Find the rank of matrix $B$












1














If $A_{n times n}$ is a non-singular matrix, such that $A times B$ is the zero matrix ($ntimes n$), then the rank of $B$ is?



(A) $n-1$



(B) $n-2$



(C) $1$



(D) $0$



The answer to this question was given to be (D).



But I am confused why only (D).



$|AB|=0$



$|A|.|B|=0, ;|A| neq 0 rightarrow |B| = 0$.



Surely rank of $B$ is less than $n.$ But why exactly $0$?






linear-algebra







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  • 1




    Your argument only works if $B$ is square, otherwise it doesn't make sense to talk about its determinant.
    – pwerth
    yesterday










  • Yes, please guide now.
    – user3767495
    yesterday
















1














If $A_{n times n}$ is a non-singular matrix, such that $A times B$ is the zero matrix ($ntimes n$), then the rank of $B$ is?



(A) $n-1$



(B) $n-2$



(C) $1$



(D) $0$



The answer to this question was given to be (D).



But I am confused why only (D).



$|AB|=0$



$|A|.|B|=0, ;|A| neq 0 rightarrow |B| = 0$.



Surely rank of $B$ is less than $n.$ But why exactly $0$?






linear-algebra







share|cite|improve this question




















  • 1




    Your argument only works if $B$ is square, otherwise it doesn't make sense to talk about its determinant.
    – pwerth
    yesterday










  • Yes, please guide now.
    – user3767495
    yesterday














1












1








1







If $A_{n times n}$ is a non-singular matrix, such that $A times B$ is the zero matrix ($ntimes n$), then the rank of $B$ is?



(A) $n-1$



(B) $n-2$



(C) $1$



(D) $0$



The answer to this question was given to be (D).



But I am confused why only (D).



$|AB|=0$



$|A|.|B|=0, ;|A| neq 0 rightarrow |B| = 0$.



Surely rank of $B$ is less than $n.$ But why exactly $0$?






linear-algebra







share|cite|improve this question















If $A_{n times n}$ is a non-singular matrix, such that $A times B$ is the zero matrix ($ntimes n$), then the rank of $B$ is?



(A) $n-1$



(B) $n-2$



(C) $1$



(D) $0$



The answer to this question was given to be (D).



But I am confused why only (D).



$|AB|=0$



$|A|.|B|=0, ;|A| neq 0 rightarrow |B| = 0$.



Surely rank of $B$ is less than $n.$ But why exactly $0$?






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









amWhy

191k28224439




191k28224439










asked yesterday









user3767495

3348




3348








  • 1




    Your argument only works if $B$ is square, otherwise it doesn't make sense to talk about its determinant.
    – pwerth
    yesterday










  • Yes, please guide now.
    – user3767495
    yesterday














  • 1




    Your argument only works if $B$ is square, otherwise it doesn't make sense to talk about its determinant.
    – pwerth
    yesterday










  • Yes, please guide now.
    – user3767495
    yesterday








1




1




Your argument only works if $B$ is square, otherwise it doesn't make sense to talk about its determinant.
– pwerth
yesterday




Your argument only works if $B$ is square, otherwise it doesn't make sense to talk about its determinant.
– pwerth
yesterday












Yes, please guide now.
– user3767495
yesterday




Yes, please guide now.
– user3767495
yesterday










2 Answers
2






active

oldest

votes


















3














If $A$ is non-singular, $A^{-1}$ exists. Then, $B = A^{-1}AB = A^{-1}0 = 0$.



Since $B$ is the zero-matrix (of whatever size), $text{rank}(B) = 0$.






share|cite|improve this answer





























    1














    Multiplication $B$ with a non-singular matrix doesn't change its rank.



    This can be seen from a nonsingular matrix can be decomposed into a product of elementary matrices, and we know that performing elementary operation doesn't change the rank of a matrix.



    Hence if $A$ is nonsingular and $AB=C$, we have $rank(B)=rank(C)$.






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

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      votes









      active

      oldest

      votes






      active

      oldest

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      3














      If $A$ is non-singular, $A^{-1}$ exists. Then, $B = A^{-1}AB = A^{-1}0 = 0$.



      Since $B$ is the zero-matrix (of whatever size), $text{rank}(B) = 0$.






      share|cite|improve this answer


























        3














        If $A$ is non-singular, $A^{-1}$ exists. Then, $B = A^{-1}AB = A^{-1}0 = 0$.



        Since $B$ is the zero-matrix (of whatever size), $text{rank}(B) = 0$.






        share|cite|improve this answer
























          3












          3








          3






          If $A$ is non-singular, $A^{-1}$ exists. Then, $B = A^{-1}AB = A^{-1}0 = 0$.



          Since $B$ is the zero-matrix (of whatever size), $text{rank}(B) = 0$.






          share|cite|improve this answer












          If $A$ is non-singular, $A^{-1}$ exists. Then, $B = A^{-1}AB = A^{-1}0 = 0$.



          Since $B$ is the zero-matrix (of whatever size), $text{rank}(B) = 0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          JimmyK4542

          40.4k245105




          40.4k245105























              1














              Multiplication $B$ with a non-singular matrix doesn't change its rank.



              This can be seen from a nonsingular matrix can be decomposed into a product of elementary matrices, and we know that performing elementary operation doesn't change the rank of a matrix.



              Hence if $A$ is nonsingular and $AB=C$, we have $rank(B)=rank(C)$.






              share|cite|improve this answer


























                1














                Multiplication $B$ with a non-singular matrix doesn't change its rank.



                This can be seen from a nonsingular matrix can be decomposed into a product of elementary matrices, and we know that performing elementary operation doesn't change the rank of a matrix.



                Hence if $A$ is nonsingular and $AB=C$, we have $rank(B)=rank(C)$.






                share|cite|improve this answer
























                  1












                  1








                  1






                  Multiplication $B$ with a non-singular matrix doesn't change its rank.



                  This can be seen from a nonsingular matrix can be decomposed into a product of elementary matrices, and we know that performing elementary operation doesn't change the rank of a matrix.



                  Hence if $A$ is nonsingular and $AB=C$, we have $rank(B)=rank(C)$.






                  share|cite|improve this answer












                  Multiplication $B$ with a non-singular matrix doesn't change its rank.



                  This can be seen from a nonsingular matrix can be decomposed into a product of elementary matrices, and we know that performing elementary operation doesn't change the rank of a matrix.



                  Hence if $A$ is nonsingular and $AB=C$, we have $rank(B)=rank(C)$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  Siong Thye Goh

                  98.9k1464116




                  98.9k1464116






























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