If a matrix and its determinant given and another matrix also given how to obtain the second matrix...












0












$begingroup$


If the matrix $$ A =
pmatrix{row1 \ row2\row3} and left|begin{array}[ccc]\ A end{array}right|
=10$$



And matrix



$$ B =
pmatrix{2row1+row2-row3 \ 2row3\5row2}
$$
then find
$$
\left|begin{array}[ccc]\ B end{array}right|
= ?$$



iam stuck at this, i know that subtracting a multiple of one row from another row does not change determinate (A), and also if we do permutation of rows 1 time the sign will be negative, but i do not know if this information is useful here or not, i need help with this one and



after reading StackTD hints B, will be:
$$ B =
pmatrix{2row1\2row3\5row2}
$$
and i know the determinant of B will in negative sign because of rows swap but i couldn't obtain B determinant






linear-algebra matrices







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    If the matrix $$ A =
    pmatrix{row1 \ row2\row3} and left|begin{array}[ccc]\ A end{array}right|
    =10$$



    And matrix



    $$ B =
    pmatrix{2row1+row2-row3 \ 2row3\5row2}
    $$
    then find
    $$
    \left|begin{array}[ccc]\ B end{array}right|
    = ?$$



    iam stuck at this, i know that subtracting a multiple of one row from another row does not change determinate (A), and also if we do permutation of rows 1 time the sign will be negative, but i do not know if this information is useful here or not, i need help with this one and



    after reading StackTD hints B, will be:
    $$ B =
    pmatrix{2row1\2row3\5row2}
    $$
    and i know the determinant of B will in negative sign because of rows swap but i couldn't obtain B determinant






    linear-algebra matrices







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      If the matrix $$ A =
      pmatrix{row1 \ row2\row3} and left|begin{array}[ccc]\ A end{array}right|
      =10$$



      And matrix



      $$ B =
      pmatrix{2row1+row2-row3 \ 2row3\5row2}
      $$
      then find
      $$
      \left|begin{array}[ccc]\ B end{array}right|
      = ?$$



      iam stuck at this, i know that subtracting a multiple of one row from another row does not change determinate (A), and also if we do permutation of rows 1 time the sign will be negative, but i do not know if this information is useful here or not, i need help with this one and



      after reading StackTD hints B, will be:
      $$ B =
      pmatrix{2row1\2row3\5row2}
      $$
      and i know the determinant of B will in negative sign because of rows swap but i couldn't obtain B determinant






      linear-algebra matrices







      share|cite|improve this question











      $endgroup$




      If the matrix $$ A =
      pmatrix{row1 \ row2\row3} and left|begin{array}[ccc]\ A end{array}right|
      =10$$



      And matrix



      $$ B =
      pmatrix{2row1+row2-row3 \ 2row3\5row2}
      $$
      then find
      $$
      \left|begin{array}[ccc]\ B end{array}right|
      = ?$$



      iam stuck at this, i know that subtracting a multiple of one row from another row does not change determinate (A), and also if we do permutation of rows 1 time the sign will be negative, but i do not know if this information is useful here or not, i need help with this one and



      after reading StackTD hints B, will be:
      $$ B =
      pmatrix{2row1\2row3\5row2}
      $$
      and i know the determinant of B will in negative sign because of rows swap but i couldn't obtain B determinant






      linear-algebra matrices




      linear-algebra matrices






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 12 '18 at 12:58









      Ethan Bolker

      43.6k551116




      43.6k551116










      asked Dec 12 '18 at 12:18









      The BeardThe Beard

      54




      54






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Use properties of determinants:




          • the determinant is linear in each row/column;

          • a determinant with two identical rows is $0$;

          • swapping two rows changes the sign of the determinant.


          Now start with linearity and follow up (I write $A_i$ for the $i$th row of the original matrix $A$):
          $$begin{vmatrix}
          2A_1+A_2-A_3 \
          2A_3 \
          5A_2
          end{vmatrix} = begin{vmatrix}
          2A_1 \
          2A_3 \
          5A_2
          end{vmatrix}+begin{vmatrix}
          A_2 \
          2A_3 \
          5A_2
          end{vmatrix}+begin{vmatrix}
          -A_3 \
          2A_3 \
          5A_2
          end{vmatrix} = ldots$$




          Addition after comment:
          $$begin{vmatrix}
          color{blue}{2}A_1 \
          color{green}{2}A_3 \
          color{red}{5}A_2
          end{vmatrix}=color{blue}{2}cdotcolor{green}{2}cdotcolor{red}{5}cdotbegin{vmatrix}
          A_1 \
          color{purple}{A_3} \
          color{purple}{A_2}
          end{vmatrix}=ldots$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            thanks for explaining that to me, now i updated the subject with matrix B but i couldn't obtain B determinant is that right ?, and how we obtain b determinant ?
            $endgroup$
            – The Beard
            Dec 12 '18 at 12:50










          • $begingroup$
            Linearity also allows you to 'extract' the multiples (in each row!); then notice the swap of two rows. See updated answer.
            $endgroup$
            – StackTD
            Dec 12 '18 at 12:53












          • $begingroup$
            thank you again for explaining to me now i have more understading of determinant properties because of your help, and from what you have told me the determinant of B will be -20 right ?
            $endgroup$
            – The Beard
            Dec 12 '18 at 13:00












          • $begingroup$
            Almost, $-20$ times the determinant $A$, so...?
            $endgroup$
            – StackTD
            Dec 12 '18 at 13:27










          • $begingroup$
            so then the determinant of b will be -10 ?
            $endgroup$
            – The Beard
            Dec 12 '18 at 13:56











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Use properties of determinants:




          • the determinant is linear in each row/column;

          • a determinant with two identical rows is $0$;

          • swapping two rows changes the sign of the determinant.


          Now start with linearity and follow up (I write $A_i$ for the $i$th row of the original matrix $A$):
          $$begin{vmatrix}
          2A_1+A_2-A_3 \
          2A_3 \
          5A_2
          end{vmatrix} = begin{vmatrix}
          2A_1 \
          2A_3 \
          5A_2
          end{vmatrix}+begin{vmatrix}
          A_2 \
          2A_3 \
          5A_2
          end{vmatrix}+begin{vmatrix}
          -A_3 \
          2A_3 \
          5A_2
          end{vmatrix} = ldots$$




          Addition after comment:
          $$begin{vmatrix}
          color{blue}{2}A_1 \
          color{green}{2}A_3 \
          color{red}{5}A_2
          end{vmatrix}=color{blue}{2}cdotcolor{green}{2}cdotcolor{red}{5}cdotbegin{vmatrix}
          A_1 \
          color{purple}{A_3} \
          color{purple}{A_2}
          end{vmatrix}=ldots$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            thanks for explaining that to me, now i updated the subject with matrix B but i couldn't obtain B determinant is that right ?, and how we obtain b determinant ?
            $endgroup$
            – The Beard
            Dec 12 '18 at 12:50










          • $begingroup$
            Linearity also allows you to 'extract' the multiples (in each row!); then notice the swap of two rows. See updated answer.
            $endgroup$
            – StackTD
            Dec 12 '18 at 12:53












          • $begingroup$
            thank you again for explaining to me now i have more understading of determinant properties because of your help, and from what you have told me the determinant of B will be -20 right ?
            $endgroup$
            – The Beard
            Dec 12 '18 at 13:00












          • $begingroup$
            Almost, $-20$ times the determinant $A$, so...?
            $endgroup$
            – StackTD
            Dec 12 '18 at 13:27










          • $begingroup$
            so then the determinant of b will be -10 ?
            $endgroup$
            – The Beard
            Dec 12 '18 at 13:56
















          2












          $begingroup$

          Use properties of determinants:




          • the determinant is linear in each row/column;

          • a determinant with two identical rows is $0$;

          • swapping two rows changes the sign of the determinant.


          Now start with linearity and follow up (I write $A_i$ for the $i$th row of the original matrix $A$):
          $$begin{vmatrix}
          2A_1+A_2-A_3 \
          2A_3 \
          5A_2
          end{vmatrix} = begin{vmatrix}
          2A_1 \
          2A_3 \
          5A_2
          end{vmatrix}+begin{vmatrix}
          A_2 \
          2A_3 \
          5A_2
          end{vmatrix}+begin{vmatrix}
          -A_3 \
          2A_3 \
          5A_2
          end{vmatrix} = ldots$$




          Addition after comment:
          $$begin{vmatrix}
          color{blue}{2}A_1 \
          color{green}{2}A_3 \
          color{red}{5}A_2
          end{vmatrix}=color{blue}{2}cdotcolor{green}{2}cdotcolor{red}{5}cdotbegin{vmatrix}
          A_1 \
          color{purple}{A_3} \
          color{purple}{A_2}
          end{vmatrix}=ldots$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            thanks for explaining that to me, now i updated the subject with matrix B but i couldn't obtain B determinant is that right ?, and how we obtain b determinant ?
            $endgroup$
            – The Beard
            Dec 12 '18 at 12:50










          • $begingroup$
            Linearity also allows you to 'extract' the multiples (in each row!); then notice the swap of two rows. See updated answer.
            $endgroup$
            – StackTD
            Dec 12 '18 at 12:53












          • $begingroup$
            thank you again for explaining to me now i have more understading of determinant properties because of your help, and from what you have told me the determinant of B will be -20 right ?
            $endgroup$
            – The Beard
            Dec 12 '18 at 13:00












          • $begingroup$
            Almost, $-20$ times the determinant $A$, so...?
            $endgroup$
            – StackTD
            Dec 12 '18 at 13:27










          • $begingroup$
            so then the determinant of b will be -10 ?
            $endgroup$
            – The Beard
            Dec 12 '18 at 13:56














          2












          2








          2





          $begingroup$

          Use properties of determinants:




          • the determinant is linear in each row/column;

          • a determinant with two identical rows is $0$;

          • swapping two rows changes the sign of the determinant.


          Now start with linearity and follow up (I write $A_i$ for the $i$th row of the original matrix $A$):
          $$begin{vmatrix}
          2A_1+A_2-A_3 \
          2A_3 \
          5A_2
          end{vmatrix} = begin{vmatrix}
          2A_1 \
          2A_3 \
          5A_2
          end{vmatrix}+begin{vmatrix}
          A_2 \
          2A_3 \
          5A_2
          end{vmatrix}+begin{vmatrix}
          -A_3 \
          2A_3 \
          5A_2
          end{vmatrix} = ldots$$




          Addition after comment:
          $$begin{vmatrix}
          color{blue}{2}A_1 \
          color{green}{2}A_3 \
          color{red}{5}A_2
          end{vmatrix}=color{blue}{2}cdotcolor{green}{2}cdotcolor{red}{5}cdotbegin{vmatrix}
          A_1 \
          color{purple}{A_3} \
          color{purple}{A_2}
          end{vmatrix}=ldots$$






          share|cite|improve this answer











          $endgroup$



          Use properties of determinants:




          • the determinant is linear in each row/column;

          • a determinant with two identical rows is $0$;

          • swapping two rows changes the sign of the determinant.


          Now start with linearity and follow up (I write $A_i$ for the $i$th row of the original matrix $A$):
          $$begin{vmatrix}
          2A_1+A_2-A_3 \
          2A_3 \
          5A_2
          end{vmatrix} = begin{vmatrix}
          2A_1 \
          2A_3 \
          5A_2
          end{vmatrix}+begin{vmatrix}
          A_2 \
          2A_3 \
          5A_2
          end{vmatrix}+begin{vmatrix}
          -A_3 \
          2A_3 \
          5A_2
          end{vmatrix} = ldots$$




          Addition after comment:
          $$begin{vmatrix}
          color{blue}{2}A_1 \
          color{green}{2}A_3 \
          color{red}{5}A_2
          end{vmatrix}=color{blue}{2}cdotcolor{green}{2}cdotcolor{red}{5}cdotbegin{vmatrix}
          A_1 \
          color{purple}{A_3} \
          color{purple}{A_2}
          end{vmatrix}=ldots$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 12 '18 at 12:55

























          answered Dec 12 '18 at 12:22









          StackTDStackTD

          22.9k2152




          22.9k2152












          • $begingroup$
            thanks for explaining that to me, now i updated the subject with matrix B but i couldn't obtain B determinant is that right ?, and how we obtain b determinant ?
            $endgroup$
            – The Beard
            Dec 12 '18 at 12:50










          • $begingroup$
            Linearity also allows you to 'extract' the multiples (in each row!); then notice the swap of two rows. See updated answer.
            $endgroup$
            – StackTD
            Dec 12 '18 at 12:53












          • $begingroup$
            thank you again for explaining to me now i have more understading of determinant properties because of your help, and from what you have told me the determinant of B will be -20 right ?
            $endgroup$
            – The Beard
            Dec 12 '18 at 13:00












          • $begingroup$
            Almost, $-20$ times the determinant $A$, so...?
            $endgroup$
            – StackTD
            Dec 12 '18 at 13:27










          • $begingroup$
            so then the determinant of b will be -10 ?
            $endgroup$
            – The Beard
            Dec 12 '18 at 13:56


















          • $begingroup$
            thanks for explaining that to me, now i updated the subject with matrix B but i couldn't obtain B determinant is that right ?, and how we obtain b determinant ?
            $endgroup$
            – The Beard
            Dec 12 '18 at 12:50










          • $begingroup$
            Linearity also allows you to 'extract' the multiples (in each row!); then notice the swap of two rows. See updated answer.
            $endgroup$
            – StackTD
            Dec 12 '18 at 12:53












          • $begingroup$
            thank you again for explaining to me now i have more understading of determinant properties because of your help, and from what you have told me the determinant of B will be -20 right ?
            $endgroup$
            – The Beard
            Dec 12 '18 at 13:00












          • $begingroup$
            Almost, $-20$ times the determinant $A$, so...?
            $endgroup$
            – StackTD
            Dec 12 '18 at 13:27










          • $begingroup$
            so then the determinant of b will be -10 ?
            $endgroup$
            – The Beard
            Dec 12 '18 at 13:56
















          $begingroup$
          thanks for explaining that to me, now i updated the subject with matrix B but i couldn't obtain B determinant is that right ?, and how we obtain b determinant ?
          $endgroup$
          – The Beard
          Dec 12 '18 at 12:50




          $begingroup$
          thanks for explaining that to me, now i updated the subject with matrix B but i couldn't obtain B determinant is that right ?, and how we obtain b determinant ?
          $endgroup$
          – The Beard
          Dec 12 '18 at 12:50












          $begingroup$
          Linearity also allows you to 'extract' the multiples (in each row!); then notice the swap of two rows. See updated answer.
          $endgroup$
          – StackTD
          Dec 12 '18 at 12:53






          $begingroup$
          Linearity also allows you to 'extract' the multiples (in each row!); then notice the swap of two rows. See updated answer.
          $endgroup$
          – StackTD
          Dec 12 '18 at 12:53














          $begingroup$
          thank you again for explaining to me now i have more understading of determinant properties because of your help, and from what you have told me the determinant of B will be -20 right ?
          $endgroup$
          – The Beard
          Dec 12 '18 at 13:00






          $begingroup$
          thank you again for explaining to me now i have more understading of determinant properties because of your help, and from what you have told me the determinant of B will be -20 right ?
          $endgroup$
          – The Beard
          Dec 12 '18 at 13:00














          $begingroup$
          Almost, $-20$ times the determinant $A$, so...?
          $endgroup$
          – StackTD
          Dec 12 '18 at 13:27




          $begingroup$
          Almost, $-20$ times the determinant $A$, so...?
          $endgroup$
          – StackTD
          Dec 12 '18 at 13:27












          $begingroup$
          so then the determinant of b will be -10 ?
          $endgroup$
          – The Beard
          Dec 12 '18 at 13:56




          $begingroup$
          so then the determinant of b will be -10 ?
          $endgroup$
          – The Beard
          Dec 12 '18 at 13:56


















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