Determinant properties with row reduction












0












$begingroup$


I have the following question here.




Let $A$ and $B$ be $3 × 3$ matrices with $det(A) = 3$ and $det(B) = 2$. Let $C = frac{1}{2}A^{-1}B^3$ and let $D$ be the reduced row echelon form of $C$. Then:



$(a)$ $det(C)=frac{4}{3}$, $det(D)=1$



$(b)$ $det(C)=frac{1}{3}$, $det(D)=1$



$(c)$ $det(C)=frac{4}{3}$, $det(D)=frac{4}{3}$



$(d)$ $det(C)=frac{1}{3}$, $det(D)=3$



$(e)$ $det(C)=frac{1}{3}$, $det(D)=frac{1}{3}$




The answer is supposed to be $b$. I know $det(C)=frac{1}{3}$ just because of determinant properties. That was easy. I'm not 100% sure how the $RREF$ of $D$ comes into play here. I know that elementary row operations affect the determinant but HOW does that affects the determinant here.



Can someone provide any guidance as to how I would calculate $det(D)$?






linear-algebra determinant







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I have the following question here.




    Let $A$ and $B$ be $3 × 3$ matrices with $det(A) = 3$ and $det(B) = 2$. Let $C = frac{1}{2}A^{-1}B^3$ and let $D$ be the reduced row echelon form of $C$. Then:



    $(a)$ $det(C)=frac{4}{3}$, $det(D)=1$



    $(b)$ $det(C)=frac{1}{3}$, $det(D)=1$



    $(c)$ $det(C)=frac{4}{3}$, $det(D)=frac{4}{3}$



    $(d)$ $det(C)=frac{1}{3}$, $det(D)=3$



    $(e)$ $det(C)=frac{1}{3}$, $det(D)=frac{1}{3}$




    The answer is supposed to be $b$. I know $det(C)=frac{1}{3}$ just because of determinant properties. That was easy. I'm not 100% sure how the $RREF$ of $D$ comes into play here. I know that elementary row operations affect the determinant but HOW does that affects the determinant here.



    Can someone provide any guidance as to how I would calculate $det(D)$?






    linear-algebra determinant







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I have the following question here.




      Let $A$ and $B$ be $3 × 3$ matrices with $det(A) = 3$ and $det(B) = 2$. Let $C = frac{1}{2}A^{-1}B^3$ and let $D$ be the reduced row echelon form of $C$. Then:



      $(a)$ $det(C)=frac{4}{3}$, $det(D)=1$



      $(b)$ $det(C)=frac{1}{3}$, $det(D)=1$



      $(c)$ $det(C)=frac{4}{3}$, $det(D)=frac{4}{3}$



      $(d)$ $det(C)=frac{1}{3}$, $det(D)=3$



      $(e)$ $det(C)=frac{1}{3}$, $det(D)=frac{1}{3}$




      The answer is supposed to be $b$. I know $det(C)=frac{1}{3}$ just because of determinant properties. That was easy. I'm not 100% sure how the $RREF$ of $D$ comes into play here. I know that elementary row operations affect the determinant but HOW does that affects the determinant here.



      Can someone provide any guidance as to how I would calculate $det(D)$?






      linear-algebra determinant







      share|cite|improve this question









      $endgroup$




      I have the following question here.




      Let $A$ and $B$ be $3 × 3$ matrices with $det(A) = 3$ and $det(B) = 2$. Let $C = frac{1}{2}A^{-1}B^3$ and let $D$ be the reduced row echelon form of $C$. Then:



      $(a)$ $det(C)=frac{4}{3}$, $det(D)=1$



      $(b)$ $det(C)=frac{1}{3}$, $det(D)=1$



      $(c)$ $det(C)=frac{4}{3}$, $det(D)=frac{4}{3}$



      $(d)$ $det(C)=frac{1}{3}$, $det(D)=3$



      $(e)$ $det(C)=frac{1}{3}$, $det(D)=frac{1}{3}$




      The answer is supposed to be $b$. I know $det(C)=frac{1}{3}$ just because of determinant properties. That was easy. I'm not 100% sure how the $RREF$ of $D$ comes into play here. I know that elementary row operations affect the determinant but HOW does that affects the determinant here.



      Can someone provide any guidance as to how I would calculate $det(D)$?






      linear-algebra determinant




      linear-algebra determinant






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 16 '18 at 9:46









      Future Math personFuture Math person

      972817




      972817






















          2 Answers
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          $begingroup$

          Since the matrix $C$ is non-singular, its row reduced echelon form is just $I$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ohh. I overthought that way to much. Thanks.
            $endgroup$
            – Future Math person
            Dec 16 '18 at 9:56



















          1












          $begingroup$

          $det(C)=det(frac12A^{-1}B^3)=frac1{2^3}det(A^{-1}B^3)=frac18cdotfrac1{|A|}cdot|B|^3=1/3$



          $C$ is invertible. The row-reduced echelon form of $C$ is the identity matrix. So $D=I_3,det(D)=1$.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            Since the matrix $C$ is non-singular, its row reduced echelon form is just $I$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Ohh. I overthought that way to much. Thanks.
              $endgroup$
              – Future Math person
              Dec 16 '18 at 9:56
















            1












            $begingroup$

            Since the matrix $C$ is non-singular, its row reduced echelon form is just $I$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Ohh. I overthought that way to much. Thanks.
              $endgroup$
              – Future Math person
              Dec 16 '18 at 9:56














            1












            1








            1





            $begingroup$

            Since the matrix $C$ is non-singular, its row reduced echelon form is just $I$.






            share|cite|improve this answer









            $endgroup$



            Since the matrix $C$ is non-singular, its row reduced echelon form is just $I$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 16 '18 at 9:56









            eranrecheseranreches

            3,816925




            3,816925












            • $begingroup$
              Ohh. I overthought that way to much. Thanks.
              $endgroup$
              – Future Math person
              Dec 16 '18 at 9:56


















            • $begingroup$
              Ohh. I overthought that way to much. Thanks.
              $endgroup$
              – Future Math person
              Dec 16 '18 at 9:56
















            $begingroup$
            Ohh. I overthought that way to much. Thanks.
            $endgroup$
            – Future Math person
            Dec 16 '18 at 9:56




            $begingroup$
            Ohh. I overthought that way to much. Thanks.
            $endgroup$
            – Future Math person
            Dec 16 '18 at 9:56











            1












            $begingroup$

            $det(C)=det(frac12A^{-1}B^3)=frac1{2^3}det(A^{-1}B^3)=frac18cdotfrac1{|A|}cdot|B|^3=1/3$



            $C$ is invertible. The row-reduced echelon form of $C$ is the identity matrix. So $D=I_3,det(D)=1$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              $det(C)=det(frac12A^{-1}B^3)=frac1{2^3}det(A^{-1}B^3)=frac18cdotfrac1{|A|}cdot|B|^3=1/3$



              $C$ is invertible. The row-reduced echelon form of $C$ is the identity matrix. So $D=I_3,det(D)=1$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                $det(C)=det(frac12A^{-1}B^3)=frac1{2^3}det(A^{-1}B^3)=frac18cdotfrac1{|A|}cdot|B|^3=1/3$



                $C$ is invertible. The row-reduced echelon form of $C$ is the identity matrix. So $D=I_3,det(D)=1$.






                share|cite|improve this answer









                $endgroup$



                $det(C)=det(frac12A^{-1}B^3)=frac1{2^3}det(A^{-1}B^3)=frac18cdotfrac1{|A|}cdot|B|^3=1/3$



                $C$ is invertible. The row-reduced echelon form of $C$ is the identity matrix. So $D=I_3,det(D)=1$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 16 '18 at 9:57









                Shubham JohriShubham Johri

                5,172717




                5,172717






























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