Sign rule for finding the adjugate of a 3x3 matrix?












1












$begingroup$


So i have this matrix



A= $$
begin{pmatrix}
1 & 3 & 0 \
-2 & -5 & 2 \
1 & 4 & 3 \
end{pmatrix}
$$



And i want to find the inverse of it. Following all the calculations, i get that the determinant is 1 and that the adjucate of the matrix by creating the matrix of co factors is



$$
begin{pmatrix}
-23 & -9 & 6 \
8 & 3 & -2 \
-3 & -1 & 1 \
end{pmatrix}
$$
I checked the answer on the answer sheet and it's right.



And i would be happy to end the excercise here BUT if i remember correctly shouldn't i apply the sign rule?



I always thought that after having calculated the matrix, i apply the sign rule which would be



$$
begin{pmatrix}
+ & - & + \
- & + & - \
+ & - & + \
end{pmatrix}
$$



If i apply the sign rule, then my matrix would look different:



$$
begin{pmatrix}
-23 & 9 & 6 \
-8 & 3 & 2 \
-3 & 1 & 1 \
end{pmatrix}
$$



and ,in order to finish the calculations, i would need to multiply the adjucate by 1 over the determinant of A which is 1.



So i would multiply 1 times each entry in the adjugate and the matrix would still be



$$
begin{pmatrix}
-23 & 9 & 6 \
-8 & 3 & 2 \
-3 & 1 & 1 \
end{pmatrix}
$$



Which, as you can clearly see, is not the correct answer.



So my question is , should i apply the sign rule? Should i not? What is it that i'm doing wrong? SHould i stop when i just find the adjugate and multiply the entries by the determinant ?



Thanks for the help!



(please do not tell me to find the inverse another way , thanks)!






matrices determinant inverse matrix-calculus







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    So i have this matrix



    A= $$
    begin{pmatrix}
    1 & 3 & 0 \
    -2 & -5 & 2 \
    1 & 4 & 3 \
    end{pmatrix}
    $$



    And i want to find the inverse of it. Following all the calculations, i get that the determinant is 1 and that the adjucate of the matrix by creating the matrix of co factors is



    $$
    begin{pmatrix}
    -23 & -9 & 6 \
    8 & 3 & -2 \
    -3 & -1 & 1 \
    end{pmatrix}
    $$
    I checked the answer on the answer sheet and it's right.



    And i would be happy to end the excercise here BUT if i remember correctly shouldn't i apply the sign rule?



    I always thought that after having calculated the matrix, i apply the sign rule which would be



    $$
    begin{pmatrix}
    + & - & + \
    - & + & - \
    + & - & + \
    end{pmatrix}
    $$



    If i apply the sign rule, then my matrix would look different:



    $$
    begin{pmatrix}
    -23 & 9 & 6 \
    -8 & 3 & 2 \
    -3 & 1 & 1 \
    end{pmatrix}
    $$



    and ,in order to finish the calculations, i would need to multiply the adjucate by 1 over the determinant of A which is 1.



    So i would multiply 1 times each entry in the adjugate and the matrix would still be



    $$
    begin{pmatrix}
    -23 & 9 & 6 \
    -8 & 3 & 2 \
    -3 & 1 & 1 \
    end{pmatrix}
    $$



    Which, as you can clearly see, is not the correct answer.



    So my question is , should i apply the sign rule? Should i not? What is it that i'm doing wrong? SHould i stop when i just find the adjugate and multiply the entries by the determinant ?



    Thanks for the help!



    (please do not tell me to find the inverse another way , thanks)!






    matrices determinant inverse matrix-calculus







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      So i have this matrix



      A= $$
      begin{pmatrix}
      1 & 3 & 0 \
      -2 & -5 & 2 \
      1 & 4 & 3 \
      end{pmatrix}
      $$



      And i want to find the inverse of it. Following all the calculations, i get that the determinant is 1 and that the adjucate of the matrix by creating the matrix of co factors is



      $$
      begin{pmatrix}
      -23 & -9 & 6 \
      8 & 3 & -2 \
      -3 & -1 & 1 \
      end{pmatrix}
      $$
      I checked the answer on the answer sheet and it's right.



      And i would be happy to end the excercise here BUT if i remember correctly shouldn't i apply the sign rule?



      I always thought that after having calculated the matrix, i apply the sign rule which would be



      $$
      begin{pmatrix}
      + & - & + \
      - & + & - \
      + & - & + \
      end{pmatrix}
      $$



      If i apply the sign rule, then my matrix would look different:



      $$
      begin{pmatrix}
      -23 & 9 & 6 \
      -8 & 3 & 2 \
      -3 & 1 & 1 \
      end{pmatrix}
      $$



      and ,in order to finish the calculations, i would need to multiply the adjucate by 1 over the determinant of A which is 1.



      So i would multiply 1 times each entry in the adjugate and the matrix would still be



      $$
      begin{pmatrix}
      -23 & 9 & 6 \
      -8 & 3 & 2 \
      -3 & 1 & 1 \
      end{pmatrix}
      $$



      Which, as you can clearly see, is not the correct answer.



      So my question is , should i apply the sign rule? Should i not? What is it that i'm doing wrong? SHould i stop when i just find the adjugate and multiply the entries by the determinant ?



      Thanks for the help!



      (please do not tell me to find the inverse another way , thanks)!






      matrices determinant inverse matrix-calculus







      share|cite|improve this question











      $endgroup$




      So i have this matrix



      A= $$
      begin{pmatrix}
      1 & 3 & 0 \
      -2 & -5 & 2 \
      1 & 4 & 3 \
      end{pmatrix}
      $$



      And i want to find the inverse of it. Following all the calculations, i get that the determinant is 1 and that the adjucate of the matrix by creating the matrix of co factors is



      $$
      begin{pmatrix}
      -23 & -9 & 6 \
      8 & 3 & -2 \
      -3 & -1 & 1 \
      end{pmatrix}
      $$
      I checked the answer on the answer sheet and it's right.



      And i would be happy to end the excercise here BUT if i remember correctly shouldn't i apply the sign rule?



      I always thought that after having calculated the matrix, i apply the sign rule which would be



      $$
      begin{pmatrix}
      + & - & + \
      - & + & - \
      + & - & + \
      end{pmatrix}
      $$



      If i apply the sign rule, then my matrix would look different:



      $$
      begin{pmatrix}
      -23 & 9 & 6 \
      -8 & 3 & 2 \
      -3 & 1 & 1 \
      end{pmatrix}
      $$



      and ,in order to finish the calculations, i would need to multiply the adjucate by 1 over the determinant of A which is 1.



      So i would multiply 1 times each entry in the adjugate and the matrix would still be



      $$
      begin{pmatrix}
      -23 & 9 & 6 \
      -8 & 3 & 2 \
      -3 & 1 & 1 \
      end{pmatrix}
      $$



      Which, as you can clearly see, is not the correct answer.



      So my question is , should i apply the sign rule? Should i not? What is it that i'm doing wrong? SHould i stop when i just find the adjugate and multiply the entries by the determinant ?



      Thanks for the help!



      (please do not tell me to find the inverse another way , thanks)!






      matrices determinant inverse matrix-calculus




      matrices determinant inverse matrix-calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 4 '18 at 12:14







      BM97

















      asked Dec 4 '18 at 12:03









      BM97BM97

      758




      758






















          1 Answer
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          $begingroup$

          You’ve already applied the sign rule when you calculated the cofactors.



          For example when you calculate the $(1,1)$ element, you do:



          $$M_{1,1}=detbegin{pmatrix}-5&2\ 4&3end{pmatrix}=-23$$



          And when you calculate the $(1,2)$ element, you do (note the negative sign!!):



          $$M_{1,2}=-detbegin{pmatrix}-2&2\ 1&3end{pmatrix}=-((-2*3)-(1*2))=8$$



          Then to get the adjugate you take the transpose of the cofactors.



          You’ve done everything correctly and incorporated the sign rule. No need to apply it twice.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you !! I greatly appreciate your simple, straightforward answer .
            $endgroup$
            – BM97
            Dec 4 '18 at 13:18











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          1 Answer
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          active

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          0












          $begingroup$

          You’ve already applied the sign rule when you calculated the cofactors.



          For example when you calculate the $(1,1)$ element, you do:



          $$M_{1,1}=detbegin{pmatrix}-5&2\ 4&3end{pmatrix}=-23$$



          And when you calculate the $(1,2)$ element, you do (note the negative sign!!):



          $$M_{1,2}=-detbegin{pmatrix}-2&2\ 1&3end{pmatrix}=-((-2*3)-(1*2))=8$$



          Then to get the adjugate you take the transpose of the cofactors.



          You’ve done everything correctly and incorporated the sign rule. No need to apply it twice.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you !! I greatly appreciate your simple, straightforward answer .
            $endgroup$
            – BM97
            Dec 4 '18 at 13:18
















          0












          $begingroup$

          You’ve already applied the sign rule when you calculated the cofactors.



          For example when you calculate the $(1,1)$ element, you do:



          $$M_{1,1}=detbegin{pmatrix}-5&2\ 4&3end{pmatrix}=-23$$



          And when you calculate the $(1,2)$ element, you do (note the negative sign!!):



          $$M_{1,2}=-detbegin{pmatrix}-2&2\ 1&3end{pmatrix}=-((-2*3)-(1*2))=8$$



          Then to get the adjugate you take the transpose of the cofactors.



          You’ve done everything correctly and incorporated the sign rule. No need to apply it twice.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you !! I greatly appreciate your simple, straightforward answer .
            $endgroup$
            – BM97
            Dec 4 '18 at 13:18














          0












          0








          0





          $begingroup$

          You’ve already applied the sign rule when you calculated the cofactors.



          For example when you calculate the $(1,1)$ element, you do:



          $$M_{1,1}=detbegin{pmatrix}-5&2\ 4&3end{pmatrix}=-23$$



          And when you calculate the $(1,2)$ element, you do (note the negative sign!!):



          $$M_{1,2}=-detbegin{pmatrix}-2&2\ 1&3end{pmatrix}=-((-2*3)-(1*2))=8$$



          Then to get the adjugate you take the transpose of the cofactors.



          You’ve done everything correctly and incorporated the sign rule. No need to apply it twice.






          share|cite|improve this answer









          $endgroup$



          You’ve already applied the sign rule when you calculated the cofactors.



          For example when you calculate the $(1,1)$ element, you do:



          $$M_{1,1}=detbegin{pmatrix}-5&2\ 4&3end{pmatrix}=-23$$



          And when you calculate the $(1,2)$ element, you do (note the negative sign!!):



          $$M_{1,2}=-detbegin{pmatrix}-2&2\ 1&3end{pmatrix}=-((-2*3)-(1*2))=8$$



          Then to get the adjugate you take the transpose of the cofactors.



          You’ve done everything correctly and incorporated the sign rule. No need to apply it twice.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 4 '18 at 12:56









          ip6ip6

          54839




          54839












          • $begingroup$
            Thank you !! I greatly appreciate your simple, straightforward answer .
            $endgroup$
            – BM97
            Dec 4 '18 at 13:18


















          • $begingroup$
            Thank you !! I greatly appreciate your simple, straightforward answer .
            $endgroup$
            – BM97
            Dec 4 '18 at 13:18
















          $begingroup$
          Thank you !! I greatly appreciate your simple, straightforward answer .
          $endgroup$
          – BM97
          Dec 4 '18 at 13:18




          $begingroup$
          Thank you !! I greatly appreciate your simple, straightforward answer .
          $endgroup$
          – BM97
          Dec 4 '18 at 13:18


















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