What is $A^{31}+B^{35}$












2














If $AB=B$ and $BA=A$ then $A^{31}+B^{35}$ is?



(A)$A^2+B^2$



(B)$A+B$



(C)$A^4+B^4$



(D)$A^3+B^3$



My work



$AB=B rightarrow ABB^{-1}=BB^{-1}$



So, $A=I$



Similarly $B=I$



All option seem valid to me :O






linear-algebra







share|cite|improve this question


















  • 4




    What if $B$ is singular?
    – Umberto P.
    yesterday










  • Then $B^{-1}$ won't exist.
    – user3767495
    yesterday






  • 1




    @user3767495: And so your entire argument breaks down ...
    – Henning Makholm
    yesterday
















2














If $AB=B$ and $BA=A$ then $A^{31}+B^{35}$ is?



(A)$A^2+B^2$



(B)$A+B$



(C)$A^4+B^4$



(D)$A^3+B^3$



My work



$AB=B rightarrow ABB^{-1}=BB^{-1}$



So, $A=I$



Similarly $B=I$



All option seem valid to me :O






linear-algebra







share|cite|improve this question


















  • 4




    What if $B$ is singular?
    – Umberto P.
    yesterday










  • Then $B^{-1}$ won't exist.
    – user3767495
    yesterday






  • 1




    @user3767495: And so your entire argument breaks down ...
    – Henning Makholm
    yesterday














2












2








2







If $AB=B$ and $BA=A$ then $A^{31}+B^{35}$ is?



(A)$A^2+B^2$



(B)$A+B$



(C)$A^4+B^4$



(D)$A^3+B^3$



My work



$AB=B rightarrow ABB^{-1}=BB^{-1}$



So, $A=I$



Similarly $B=I$



All option seem valid to me :O






linear-algebra







share|cite|improve this question













If $AB=B$ and $BA=A$ then $A^{31}+B^{35}$ is?



(A)$A^2+B^2$



(B)$A+B$



(C)$A^4+B^4$



(D)$A^3+B^3$



My work



$AB=B rightarrow ABB^{-1}=BB^{-1}$



So, $A=I$



Similarly $B=I$



All option seem valid to me :O






linear-algebra




linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked yesterday









user3767495

3348




3348








  • 4




    What if $B$ is singular?
    – Umberto P.
    yesterday










  • Then $B^{-1}$ won't exist.
    – user3767495
    yesterday






  • 1




    @user3767495: And so your entire argument breaks down ...
    – Henning Makholm
    yesterday














  • 4




    What if $B$ is singular?
    – Umberto P.
    yesterday










  • Then $B^{-1}$ won't exist.
    – user3767495
    yesterday






  • 1




    @user3767495: And so your entire argument breaks down ...
    – Henning Makholm
    yesterday








4




4




What if $B$ is singular?
– Umberto P.
yesterday




What if $B$ is singular?
– Umberto P.
yesterday












Then $B^{-1}$ won't exist.
– user3767495
yesterday




Then $B^{-1}$ won't exist.
– user3767495
yesterday




1




1




@user3767495: And so your entire argument breaks down ...
– Henning Makholm
yesterday




@user3767495: And so your entire argument breaks down ...
– Henning Makholm
yesterday










1 Answer
1






active

oldest

votes


















9














You are given $AB = B$ so that $ABA = BA$. You are also given $BA = A$ so that $A^2 = A$.



From here $A^{31} = A$ is a friendly exercise.



For that matter, so is $B^{35} = B$.



Happy holidays.






share|cite|improve this answer





















  • So, these are idempotent matrices.But why other options are invalid?
    – user3767495
    yesterday






  • 3




    The other options are not invalid. If $A^2 = A$, then $A^4 = A^3 = A^2 = A$. Similarly for $B$. Thus, all four options are equal. If you can only pick one though, I'd go with (B) as it is the "simplest" answer in some sense.
    – JimmyK4542
    yesterday












  • Thank you so much Guys :)
    – user3767495
    yesterday











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









9














You are given $AB = B$ so that $ABA = BA$. You are also given $BA = A$ so that $A^2 = A$.



From here $A^{31} = A$ is a friendly exercise.



For that matter, so is $B^{35} = B$.



Happy holidays.






share|cite|improve this answer





















  • So, these are idempotent matrices.But why other options are invalid?
    – user3767495
    yesterday






  • 3




    The other options are not invalid. If $A^2 = A$, then $A^4 = A^3 = A^2 = A$. Similarly for $B$. Thus, all four options are equal. If you can only pick one though, I'd go with (B) as it is the "simplest" answer in some sense.
    – JimmyK4542
    yesterday












  • Thank you so much Guys :)
    – user3767495
    yesterday
















9














You are given $AB = B$ so that $ABA = BA$. You are also given $BA = A$ so that $A^2 = A$.



From here $A^{31} = A$ is a friendly exercise.



For that matter, so is $B^{35} = B$.



Happy holidays.






share|cite|improve this answer





















  • So, these are idempotent matrices.But why other options are invalid?
    – user3767495
    yesterday






  • 3




    The other options are not invalid. If $A^2 = A$, then $A^4 = A^3 = A^2 = A$. Similarly for $B$. Thus, all four options are equal. If you can only pick one though, I'd go with (B) as it is the "simplest" answer in some sense.
    – JimmyK4542
    yesterday












  • Thank you so much Guys :)
    – user3767495
    yesterday














9












9








9






You are given $AB = B$ so that $ABA = BA$. You are also given $BA = A$ so that $A^2 = A$.



From here $A^{31} = A$ is a friendly exercise.



For that matter, so is $B^{35} = B$.



Happy holidays.






share|cite|improve this answer












You are given $AB = B$ so that $ABA = BA$. You are also given $BA = A$ so that $A^2 = A$.



From here $A^{31} = A$ is a friendly exercise.



For that matter, so is $B^{35} = B$.



Happy holidays.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered yesterday









Umberto P.

38.4k13064




38.4k13064












  • So, these are idempotent matrices.But why other options are invalid?
    – user3767495
    yesterday






  • 3




    The other options are not invalid. If $A^2 = A$, then $A^4 = A^3 = A^2 = A$. Similarly for $B$. Thus, all four options are equal. If you can only pick one though, I'd go with (B) as it is the "simplest" answer in some sense.
    – JimmyK4542
    yesterday












  • Thank you so much Guys :)
    – user3767495
    yesterday


















  • So, these are idempotent matrices.But why other options are invalid?
    – user3767495
    yesterday






  • 3




    The other options are not invalid. If $A^2 = A$, then $A^4 = A^3 = A^2 = A$. Similarly for $B$. Thus, all four options are equal. If you can only pick one though, I'd go with (B) as it is the "simplest" answer in some sense.
    – JimmyK4542
    yesterday












  • Thank you so much Guys :)
    – user3767495
    yesterday
















So, these are idempotent matrices.But why other options are invalid?
– user3767495
yesterday




So, these are idempotent matrices.But why other options are invalid?
– user3767495
yesterday




3




3




The other options are not invalid. If $A^2 = A$, then $A^4 = A^3 = A^2 = A$. Similarly for $B$. Thus, all four options are equal. If you can only pick one though, I'd go with (B) as it is the "simplest" answer in some sense.
– JimmyK4542
yesterday






The other options are not invalid. If $A^2 = A$, then $A^4 = A^3 = A^2 = A$. Similarly for $B$. Thus, all four options are equal. If you can only pick one though, I'd go with (B) as it is the "simplest" answer in some sense.
– JimmyK4542
yesterday














Thank you so much Guys :)
– user3767495
yesterday




Thank you so much Guys :)
– user3767495
yesterday


















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