$$ From here we get $$z=-6,x=10,y=13$$






                share|cite|improve this answer


























                  2














                  Multiplying the second by $2$ and adding to the third we get
                  $$2y+3z=8$$
                  $$y+2z=1$$ From here we get $$z=-6,x=10,y=13$$






                  share|cite|improve this answer
























                    2












                    2








                    2






                    Multiplying the second by $2$ and adding to the third we get
                    $$2y+3z=8$$
                    $$y+2z=1$$ From here we get $$z=-6,x=10,y=13$$






                    share|cite|improve this answer












                    Multiplying the second by $2$ and adding to the third we get
                    $$2y+3z=8$$
                    $$y+2z=1$$ From here we get $$z=-6,x=10,y=13$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour ago









                    Dr. Sonnhard Graubner

                    73.1k42865




                    73.1k42865























                        0














                        The general criterion for the existence of a solution of a non-homogeneous linear system is the matrix of the homogeneous part and the augmented matrix have the same rank, i.e. here
                        $$operatorname{rank}begin{bmatrix}0&1&2\-1&1&0\2&0&3end{bmatrix}=operatorname{rank}begin{bmatrix}0&1&2&1\-1&1&0&3\2&0&3&2end{bmatrix}.$$
                        Furthermore, this rank is the codimension of the (affine) subspace of solutions.



                        The homogeneous linear system has maximal rank ($3$) as its determinant is non-zero. Hence the augmented matrix also has rank $3$, and there exists a unique solution.






                        share|cite|improve this answer


























                          0














                          The general criterion for the existence of a solution of a non-homogeneous linear system is the matrix of the homogeneous part and the augmented matrix have the same rank, i.e. here
                          $$operatorname{rank}begin{bmatrix}0&1&2\-1&1&0\2&0&3end{bmatrix}=operatorname{rank}begin{bmatrix}0&1&2&1\-1&1&0&3\2&0&3&2end{bmatrix}.$$
                          Furthermore, this rank is the codimension of the (affine) subspace of solutions.



                          The homogeneous linear system has maximal rank ($3$) as its determinant is non-zero. Hence the augmented matrix also has rank $3$, and there exists a unique solution.






                          share|cite|improve this answer
























                            0












                            0








                            0






                            The general criterion for the existence of a solution of a non-homogeneous linear system is the matrix of the homogeneous part and the augmented matrix have the same rank, i.e. here
                            $$operatorname{rank}begin{bmatrix}0&1&2\-1&1&0\2&0&3end{bmatrix}=operatorname{rank}begin{bmatrix}0&1&2&1\-1&1&0&3\2&0&3&2end{bmatrix}.$$
                            Furthermore, this rank is the codimension of the (affine) subspace of solutions.



                            The homogeneous linear system has maximal rank ($3$) as its determinant is non-zero. Hence the augmented matrix also has rank $3$, and there exists a unique solution.






                            share|cite|improve this answer












                            The general criterion for the existence of a solution of a non-homogeneous linear system is the matrix of the homogeneous part and the augmented matrix have the same rank, i.e. here
                            $$operatorname{rank}begin{bmatrix}0&1&2\-1&1&0\2&0&3end{bmatrix}=operatorname{rank}begin{bmatrix}0&1&2&1\-1&1&0&3\2&0&3&2end{bmatrix}.$$
                            Furthermore, this rank is the codimension of the (affine) subspace of solutions.



                            The homogeneous linear system has maximal rank ($3$) as its determinant is non-zero. Hence the augmented matrix also has rank $3$, and there exists a unique solution.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 1 hour ago









                            Bernard

                            118k639112




                            118k639112






























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