Inconsistent solutions and their image/kernel

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We have an inconsistent system given by Ax = b with matrix A is a 3x3 matrix.



Which of the following is true?
(a) rref(A) has at least one row full of zeroes
(b) im(A) is not R^3
(c) ker(A) does not equal to 0



I can now conclude that rref(A) has at least one row full of zeroes by identification of inconsistent system.



Then by extension, b is true since the column space will be missing one leading one.



I'm uncertain about part C




linear-algebra linear-transformations matrix-rank




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    We have an inconsistent system given by Ax = b with matrix A is a 3x3 matrix.



    Which of the following is true?
    (a) rref(A) has at least one row full of zeroes
    (b) im(A) is not R^3
    (c) ker(A) does not equal to 0



    I can now conclude that rref(A) has at least one row full of zeroes by identification of inconsistent system.



    Then by extension, b is true since the column space will be missing one leading one.



    I'm uncertain about part C




    linear-algebra linear-transformations matrix-rank




    share|cite|improve this question





















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      We have an inconsistent system given by Ax = b with matrix A is a 3x3 matrix.



      Which of the following is true?
      (a) rref(A) has at least one row full of zeroes
      (b) im(A) is not R^3
      (c) ker(A) does not equal to 0



      I can now conclude that rref(A) has at least one row full of zeroes by identification of inconsistent system.



      Then by extension, b is true since the column space will be missing one leading one.



      I'm uncertain about part C




      linear-algebra linear-transformations matrix-rank




      share|cite|improve this question











      We have an inconsistent system given by Ax = b with matrix A is a 3x3 matrix.



      Which of the following is true?
      (a) rref(A) has at least one row full of zeroes
      (b) im(A) is not R^3
      (c) ker(A) does not equal to 0



      I can now conclude that rref(A) has at least one row full of zeroes by identification of inconsistent system.



      Then by extension, b is true since the column space will be missing one leading one.



      I'm uncertain about part C





      linear-algebra linear-transformations matrix-rank





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      share|cite|improve this question




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      asked Aug 3 at 14:28









      seekingalpha23

      156




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          The fact that $Ax= b$ does not have a solution, means that $b$ does not lie in the range of $A$. Consequently, $mboxim A$ cannot equal $mathbb R^3$ as it does not contain $b$.



          Since $A$ is not a surjective map, then by the rank-nullity theorem, $A$ is not an injective map, because $A : mathbb R^3 to mboxim A$ mut satisfy $dim ker A + dim mboxim A = 3$, but $dim mboxim A < 3$ so $dim ker A > 1$, which means that the kernel is non-trivial.



          Finally, the row rank of $A$, equals the column rank of $A$, equals the dimension of the image of $A$, which is not equal to $3$. Since there are three rows, and the image space has smaller dimension, row reduction will lead to a row of zeroes.



          So the answer to all three is yes.






          share|cite|improve this answer





















          • Thank you for this response. I can see why those solutions are all related. I am also struggling with this problem as well. Can you please take a look if it does not inconvenience you sir. Thank You math.stackexchange.com/questions/2871079/…
            – seekingalpha23
            2 days ago










          • I will look at the question. Also, thank you.
            – Ð°ÑÑ‚он вілла олоф мэллбэрг
            2 days ago










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          accepted










          The fact that $Ax= b$ does not have a solution, means that $b$ does not lie in the range of $A$. Consequently, $mboxim A$ cannot equal $mathbb R^3$ as it does not contain $b$.



          Since $A$ is not a surjective map, then by the rank-nullity theorem, $A$ is not an injective map, because $A : mathbb R^3 to mboxim A$ mut satisfy $dim ker A + dim mboxim A = 3$, but $dim mboxim A < 3$ so $dim ker A > 1$, which means that the kernel is non-trivial.



          Finally, the row rank of $A$, equals the column rank of $A$, equals the dimension of the image of $A$, which is not equal to $3$. Since there are three rows, and the image space has smaller dimension, row reduction will lead to a row of zeroes.



          So the answer to all three is yes.






          share|cite|improve this answer





















          • Thank you for this response. I can see why those solutions are all related. I am also struggling with this problem as well. Can you please take a look if it does not inconvenience you sir. Thank You math.stackexchange.com/questions/2871079/…
            – seekingalpha23
            2 days ago










          • I will look at the question. Also, thank you.
            – Ð°ÑÑ‚он вілла олоф мэллбэрг
            2 days ago














          up vote
          1
          down vote



          accepted










          The fact that $Ax= b$ does not have a solution, means that $b$ does not lie in the range of $A$. Consequently, $mboxim A$ cannot equal $mathbb R^3$ as it does not contain $b$.



          Since $A$ is not a surjective map, then by the rank-nullity theorem, $A$ is not an injective map, because $A : mathbb R^3 to mboxim A$ mut satisfy $dim ker A + dim mboxim A = 3$, but $dim mboxim A < 3$ so $dim ker A > 1$, which means that the kernel is non-trivial.



          Finally, the row rank of $A$, equals the column rank of $A$, equals the dimension of the image of $A$, which is not equal to $3$. Since there are three rows, and the image space has smaller dimension, row reduction will lead to a row of zeroes.



          So the answer to all three is yes.






          share|cite|improve this answer





















          • Thank you for this response. I can see why those solutions are all related. I am also struggling with this problem as well. Can you please take a look if it does not inconvenience you sir. Thank You math.stackexchange.com/questions/2871079/…
            – seekingalpha23
            2 days ago










          • I will look at the question. Also, thank you.
            – Ð°ÑÑ‚он вілла олоф мэллбэрг
            2 days ago












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          The fact that $Ax= b$ does not have a solution, means that $b$ does not lie in the range of $A$. Consequently, $mboxim A$ cannot equal $mathbb R^3$ as it does not contain $b$.



          Since $A$ is not a surjective map, then by the rank-nullity theorem, $A$ is not an injective map, because $A : mathbb R^3 to mboxim A$ mut satisfy $dim ker A + dim mboxim A = 3$, but $dim mboxim A < 3$ so $dim ker A > 1$, which means that the kernel is non-trivial.



          Finally, the row rank of $A$, equals the column rank of $A$, equals the dimension of the image of $A$, which is not equal to $3$. Since there are three rows, and the image space has smaller dimension, row reduction will lead to a row of zeroes.



          So the answer to all three is yes.






          share|cite|improve this answer













          The fact that $Ax= b$ does not have a solution, means that $b$ does not lie in the range of $A$. Consequently, $mboxim A$ cannot equal $mathbb R^3$ as it does not contain $b$.



          Since $A$ is not a surjective map, then by the rank-nullity theorem, $A$ is not an injective map, because $A : mathbb R^3 to mboxim A$ mut satisfy $dim ker A + dim mboxim A = 3$, but $dim mboxim A < 3$ so $dim ker A > 1$, which means that the kernel is non-trivial.



          Finally, the row rank of $A$, equals the column rank of $A$, equals the dimension of the image of $A$, which is not equal to $3$. Since there are three rows, and the image space has smaller dimension, row reduction will lead to a row of zeroes.



          So the answer to all three is yes.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Aug 3 at 14:40









          астон вілла олоф мэллбэрг

          31.5k22363




          31.5k22363











          • Thank you for this response. I can see why those solutions are all related. I am also struggling with this problem as well. Can you please take a look if it does not inconvenience you sir. Thank You math.stackexchange.com/questions/2871079/…
            – seekingalpha23
            2 days ago










          • I will look at the question. Also, thank you.
            – Ð°ÑÑ‚он вілла олоф мэллбэрг
            2 days ago
















          • Thank you for this response. I can see why those solutions are all related. I am also struggling with this problem as well. Can you please take a look if it does not inconvenience you sir. Thank You math.stackexchange.com/questions/2871079/…
            – seekingalpha23
            2 days ago










          • I will look at the question. Also, thank you.
            – Ð°ÑÑ‚он вілла олоф мэллбэрг
            2 days ago















          Thank you for this response. I can see why those solutions are all related. I am also struggling with this problem as well. Can you please take a look if it does not inconvenience you sir. Thank You math.stackexchange.com/questions/2871079/…
          – seekingalpha23
          2 days ago




          Thank you for this response. I can see why those solutions are all related. I am also struggling with this problem as well. Can you please take a look if it does not inconvenience you sir. Thank You math.stackexchange.com/questions/2871079/…
          – seekingalpha23
          2 days ago












          I will look at the question. Also, thank you.
          – Ð°ÑÑ‚он вілла олоф мэллбэрг
          2 days ago




          I will look at the question. Also, thank you.
          – Ð°ÑÑ‚он вілла олоф мэллбэрг
          2 days ago












           

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