Matrix equation where some entries are solution to a polynome












1















Let $z$ be a solution to $z^2+z+1=0$. Find a solution to
begin{bmatrix}1&1&1&3\1&1&1&-1\1&z&z^2&0\1&z^2&z&0end
{bmatrix}begin{bmatrix}x_1\x_2\x_3\x_4end
{bmatrix} =begin{bmatrix}9\1\0\0end{bmatrix}




This an old exam question in my linear algebra course. I initially just tried solving the equation, but setting $z$ and $z^2$ in the matrix becomes messy quickly when I try to row reduce. The solution is elegant, but I don't understand it;




1) The equation $z^2+z+1=0$ ensures that every vector of the form
begin{bmatrix}x_1\x_2\x_3\x_4end
{bmatrix} = begin{bmatrix}a\a\a\bend{bmatrix}
solves the last two equations. It therefore suffices to find such $a$ and $b$ that the first two equations are also satisified. These read $3a+3b=9$ and $3a-b=1$. This forces $a=1$ and $b=2$ and
begin{bmatrix}x_1\x_2\x_3\x_4end
{bmatrix} = begin{bmatrix}1\1\1\2end{bmatrix} is indeed a solution




I don't understand 1), the rest follows naturally I guess. Could someone elaborate on why 1) is true? Where do the 3 $a$s and 1 $b$ come from?






linear-algebra matrix-equations







share|cite|improve this question



























    1















    Let $z$ be a solution to $z^2+z+1=0$. Find a solution to
    begin{bmatrix}1&1&1&3\1&1&1&-1\1&z&z^2&0\1&z^2&z&0end
    {bmatrix}begin{bmatrix}x_1\x_2\x_3\x_4end
    {bmatrix} =begin{bmatrix}9\1\0\0end{bmatrix}




    This an old exam question in my linear algebra course. I initially just tried solving the equation, but setting $z$ and $z^2$ in the matrix becomes messy quickly when I try to row reduce. The solution is elegant, but I don't understand it;




    1) The equation $z^2+z+1=0$ ensures that every vector of the form
    begin{bmatrix}x_1\x_2\x_3\x_4end
    {bmatrix} = begin{bmatrix}a\a\a\bend{bmatrix}
    solves the last two equations. It therefore suffices to find such $a$ and $b$ that the first two equations are also satisified. These read $3a+3b=9$ and $3a-b=1$. This forces $a=1$ and $b=2$ and
    begin{bmatrix}x_1\x_2\x_3\x_4end
    {bmatrix} = begin{bmatrix}1\1\1\2end{bmatrix}     is indeed a solution




    I don't understand 1), the rest follows naturally I guess. Could someone elaborate on why 1) is true? Where do the 3 $a$s and 1 $b$ come from?






    linear-algebra matrix-equations







    share|cite|improve this question

























      1












      1








      1








      Let $z$ be a solution to $z^2+z+1=0$. Find a solution to
      begin{bmatrix}1&1&1&3\1&1&1&-1\1&z&z^2&0\1&z^2&z&0end
      {bmatrix}begin{bmatrix}x_1\x_2\x_3\x_4end
      {bmatrix} =begin{bmatrix}9\1\0\0end{bmatrix}




      This an old exam question in my linear algebra course. I initially just tried solving the equation, but setting $z$ and $z^2$ in the matrix becomes messy quickly when I try to row reduce. The solution is elegant, but I don't understand it;




      1) The equation $z^2+z+1=0$ ensures that every vector of the form
      begin{bmatrix}x_1\x_2\x_3\x_4end
      {bmatrix} = begin{bmatrix}a\a\a\bend{bmatrix}
      solves the last two equations. It therefore suffices to find such $a$ and $b$ that the first two equations are also satisified. These read $3a+3b=9$ and $3a-b=1$. This forces $a=1$ and $b=2$ and
      begin{bmatrix}x_1\x_2\x_3\x_4end
      {bmatrix} = begin{bmatrix}1\1\1\2end{bmatrix}       is indeed a solution




      I don't understand 1), the rest follows naturally I guess. Could someone elaborate on why 1) is true? Where do the 3 $a$s and 1 $b$ come from?






      linear-algebra matrix-equations







      share|cite|improve this question














      Let $z$ be a solution to $z^2+z+1=0$. Find a solution to
      begin{bmatrix}1&1&1&3\1&1&1&-1\1&z&z^2&0\1&z^2&z&0end
      {bmatrix}begin{bmatrix}x_1\x_2\x_3\x_4end
      {bmatrix} =begin{bmatrix}9\1\0\0end{bmatrix}




      This an old exam question in my linear algebra course. I initially just tried solving the equation, but setting $z$ and $z^2$ in the matrix becomes messy quickly when I try to row reduce. The solution is elegant, but I don't understand it;




      1) The equation $z^2+z+1=0$ ensures that every vector of the form
      begin{bmatrix}x_1\x_2\x_3\x_4end
      {bmatrix} = begin{bmatrix}a\a\a\bend{bmatrix}
      solves the last two equations. It therefore suffices to find such $a$ and $b$ that the first two equations are also satisified. These read $3a+3b=9$ and $3a-b=1$. This forces $a=1$ and $b=2$ and
      begin{bmatrix}x_1\x_2\x_3\x_4end
      {bmatrix} = begin{bmatrix}1\1\1\2end{bmatrix}       is indeed a solution




      I don't understand 1), the rest follows naturally I guess. Could someone elaborate on why 1) is true? Where do the 3 $a$s and 1 $b$ come from?






      linear-algebra matrix-equations




      linear-algebra matrix-equations






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      asked Dec 2 at 9:43









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          $$begin{bmatrix}1&1&1&3\1&1&1&-1\1&z&z^2&0\1&z^2&z&0end
          {bmatrix}begin{bmatrix}a\a\a\bend
          {bmatrix} =begin{bmatrix}3a+3b\3a-b\a(1+z+z^2)\a(1+z+z^2)end{bmatrix}=begin{bmatrix}3a+3b\3a-b\0\0end{bmatrix}$$



          We just have to solve for $3a+3b=9$ and $3a-b=1$.



          We are already told that $1+z+z^2=0$, to use this information, we tend to think of how to sum up the three columns.






          share|cite|improve this answer























          • Thank you for the correction.
            – Siong Thye Goh
            Dec 2 at 10:07











          Your Answer





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          $$begin{bmatrix}1&1&1&3\1&1&1&-1\1&z&z^2&0\1&z^2&z&0end
          {bmatrix}begin{bmatrix}a\a\a\bend
          {bmatrix} =begin{bmatrix}3a+3b\3a-b\a(1+z+z^2)\a(1+z+z^2)end{bmatrix}=begin{bmatrix}3a+3b\3a-b\0\0end{bmatrix}$$



          We just have to solve for $3a+3b=9$ and $3a-b=1$.



          We are already told that $1+z+z^2=0$, to use this information, we tend to think of how to sum up the three columns.






          share|cite|improve this answer























          • Thank you for the correction.
            – Siong Thye Goh
            Dec 2 at 10:07
















          3














          $$begin{bmatrix}1&1&1&3\1&1&1&-1\1&z&z^2&0\1&z^2&z&0end
          {bmatrix}begin{bmatrix}a\a\a\bend
          {bmatrix} =begin{bmatrix}3a+3b\3a-b\a(1+z+z^2)\a(1+z+z^2)end{bmatrix}=begin{bmatrix}3a+3b\3a-b\0\0end{bmatrix}$$



          We just have to solve for $3a+3b=9$ and $3a-b=1$.



          We are already told that $1+z+z^2=0$, to use this information, we tend to think of how to sum up the three columns.






          share|cite|improve this answer























          • Thank you for the correction.
            – Siong Thye Goh
            Dec 2 at 10:07














          3












          3








          3






          $$begin{bmatrix}1&1&1&3\1&1&1&-1\1&z&z^2&0\1&z^2&z&0end
          {bmatrix}begin{bmatrix}a\a\a\bend
          {bmatrix} =begin{bmatrix}3a+3b\3a-b\a(1+z+z^2)\a(1+z+z^2)end{bmatrix}=begin{bmatrix}3a+3b\3a-b\0\0end{bmatrix}$$



          We just have to solve for $3a+3b=9$ and $3a-b=1$.



          We are already told that $1+z+z^2=0$, to use this information, we tend to think of how to sum up the three columns.






          share|cite|improve this answer














          $$begin{bmatrix}1&1&1&3\1&1&1&-1\1&z&z^2&0\1&z^2&z&0end
          {bmatrix}begin{bmatrix}a\a\a\bend
          {bmatrix} =begin{bmatrix}3a+3b\3a-b\a(1+z+z^2)\a(1+z+z^2)end{bmatrix}=begin{bmatrix}3a+3b\3a-b\0\0end{bmatrix}$$



          We just have to solve for $3a+3b=9$ and $3a-b=1$.



          We are already told that $1+z+z^2=0$, to use this information, we tend to think of how to sum up the three columns.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 2 at 10:07

























          answered Dec 2 at 10:04









          Siong Thye Goh

          99k1464117




          99k1464117












          • Thank you for the correction.
            – Siong Thye Goh
            Dec 2 at 10:07


















          • Thank you for the correction.
            – Siong Thye Goh
            Dec 2 at 10:07
















          Thank you for the correction.
          – Siong Thye Goh
          Dec 2 at 10:07




          Thank you for the correction.
          – Siong Thye Goh
          Dec 2 at 10:07


















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