Reduced row echelon form. How do get to this matrix?

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I am reading this text and I'm stuck on how they get to the reduced row echelon form of this matrix:



enter image description here



I've tried a ton of combinations but I still can't get there. Can someone show me the way? Here's one that I have but then I'm stuck:



$beginmatrix
\ 1 & 2 & 5
\ 0 & -5 & -5
endmatrix$



But then how do I get rid of the 2 in the second column of the first row to 0?




linear-algebra




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  • Start by dividing two two by $-5$...
    – JavaMan
    Jul 31 at 19:41














up vote
1
down vote

favorite












I am reading this text and I'm stuck on how they get to the reduced row echelon form of this matrix:



enter image description here



I've tried a ton of combinations but I still can't get there. Can someone show me the way? Here's one that I have but then I'm stuck:



$beginmatrix
\ 1 & 2 & 5
\ 0 & -5 & -5
endmatrix$



But then how do I get rid of the 2 in the second column of the first row to 0?




linear-algebra




share|cite|improve this question



















  • Start by dividing two two by $-5$...
    – JavaMan
    Jul 31 at 19:41












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I am reading this text and I'm stuck on how they get to the reduced row echelon form of this matrix:



enter image description here



I've tried a ton of combinations but I still can't get there. Can someone show me the way? Here's one that I have but then I'm stuck:



$beginmatrix
\ 1 & 2 & 5
\ 0 & -5 & -5
endmatrix$



But then how do I get rid of the 2 in the second column of the first row to 0?




linear-algebra




share|cite|improve this question











I am reading this text and I'm stuck on how they get to the reduced row echelon form of this matrix:



enter image description here



I've tried a ton of combinations but I still can't get there. Can someone show me the way? Here's one that I have but then I'm stuck:



$beginmatrix
\ 1 & 2 & 5
\ 0 & -5 & -5
endmatrix$



But then how do I get rid of the 2 in the second column of the first row to 0?





linear-algebra





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 31 at 19:40









Jwan622

1,60111224




1,60111224











  • Start by dividing two two by $-5$...
    – JavaMan
    Jul 31 at 19:41
















  • Start by dividing two two by $-5$...
    – JavaMan
    Jul 31 at 19:41















Start by dividing two two by $-5$...
– JavaMan
Jul 31 at 19:41




Start by dividing two two by $-5$...
– JavaMan
Jul 31 at 19:41










2 Answers
2






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oldest

votes

















up vote
1
down vote



accepted










We have
$$beginbmatrix
1 & 2 &5 \
2 & -1 & 5
endbmatrix$$
Denoting row $i$ as $r_i$ and assume that the left arrow $A leftarrow B$ means that we are putting quantity $B$ in $A$. Moreover,
we know that we can do linear operations on the rows to reach the row-echelon form, i.e.



$r_2 leftarrow r_2 - 2r_1$



$$beginbmatrix
1 & 2 &5 \
0 & -5 & -5
endbmatrix$$



$r_1 leftarrow r_1 + frac25r_2$



$$beginbmatrix
1 & 0 &3 \
0 & -5 & -5
endbmatrix$$
$r_2 leftarrow -frac15r_2$
$$beginbmatrix
1 & 0 &3 \
0 & 1 & 1
endbmatrix$$






share|cite|improve this answer




























    up vote
    1
    down vote













    Proceed systematically instead of trying to guess at combinations that might work. You’ve gotten a pivot in the first row and everything else in the pivot column is zero, so you move on to the second row. Find the leftmost nonzero element and divide the row by that number to make its first nonzero element a $1$. Now it should be pretty obvious what multiple of the second row to subtract from the first to zero out the element above this new pivot. If there were more to the matrix, you’d also zero out the rest of the second column, then move on to the next nonzero row. Lather, rinse, repeat.






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      2 Answers
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      2 Answers
      2






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      active

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      up vote
      1
      down vote



      accepted










      We have
      $$beginbmatrix
      1 & 2 &5 \
      2 & -1 & 5
      endbmatrix$$
      Denoting row $i$ as $r_i$ and assume that the left arrow $A leftarrow B$ means that we are putting quantity $B$ in $A$. Moreover,
      we know that we can do linear operations on the rows to reach the row-echelon form, i.e.



      $r_2 leftarrow r_2 - 2r_1$



      $$beginbmatrix
      1 & 2 &5 \
      0 & -5 & -5
      endbmatrix$$



      $r_1 leftarrow r_1 + frac25r_2$



      $$beginbmatrix
      1 & 0 &3 \
      0 & -5 & -5
      endbmatrix$$
      $r_2 leftarrow -frac15r_2$
      $$beginbmatrix
      1 & 0 &3 \
      0 & 1 & 1
      endbmatrix$$






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted










        We have
        $$beginbmatrix
        1 & 2 &5 \
        2 & -1 & 5
        endbmatrix$$
        Denoting row $i$ as $r_i$ and assume that the left arrow $A leftarrow B$ means that we are putting quantity $B$ in $A$. Moreover,
        we know that we can do linear operations on the rows to reach the row-echelon form, i.e.



        $r_2 leftarrow r_2 - 2r_1$



        $$beginbmatrix
        1 & 2 &5 \
        0 & -5 & -5
        endbmatrix$$



        $r_1 leftarrow r_1 + frac25r_2$



        $$beginbmatrix
        1 & 0 &3 \
        0 & -5 & -5
        endbmatrix$$
        $r_2 leftarrow -frac15r_2$
        $$beginbmatrix
        1 & 0 &3 \
        0 & 1 & 1
        endbmatrix$$






        share|cite|improve this answer























          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          We have
          $$beginbmatrix
          1 & 2 &5 \
          2 & -1 & 5
          endbmatrix$$
          Denoting row $i$ as $r_i$ and assume that the left arrow $A leftarrow B$ means that we are putting quantity $B$ in $A$. Moreover,
          we know that we can do linear operations on the rows to reach the row-echelon form, i.e.



          $r_2 leftarrow r_2 - 2r_1$



          $$beginbmatrix
          1 & 2 &5 \
          0 & -5 & -5
          endbmatrix$$



          $r_1 leftarrow r_1 + frac25r_2$



          $$beginbmatrix
          1 & 0 &3 \
          0 & -5 & -5
          endbmatrix$$
          $r_2 leftarrow -frac15r_2$
          $$beginbmatrix
          1 & 0 &3 \
          0 & 1 & 1
          endbmatrix$$






          share|cite|improve this answer













          We have
          $$beginbmatrix
          1 & 2 &5 \
          2 & -1 & 5
          endbmatrix$$
          Denoting row $i$ as $r_i$ and assume that the left arrow $A leftarrow B$ means that we are putting quantity $B$ in $A$. Moreover,
          we know that we can do linear operations on the rows to reach the row-echelon form, i.e.



          $r_2 leftarrow r_2 - 2r_1$



          $$beginbmatrix
          1 & 2 &5 \
          0 & -5 & -5
          endbmatrix$$



          $r_1 leftarrow r_1 + frac25r_2$



          $$beginbmatrix
          1 & 0 &3 \
          0 & -5 & -5
          endbmatrix$$
          $r_2 leftarrow -frac15r_2$
          $$beginbmatrix
          1 & 0 &3 \
          0 & 1 & 1
          endbmatrix$$







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 31 at 19:51









          Ahmad Bazzi

          2,170417




          2,170417




















              up vote
              1
              down vote













              Proceed systematically instead of trying to guess at combinations that might work. You’ve gotten a pivot in the first row and everything else in the pivot column is zero, so you move on to the second row. Find the leftmost nonzero element and divide the row by that number to make its first nonzero element a $1$. Now it should be pretty obvious what multiple of the second row to subtract from the first to zero out the element above this new pivot. If there were more to the matrix, you’d also zero out the rest of the second column, then move on to the next nonzero row. Lather, rinse, repeat.






              share|cite|improve this answer

























                up vote
                1
                down vote













                Proceed systematically instead of trying to guess at combinations that might work. You’ve gotten a pivot in the first row and everything else in the pivot column is zero, so you move on to the second row. Find the leftmost nonzero element and divide the row by that number to make its first nonzero element a $1$. Now it should be pretty obvious what multiple of the second row to subtract from the first to zero out the element above this new pivot. If there were more to the matrix, you’d also zero out the rest of the second column, then move on to the next nonzero row. Lather, rinse, repeat.






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Proceed systematically instead of trying to guess at combinations that might work. You’ve gotten a pivot in the first row and everything else in the pivot column is zero, so you move on to the second row. Find the leftmost nonzero element and divide the row by that number to make its first nonzero element a $1$. Now it should be pretty obvious what multiple of the second row to subtract from the first to zero out the element above this new pivot. If there were more to the matrix, you’d also zero out the rest of the second column, then move on to the next nonzero row. Lather, rinse, repeat.






                  share|cite|improve this answer













                  Proceed systematically instead of trying to guess at combinations that might work. You’ve gotten a pivot in the first row and everything else in the pivot column is zero, so you move on to the second row. Find the leftmost nonzero element and divide the row by that number to make its first nonzero element a $1$. Now it should be pretty obvious what multiple of the second row to subtract from the first to zero out the element above this new pivot. If there were more to the matrix, you’d also zero out the rest of the second column, then move on to the next nonzero row. Lather, rinse, repeat.







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Jul 31 at 19:48









                  amd

                  25.7k2943




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