Linear Algerba - translation of space $V$ by a six-dimensional vector $b$











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The task I have has 2 parts, I did the first one but now I'm struggling with the second one.



In the first part, I was supposed to find all $2times3$ matrices $A$ that satisfy the equation: $Atimes[1;1;1]^T=[0;0]^T$. I did that, the $A$ matrices are in the form:
$$begin{bmatrix}x&y&-x-y\z&w&-z-wend{bmatrix}$$



Then I was supposed to find the basis and dimension for the linear space $V$ that these matrices create, which I also did - dimension is $4$ and the basis is:



$${begin{bmatrix}1&0&-1\0&0&0end{bmatrix},begin{bmatrix}-1&1&0\0&0&0end{bmatrix},begin{bmatrix}0&0&0\1&-1&0end{bmatrix},begin{bmatrix}0&0&0\-1&0&1end{bmatrix}}$$



Now in the second part, I'm looking for all $B$ matrices that satisfy this equation: $Btimes[1;1;1]^T=[6;6]^T$ and I'm supposed to present the set of all the $B's$ as as a translation of the space $V$ by a six-dimensional vector $b$.



How would this vector b look?






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    The task I have has 2 parts, I did the first one but now I'm struggling with the second one.



    In the first part, I was supposed to find all $2times3$ matrices $A$ that satisfy the equation: $Atimes[1;1;1]^T=[0;0]^T$. I did that, the $A$ matrices are in the form:
    $$begin{bmatrix}x&y&-x-y\z&w&-z-wend{bmatrix}$$



    Then I was supposed to find the basis and dimension for the linear space $V$ that these matrices create, which I also did - dimension is $4$ and the basis is:



    $${begin{bmatrix}1&0&-1\0&0&0end{bmatrix},begin{bmatrix}-1&1&0\0&0&0end{bmatrix},begin{bmatrix}0&0&0\1&-1&0end{bmatrix},begin{bmatrix}0&0&0\-1&0&1end{bmatrix}}$$



    Now in the second part, I'm looking for all $B$ matrices that satisfy this equation: $Btimes[1;1;1]^T=[6;6]^T$ and I'm supposed to present the set of all the $B's$ as as a translation of the space $V$ by a six-dimensional vector $b$.



    How would this vector b look?






    matrices







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      The task I have has 2 parts, I did the first one but now I'm struggling with the second one.



      In the first part, I was supposed to find all $2times3$ matrices $A$ that satisfy the equation: $Atimes[1;1;1]^T=[0;0]^T$. I did that, the $A$ matrices are in the form:
      $$begin{bmatrix}x&y&-x-y\z&w&-z-wend{bmatrix}$$



      Then I was supposed to find the basis and dimension for the linear space $V$ that these matrices create, which I also did - dimension is $4$ and the basis is:



      $${begin{bmatrix}1&0&-1\0&0&0end{bmatrix},begin{bmatrix}-1&1&0\0&0&0end{bmatrix},begin{bmatrix}0&0&0\1&-1&0end{bmatrix},begin{bmatrix}0&0&0\-1&0&1end{bmatrix}}$$



      Now in the second part, I'm looking for all $B$ matrices that satisfy this equation: $Btimes[1;1;1]^T=[6;6]^T$ and I'm supposed to present the set of all the $B's$ as as a translation of the space $V$ by a six-dimensional vector $b$.



      How would this vector b look?






      matrices







      share|cite|improve this question













      The task I have has 2 parts, I did the first one but now I'm struggling with the second one.



      In the first part, I was supposed to find all $2times3$ matrices $A$ that satisfy the equation: $Atimes[1;1;1]^T=[0;0]^T$. I did that, the $A$ matrices are in the form:
      $$begin{bmatrix}x&y&-x-y\z&w&-z-wend{bmatrix}$$



      Then I was supposed to find the basis and dimension for the linear space $V$ that these matrices create, which I also did - dimension is $4$ and the basis is:



      $${begin{bmatrix}1&0&-1\0&0&0end{bmatrix},begin{bmatrix}-1&1&0\0&0&0end{bmatrix},begin{bmatrix}0&0&0\1&-1&0end{bmatrix},begin{bmatrix}0&0&0\-1&0&1end{bmatrix}}$$



      Now in the second part, I'm looking for all $B$ matrices that satisfy this equation: $Btimes[1;1;1]^T=[6;6]^T$ and I'm supposed to present the set of all the $B's$ as as a translation of the space $V$ by a six-dimensional vector $b$.



      How would this vector b look?






      matrices




      matrices






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      share|cite|improve this question











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      share|cite|improve this question










      asked Nov 24 at 15:52









      agromek

      345




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          1 Answer
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          $b$ would look like $begin{bmatrix}b_1&b_2&b_3\b_4&b_5&b_6end{bmatrix}$. Or if you like $b=(b_1,dots,b_6)$.



          You need to find the $b_i$. $b$ should be a particular solution of the second equation.






          share|cite|improve this answer























          • b would look something like this (I did the same thing as in the first part): $$begin{bmatrix}x&y&6-x-y\z&w&6-z-wend{bmatrix}$$ so now to provide the answer do I just find one particular b, for example: $$begin{bmatrix}1&2&3\5&0&1end{bmatrix}$$ and present the set of $B's$ as the translation of $V$ by vector $b$ so: $$alphatimesbegin{bmatrix}1&0&-1\0&0&0end{bmatrix}+betatimesbegin{bmatrix}-1&1&0\0&0&0end{bmatrix}+gammatimesbegin{bmatrix}0&0&0\1&-1&0end{bmatrix}+deltatimesbegin{bmatrix}0&0&0\-1&0&1end{bmatrix}+begin{bmatrix}1&2&3\5&0&1end{bmatrix}?$$
            – agromek
            Nov 24 at 17:44












          • @agromek I believe so. That is, yes.
            – Chris Custer
            Nov 24 at 17:49










          • Ok, thank you very much:)
            – agromek
            Nov 24 at 17:50










          • You're welcome. Adding any element of the kernel to a particular solution gives another solution.
            – Chris Custer
            Nov 24 at 17:53











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          1 Answer
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          oldest

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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

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          up vote
          0
          down vote



          accepted










          $b$ would look like $begin{bmatrix}b_1&b_2&b_3\b_4&b_5&b_6end{bmatrix}$. Or if you like $b=(b_1,dots,b_6)$.



          You need to find the $b_i$. $b$ should be a particular solution of the second equation.






          share|cite|improve this answer























          • b would look something like this (I did the same thing as in the first part): $$begin{bmatrix}x&y&6-x-y\z&w&6-z-wend{bmatrix}$$ so now to provide the answer do I just find one particular b, for example: $$begin{bmatrix}1&2&3\5&0&1end{bmatrix}$$ and present the set of $B's$ as the translation of $V$ by vector $b$ so: $$alphatimesbegin{bmatrix}1&0&-1\0&0&0end{bmatrix}+betatimesbegin{bmatrix}-1&1&0\0&0&0end{bmatrix}+gammatimesbegin{bmatrix}0&0&0\1&-1&0end{bmatrix}+deltatimesbegin{bmatrix}0&0&0\-1&0&1end{bmatrix}+begin{bmatrix}1&2&3\5&0&1end{bmatrix}?$$
            – agromek
            Nov 24 at 17:44












          • @agromek I believe so. That is, yes.
            – Chris Custer
            Nov 24 at 17:49










          • Ok, thank you very much:)
            – agromek
            Nov 24 at 17:50










          • You're welcome. Adding any element of the kernel to a particular solution gives another solution.
            – Chris Custer
            Nov 24 at 17:53















          up vote
          0
          down vote



          accepted










          $b$ would look like $begin{bmatrix}b_1&b_2&b_3\b_4&b_5&b_6end{bmatrix}$. Or if you like $b=(b_1,dots,b_6)$.



          You need to find the $b_i$. $b$ should be a particular solution of the second equation.






          share|cite|improve this answer























          • b would look something like this (I did the same thing as in the first part): $$begin{bmatrix}x&y&6-x-y\z&w&6-z-wend{bmatrix}$$ so now to provide the answer do I just find one particular b, for example: $$begin{bmatrix}1&2&3\5&0&1end{bmatrix}$$ and present the set of $B's$ as the translation of $V$ by vector $b$ so: $$alphatimesbegin{bmatrix}1&0&-1\0&0&0end{bmatrix}+betatimesbegin{bmatrix}-1&1&0\0&0&0end{bmatrix}+gammatimesbegin{bmatrix}0&0&0\1&-1&0end{bmatrix}+deltatimesbegin{bmatrix}0&0&0\-1&0&1end{bmatrix}+begin{bmatrix}1&2&3\5&0&1end{bmatrix}?$$
            – agromek
            Nov 24 at 17:44












          • @agromek I believe so. That is, yes.
            – Chris Custer
            Nov 24 at 17:49










          • Ok, thank you very much:)
            – agromek
            Nov 24 at 17:50










          • You're welcome. Adding any element of the kernel to a particular solution gives another solution.
            – Chris Custer
            Nov 24 at 17:53













          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          $b$ would look like $begin{bmatrix}b_1&b_2&b_3\b_4&b_5&b_6end{bmatrix}$. Or if you like $b=(b_1,dots,b_6)$.



          You need to find the $b_i$. $b$ should be a particular solution of the second equation.






          share|cite|improve this answer














          $b$ would look like $begin{bmatrix}b_1&b_2&b_3\b_4&b_5&b_6end{bmatrix}$. Or if you like $b=(b_1,dots,b_6)$.



          You need to find the $b_i$. $b$ should be a particular solution of the second equation.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 24 at 17:43

























          answered Nov 24 at 17:28









          Chris Custer

          10.4k3724




          10.4k3724












          • b would look something like this (I did the same thing as in the first part): $$begin{bmatrix}x&y&6-x-y\z&w&6-z-wend{bmatrix}$$ so now to provide the answer do I just find one particular b, for example: $$begin{bmatrix}1&2&3\5&0&1end{bmatrix}$$ and present the set of $B's$ as the translation of $V$ by vector $b$ so: $$alphatimesbegin{bmatrix}1&0&-1\0&0&0end{bmatrix}+betatimesbegin{bmatrix}-1&1&0\0&0&0end{bmatrix}+gammatimesbegin{bmatrix}0&0&0\1&-1&0end{bmatrix}+deltatimesbegin{bmatrix}0&0&0\-1&0&1end{bmatrix}+begin{bmatrix}1&2&3\5&0&1end{bmatrix}?$$
            – agromek
            Nov 24 at 17:44












          • @agromek I believe so. That is, yes.
            – Chris Custer
            Nov 24 at 17:49










          • Ok, thank you very much:)
            – agromek
            Nov 24 at 17:50










          • You're welcome. Adding any element of the kernel to a particular solution gives another solution.
            – Chris Custer
            Nov 24 at 17:53


















          • b would look something like this (I did the same thing as in the first part): $$begin{bmatrix}x&y&6-x-y\z&w&6-z-wend{bmatrix}$$ so now to provide the answer do I just find one particular b, for example: $$begin{bmatrix}1&2&3\5&0&1end{bmatrix}$$ and present the set of $B's$ as the translation of $V$ by vector $b$ so: $$alphatimesbegin{bmatrix}1&0&-1\0&0&0end{bmatrix}+betatimesbegin{bmatrix}-1&1&0\0&0&0end{bmatrix}+gammatimesbegin{bmatrix}0&0&0\1&-1&0end{bmatrix}+deltatimesbegin{bmatrix}0&0&0\-1&0&1end{bmatrix}+begin{bmatrix}1&2&3\5&0&1end{bmatrix}?$$
            – agromek
            Nov 24 at 17:44












          • @agromek I believe so. That is, yes.
            – Chris Custer
            Nov 24 at 17:49










          • Ok, thank you very much:)
            – agromek
            Nov 24 at 17:50










          • You're welcome. Adding any element of the kernel to a particular solution gives another solution.
            – Chris Custer
            Nov 24 at 17:53
















          b would look something like this (I did the same thing as in the first part): $$begin{bmatrix}x&y&6-x-y\z&w&6-z-wend{bmatrix}$$ so now to provide the answer do I just find one particular b, for example: $$begin{bmatrix}1&2&3\5&0&1end{bmatrix}$$ and present the set of $B's$ as the translation of $V$ by vector $b$ so: $$alphatimesbegin{bmatrix}1&0&-1\0&0&0end{bmatrix}+betatimesbegin{bmatrix}-1&1&0\0&0&0end{bmatrix}+gammatimesbegin{bmatrix}0&0&0\1&-1&0end{bmatrix}+deltatimesbegin{bmatrix}0&0&0\-1&0&1end{bmatrix}+begin{bmatrix}1&2&3\5&0&1end{bmatrix}?$$
          – agromek
          Nov 24 at 17:44






          b would look something like this (I did the same thing as in the first part): $$begin{bmatrix}x&y&6-x-y\z&w&6-z-wend{bmatrix}$$ so now to provide the answer do I just find one particular b, for example: $$begin{bmatrix}1&2&3\5&0&1end{bmatrix}$$ and present the set of $B's$ as the translation of $V$ by vector $b$ so: $$alphatimesbegin{bmatrix}1&0&-1\0&0&0end{bmatrix}+betatimesbegin{bmatrix}-1&1&0\0&0&0end{bmatrix}+gammatimesbegin{bmatrix}0&0&0\1&-1&0end{bmatrix}+deltatimesbegin{bmatrix}0&0&0\-1&0&1end{bmatrix}+begin{bmatrix}1&2&3\5&0&1end{bmatrix}?$$
          – agromek
          Nov 24 at 17:44














          @agromek I believe so. That is, yes.
          – Chris Custer
          Nov 24 at 17:49




          @agromek I believe so. That is, yes.
          – Chris Custer
          Nov 24 at 17:49












          Ok, thank you very much:)
          – agromek
          Nov 24 at 17:50




          Ok, thank you very much:)
          – agromek
          Nov 24 at 17:50












          You're welcome. Adding any element of the kernel to a particular solution gives another solution.
          – Chris Custer
          Nov 24 at 17:53




          You're welcome. Adding any element of the kernel to a particular solution gives another solution.
          – Chris Custer
          Nov 24 at 17:53


















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