Overdetermined linear system?












1












$begingroup$


What should be the conditions on coefficients $a_i$ and $b_i$ such that the following overdetermined linear system of equations has unique solution.
$$a_i x+y=b_i$$
where $i=1,2,3...,n$.



The system represents $n$ straight lines and it is possible to make them ins
tersect at one point, hence sytem must have a unique solution for some $a_i,b_i$.






linear-algebra systems-of-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Where is $b_i$ in your equation?
    $endgroup$
    – induction601
    Dec 17 '18 at 6:35










  • $begingroup$
    Typo! Fixed. Sorry
    $endgroup$
    – ersh
    Dec 17 '18 at 6:42
















1












$begingroup$


What should be the conditions on coefficients $a_i$ and $b_i$ such that the following overdetermined linear system of equations has unique solution.
$$a_i x+y=b_i$$
where $i=1,2,3...,n$.



The system represents $n$ straight lines and it is possible to make them ins
tersect at one point, hence sytem must have a unique solution for some $a_i,b_i$.






linear-algebra systems-of-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Where is $b_i$ in your equation?
    $endgroup$
    – induction601
    Dec 17 '18 at 6:35










  • $begingroup$
    Typo! Fixed. Sorry
    $endgroup$
    – ersh
    Dec 17 '18 at 6:42














1












1








1





$begingroup$


What should be the conditions on coefficients $a_i$ and $b_i$ such that the following overdetermined linear system of equations has unique solution.
$$a_i x+y=b_i$$
where $i=1,2,3...,n$.



The system represents $n$ straight lines and it is possible to make them ins
tersect at one point, hence sytem must have a unique solution for some $a_i,b_i$.






linear-algebra systems-of-equations







share|cite|improve this question











$endgroup$




What should be the conditions on coefficients $a_i$ and $b_i$ such that the following overdetermined linear system of equations has unique solution.
$$a_i x+y=b_i$$
where $i=1,2,3...,n$.



The system represents $n$ straight lines and it is possible to make them ins
tersect at one point, hence sytem must have a unique solution for some $a_i,b_i$.






linear-algebra systems-of-equations




linear-algebra systems-of-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 6:41







ersh

















asked Dec 17 '18 at 6:30









ershersh

436113




436113












  • $begingroup$
    Where is $b_i$ in your equation?
    $endgroup$
    – induction601
    Dec 17 '18 at 6:35










  • $begingroup$
    Typo! Fixed. Sorry
    $endgroup$
    – ersh
    Dec 17 '18 at 6:42


















  • $begingroup$
    Where is $b_i$ in your equation?
    $endgroup$
    – induction601
    Dec 17 '18 at 6:35










  • $begingroup$
    Typo! Fixed. Sorry
    $endgroup$
    – ersh
    Dec 17 '18 at 6:42
















$begingroup$
Where is $b_i$ in your equation?
$endgroup$
– induction601
Dec 17 '18 at 6:35




$begingroup$
Where is $b_i$ in your equation?
$endgroup$
– induction601
Dec 17 '18 at 6:35












$begingroup$
Typo! Fixed. Sorry
$endgroup$
– ersh
Dec 17 '18 at 6:42




$begingroup$
Typo! Fixed. Sorry
$endgroup$
– ersh
Dec 17 '18 at 6:42










1 Answer
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$begingroup$

The system is over-determined for $n>2$ but here is a general method. We can write the augmented matrix for the system $Abegin{bmatrix}x\yend{bmatrix}=B$ as under:



$begin{bmatrix}a_1&1&Big|&b_1\a_2&1&Big|&b_2\vdots&vdots&Big|&vdots\a_n&1&Big|&b_nend{bmatrix}$



For a unique solution to exist, we should have $2$ linearly-independent equations to solve for the $2$ unknowns. In other words, the rank of the coefficient and augmented matrices should be $2$. Recall that no more than $2$ vectors in $Bbb R^2$ can be linearly independent, so the rank of the coefficient matrix $A, text{rank}(A)le2$. For $text{rank}(A)=2$, we need to ensure at-least two $a_i$ are distinct. Say we have distinct $a_1ne0,a_2ne a_1$.



The point of intersection of $a_1x+y=b_1,a_2x+y=b_2$ is given by $(X,Y)=displaystyleBig(frac{b_1-b_2}{a_1-a_2},frac{a_1b_2-a_2b_1}{a_1-a_2}Big)$.



$displaystyle R_ito R_i-frac{a_i}{a_1}cdot R_1, i>1$



$displaystyle R_jto R_j-frac{1-frac{a_i}{a_1}}{1-frac{a_2}{a_1}}cdot R_2, j>2$



$simbegin{bmatrix}a_1&1&Big|&b_1\0&1-frac{a_2}{a_1}&Big|&b_2-frac{a_2}{a_1}cdot b_1\0&0&Big|&b'_3\vdots&vdots&Big|&vdots\0&0&Big|&b'_nend{bmatrix}$



$displaystyle b'_i=b_i-frac{a_i}{a_1}cdot b_1-frac{1-frac{a_i}{a_1}}{1-frac{a_2}{a_1}}cdotBig[b_2-frac{a_2}{a_1}cdot b_1Big] forall i>2$



For the rank of the augmented matrix to be $0$, we require $b'_i=0$



$displaystyletherefore b_i=frac{(a_i-a_2)b_1+(a_1-a_i)b_2}{a_1-a_2}=Big[frac{b_1-b_2}{a_1-a_2}Big]cdot a_i+Big(frac{a_1b_2-a_2b_1}{a_1-a_2}Big), forall i>2$




Therefore, if:




  • There are two lines $L_i,L_j$ not parallel to each other $(a_ine a_j)$ intersecting at $(X,Y)$;


  • $displaystyle b_k=Big[frac{b_i-b_j}{a_i-a_j}Big]cdot a_k+Big(frac{a_ib_j-a_jb_i}{a_i-a_j}Big)=Xa_k+Y, forall kne i,j$; that is, $(X,Y)$ lies on all the remaining lines;



Then, the straight lines $L_i:= a_ix+y=b_i,iin{1,2,...,n}$ intersect at the point $displaystyle(X,Y)=Big(frac{b_i-b_j}{a_i-a_j},frac{a_ib_j-a_jb_i}{a_i-a_j}Big)$ uniquely.







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    $begingroup$

    The system is over-determined for $n>2$ but here is a general method. We can write the augmented matrix for the system $Abegin{bmatrix}x\yend{bmatrix}=B$ as under:



    $begin{bmatrix}a_1&1&Big|&b_1\a_2&1&Big|&b_2\vdots&vdots&Big|&vdots\a_n&1&Big|&b_nend{bmatrix}$



    For a unique solution to exist, we should have $2$ linearly-independent equations to solve for the $2$ unknowns. In other words, the rank of the coefficient and augmented matrices should be $2$. Recall that no more than $2$ vectors in $Bbb R^2$ can be linearly independent, so the rank of the coefficient matrix $A, text{rank}(A)le2$. For $text{rank}(A)=2$, we need to ensure at-least two $a_i$ are distinct. Say we have distinct $a_1ne0,a_2ne a_1$.



    The point of intersection of $a_1x+y=b_1,a_2x+y=b_2$ is given by $(X,Y)=displaystyleBig(frac{b_1-b_2}{a_1-a_2},frac{a_1b_2-a_2b_1}{a_1-a_2}Big)$.



    $displaystyle R_ito R_i-frac{a_i}{a_1}cdot R_1, i>1$



    $displaystyle R_jto R_j-frac{1-frac{a_i}{a_1}}{1-frac{a_2}{a_1}}cdot R_2, j>2$



    $simbegin{bmatrix}a_1&1&Big|&b_1\0&1-frac{a_2}{a_1}&Big|&b_2-frac{a_2}{a_1}cdot b_1\0&0&Big|&b'_3\vdots&vdots&Big|&vdots\0&0&Big|&b'_nend{bmatrix}$



    $displaystyle b'_i=b_i-frac{a_i}{a_1}cdot b_1-frac{1-frac{a_i}{a_1}}{1-frac{a_2}{a_1}}cdotBig[b_2-frac{a_2}{a_1}cdot b_1Big] forall i>2$



    For the rank of the augmented matrix to be $0$, we require $b'_i=0$



    $displaystyletherefore b_i=frac{(a_i-a_2)b_1+(a_1-a_i)b_2}{a_1-a_2}=Big[frac{b_1-b_2}{a_1-a_2}Big]cdot a_i+Big(frac{a_1b_2-a_2b_1}{a_1-a_2}Big), forall i>2$




    Therefore, if:




    • There are two lines $L_i,L_j$ not parallel to each other $(a_ine a_j)$ intersecting at $(X,Y)$;


    • $displaystyle b_k=Big[frac{b_i-b_j}{a_i-a_j}Big]cdot a_k+Big(frac{a_ib_j-a_jb_i}{a_i-a_j}Big)=Xa_k+Y, forall kne i,j$; that is, $(X,Y)$ lies on all the remaining lines;



    Then, the straight lines $L_i:= a_ix+y=b_i,iin{1,2,...,n}$ intersect at the point $displaystyle(X,Y)=Big(frac{b_i-b_j}{a_i-a_j},frac{a_ib_j-a_jb_i}{a_i-a_j}Big)$ uniquely.







    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      The system is over-determined for $n>2$ but here is a general method. We can write the augmented matrix for the system $Abegin{bmatrix}x\yend{bmatrix}=B$ as under:



      $begin{bmatrix}a_1&1&Big|&b_1\a_2&1&Big|&b_2\vdots&vdots&Big|&vdots\a_n&1&Big|&b_nend{bmatrix}$



      For a unique solution to exist, we should have $2$ linearly-independent equations to solve for the $2$ unknowns. In other words, the rank of the coefficient and augmented matrices should be $2$. Recall that no more than $2$ vectors in $Bbb R^2$ can be linearly independent, so the rank of the coefficient matrix $A, text{rank}(A)le2$. For $text{rank}(A)=2$, we need to ensure at-least two $a_i$ are distinct. Say we have distinct $a_1ne0,a_2ne a_1$.



      The point of intersection of $a_1x+y=b_1,a_2x+y=b_2$ is given by $(X,Y)=displaystyleBig(frac{b_1-b_2}{a_1-a_2},frac{a_1b_2-a_2b_1}{a_1-a_2}Big)$.



      $displaystyle R_ito R_i-frac{a_i}{a_1}cdot R_1, i>1$



      $displaystyle R_jto R_j-frac{1-frac{a_i}{a_1}}{1-frac{a_2}{a_1}}cdot R_2, j>2$



      $simbegin{bmatrix}a_1&1&Big|&b_1\0&1-frac{a_2}{a_1}&Big|&b_2-frac{a_2}{a_1}cdot b_1\0&0&Big|&b'_3\vdots&vdots&Big|&vdots\0&0&Big|&b'_nend{bmatrix}$



      $displaystyle b'_i=b_i-frac{a_i}{a_1}cdot b_1-frac{1-frac{a_i}{a_1}}{1-frac{a_2}{a_1}}cdotBig[b_2-frac{a_2}{a_1}cdot b_1Big] forall i>2$



      For the rank of the augmented matrix to be $0$, we require $b'_i=0$



      $displaystyletherefore b_i=frac{(a_i-a_2)b_1+(a_1-a_i)b_2}{a_1-a_2}=Big[frac{b_1-b_2}{a_1-a_2}Big]cdot a_i+Big(frac{a_1b_2-a_2b_1}{a_1-a_2}Big), forall i>2$




      Therefore, if:




      • There are two lines $L_i,L_j$ not parallel to each other $(a_ine a_j)$ intersecting at $(X,Y)$;


      • $displaystyle b_k=Big[frac{b_i-b_j}{a_i-a_j}Big]cdot a_k+Big(frac{a_ib_j-a_jb_i}{a_i-a_j}Big)=Xa_k+Y, forall kne i,j$; that is, $(X,Y)$ lies on all the remaining lines;



      Then, the straight lines $L_i:= a_ix+y=b_i,iin{1,2,...,n}$ intersect at the point $displaystyle(X,Y)=Big(frac{b_i-b_j}{a_i-a_j},frac{a_ib_j-a_jb_i}{a_i-a_j}Big)$ uniquely.







      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        The system is over-determined for $n>2$ but here is a general method. We can write the augmented matrix for the system $Abegin{bmatrix}x\yend{bmatrix}=B$ as under:



        $begin{bmatrix}a_1&1&Big|&b_1\a_2&1&Big|&b_2\vdots&vdots&Big|&vdots\a_n&1&Big|&b_nend{bmatrix}$



        For a unique solution to exist, we should have $2$ linearly-independent equations to solve for the $2$ unknowns. In other words, the rank of the coefficient and augmented matrices should be $2$. Recall that no more than $2$ vectors in $Bbb R^2$ can be linearly independent, so the rank of the coefficient matrix $A, text{rank}(A)le2$. For $text{rank}(A)=2$, we need to ensure at-least two $a_i$ are distinct. Say we have distinct $a_1ne0,a_2ne a_1$.



        The point of intersection of $a_1x+y=b_1,a_2x+y=b_2$ is given by $(X,Y)=displaystyleBig(frac{b_1-b_2}{a_1-a_2},frac{a_1b_2-a_2b_1}{a_1-a_2}Big)$.



        $displaystyle R_ito R_i-frac{a_i}{a_1}cdot R_1, i>1$



        $displaystyle R_jto R_j-frac{1-frac{a_i}{a_1}}{1-frac{a_2}{a_1}}cdot R_2, j>2$



        $simbegin{bmatrix}a_1&1&Big|&b_1\0&1-frac{a_2}{a_1}&Big|&b_2-frac{a_2}{a_1}cdot b_1\0&0&Big|&b'_3\vdots&vdots&Big|&vdots\0&0&Big|&b'_nend{bmatrix}$



        $displaystyle b'_i=b_i-frac{a_i}{a_1}cdot b_1-frac{1-frac{a_i}{a_1}}{1-frac{a_2}{a_1}}cdotBig[b_2-frac{a_2}{a_1}cdot b_1Big] forall i>2$



        For the rank of the augmented matrix to be $0$, we require $b'_i=0$



        $displaystyletherefore b_i=frac{(a_i-a_2)b_1+(a_1-a_i)b_2}{a_1-a_2}=Big[frac{b_1-b_2}{a_1-a_2}Big]cdot a_i+Big(frac{a_1b_2-a_2b_1}{a_1-a_2}Big), forall i>2$




        Therefore, if:




        • There are two lines $L_i,L_j$ not parallel to each other $(a_ine a_j)$ intersecting at $(X,Y)$;


        • $displaystyle b_k=Big[frac{b_i-b_j}{a_i-a_j}Big]cdot a_k+Big(frac{a_ib_j-a_jb_i}{a_i-a_j}Big)=Xa_k+Y, forall kne i,j$; that is, $(X,Y)$ lies on all the remaining lines;



        Then, the straight lines $L_i:= a_ix+y=b_i,iin{1,2,...,n}$ intersect at the point $displaystyle(X,Y)=Big(frac{b_i-b_j}{a_i-a_j},frac{a_ib_j-a_jb_i}{a_i-a_j}Big)$ uniquely.







        share|cite|improve this answer











        $endgroup$



        The system is over-determined for $n>2$ but here is a general method. We can write the augmented matrix for the system $Abegin{bmatrix}x\yend{bmatrix}=B$ as under:



        $begin{bmatrix}a_1&1&Big|&b_1\a_2&1&Big|&b_2\vdots&vdots&Big|&vdots\a_n&1&Big|&b_nend{bmatrix}$



        For a unique solution to exist, we should have $2$ linearly-independent equations to solve for the $2$ unknowns. In other words, the rank of the coefficient and augmented matrices should be $2$. Recall that no more than $2$ vectors in $Bbb R^2$ can be linearly independent, so the rank of the coefficient matrix $A, text{rank}(A)le2$. For $text{rank}(A)=2$, we need to ensure at-least two $a_i$ are distinct. Say we have distinct $a_1ne0,a_2ne a_1$.



        The point of intersection of $a_1x+y=b_1,a_2x+y=b_2$ is given by $(X,Y)=displaystyleBig(frac{b_1-b_2}{a_1-a_2},frac{a_1b_2-a_2b_1}{a_1-a_2}Big)$.



        $displaystyle R_ito R_i-frac{a_i}{a_1}cdot R_1, i>1$



        $displaystyle R_jto R_j-frac{1-frac{a_i}{a_1}}{1-frac{a_2}{a_1}}cdot R_2, j>2$



        $simbegin{bmatrix}a_1&1&Big|&b_1\0&1-frac{a_2}{a_1}&Big|&b_2-frac{a_2}{a_1}cdot b_1\0&0&Big|&b'_3\vdots&vdots&Big|&vdots\0&0&Big|&b'_nend{bmatrix}$



        $displaystyle b'_i=b_i-frac{a_i}{a_1}cdot b_1-frac{1-frac{a_i}{a_1}}{1-frac{a_2}{a_1}}cdotBig[b_2-frac{a_2}{a_1}cdot b_1Big] forall i>2$



        For the rank of the augmented matrix to be $0$, we require $b'_i=0$



        $displaystyletherefore b_i=frac{(a_i-a_2)b_1+(a_1-a_i)b_2}{a_1-a_2}=Big[frac{b_1-b_2}{a_1-a_2}Big]cdot a_i+Big(frac{a_1b_2-a_2b_1}{a_1-a_2}Big), forall i>2$




        Therefore, if:




        • There are two lines $L_i,L_j$ not parallel to each other $(a_ine a_j)$ intersecting at $(X,Y)$;


        • $displaystyle b_k=Big[frac{b_i-b_j}{a_i-a_j}Big]cdot a_k+Big(frac{a_ib_j-a_jb_i}{a_i-a_j}Big)=Xa_k+Y, forall kne i,j$; that is, $(X,Y)$ lies on all the remaining lines;



        Then, the straight lines $L_i:= a_ix+y=b_i,iin{1,2,...,n}$ intersect at the point $displaystyle(X,Y)=Big(frac{b_i-b_j}{a_i-a_j},frac{a_ib_j-a_jb_i}{a_i-a_j}Big)$ uniquely.








        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 17 '18 at 13:02

























        answered Dec 17 '18 at 8:07









        Shubham JohriShubham Johri

        5,262718




        5,262718






























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            Musa

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            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
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