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












2












$begingroup$


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









$endgroup$








  • 4




    $begingroup$
    What if $B$ is singular?
    $endgroup$
    – Umberto P.
    Dec 25 '18 at 3:48










  • $begingroup$
    Then $B^{-1}$ won't exist.
    $endgroup$
    – user3767495
    Dec 25 '18 at 3:53






  • 1




    $begingroup$
    @user3767495: And so your entire argument breaks down ...
    $endgroup$
    – Henning Makholm
    Dec 25 '18 at 3:53
















2












$begingroup$


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









$endgroup$








  • 4




    $begingroup$
    What if $B$ is singular?
    $endgroup$
    – Umberto P.
    Dec 25 '18 at 3:48










  • $begingroup$
    Then $B^{-1}$ won't exist.
    $endgroup$
    – user3767495
    Dec 25 '18 at 3:53






  • 1




    $begingroup$
    @user3767495: And so your entire argument breaks down ...
    $endgroup$
    – Henning Makholm
    Dec 25 '18 at 3:53














2












2








2





$begingroup$


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









$endgroup$




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 Dec 25 '18 at 3:45









user3767495user3767495

40218




40218








  • 4




    $begingroup$
    What if $B$ is singular?
    $endgroup$
    – Umberto P.
    Dec 25 '18 at 3:48










  • $begingroup$
    Then $B^{-1}$ won't exist.
    $endgroup$
    – user3767495
    Dec 25 '18 at 3:53






  • 1




    $begingroup$
    @user3767495: And so your entire argument breaks down ...
    $endgroup$
    – Henning Makholm
    Dec 25 '18 at 3:53














  • 4




    $begingroup$
    What if $B$ is singular?
    $endgroup$
    – Umberto P.
    Dec 25 '18 at 3:48










  • $begingroup$
    Then $B^{-1}$ won't exist.
    $endgroup$
    – user3767495
    Dec 25 '18 at 3:53






  • 1




    $begingroup$
    @user3767495: And so your entire argument breaks down ...
    $endgroup$
    – Henning Makholm
    Dec 25 '18 at 3:53








4




4




$begingroup$
What if $B$ is singular?
$endgroup$
– Umberto P.
Dec 25 '18 at 3:48




$begingroup$
What if $B$ is singular?
$endgroup$
– Umberto P.
Dec 25 '18 at 3:48












$begingroup$
Then $B^{-1}$ won't exist.
$endgroup$
– user3767495
Dec 25 '18 at 3:53




$begingroup$
Then $B^{-1}$ won't exist.
$endgroup$
– user3767495
Dec 25 '18 at 3:53




1




1




$begingroup$
@user3767495: And so your entire argument breaks down ...
$endgroup$
– Henning Makholm
Dec 25 '18 at 3:53




$begingroup$
@user3767495: And so your entire argument breaks down ...
$endgroup$
– Henning Makholm
Dec 25 '18 at 3:53










1 Answer
1






active

oldest

votes


















9












$begingroup$

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









$endgroup$













  • $begingroup$
    So, these are idempotent matrices.But why other options are invalid?
    $endgroup$
    – user3767495
    Dec 25 '18 at 3:58






  • 3




    $begingroup$
    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.
    $endgroup$
    – JimmyK4542
    Dec 25 '18 at 3:59












  • $begingroup$
    Thank you so much Guys :)
    $endgroup$
    – user3767495
    Dec 25 '18 at 4:03












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









9












$begingroup$

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









$endgroup$













  • $begingroup$
    So, these are idempotent matrices.But why other options are invalid?
    $endgroup$
    – user3767495
    Dec 25 '18 at 3:58






  • 3




    $begingroup$
    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.
    $endgroup$
    – JimmyK4542
    Dec 25 '18 at 3:59












  • $begingroup$
    Thank you so much Guys :)
    $endgroup$
    – user3767495
    Dec 25 '18 at 4:03
















9












$begingroup$

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









$endgroup$













  • $begingroup$
    So, these are idempotent matrices.But why other options are invalid?
    $endgroup$
    – user3767495
    Dec 25 '18 at 3:58






  • 3




    $begingroup$
    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.
    $endgroup$
    – JimmyK4542
    Dec 25 '18 at 3:59












  • $begingroup$
    Thank you so much Guys :)
    $endgroup$
    – user3767495
    Dec 25 '18 at 4:03














9












9








9





$begingroup$

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









$endgroup$



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 Dec 25 '18 at 3:55









Umberto P.Umberto P.

40.9k13471




40.9k13471












  • $begingroup$
    So, these are idempotent matrices.But why other options are invalid?
    $endgroup$
    – user3767495
    Dec 25 '18 at 3:58






  • 3




    $begingroup$
    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.
    $endgroup$
    – JimmyK4542
    Dec 25 '18 at 3:59












  • $begingroup$
    Thank you so much Guys :)
    $endgroup$
    – user3767495
    Dec 25 '18 at 4:03


















  • $begingroup$
    So, these are idempotent matrices.But why other options are invalid?
    $endgroup$
    – user3767495
    Dec 25 '18 at 3:58






  • 3




    $begingroup$
    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.
    $endgroup$
    – JimmyK4542
    Dec 25 '18 at 3:59












  • $begingroup$
    Thank you so much Guys :)
    $endgroup$
    – user3767495
    Dec 25 '18 at 4:03
















$begingroup$
So, these are idempotent matrices.But why other options are invalid?
$endgroup$
– user3767495
Dec 25 '18 at 3:58




$begingroup$
So, these are idempotent matrices.But why other options are invalid?
$endgroup$
– user3767495
Dec 25 '18 at 3:58




3




3




$begingroup$
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.
$endgroup$
– JimmyK4542
Dec 25 '18 at 3:59






$begingroup$
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.
$endgroup$
– JimmyK4542
Dec 25 '18 at 3:59














$begingroup$
Thank you so much Guys :)
$endgroup$
– user3767495
Dec 25 '18 at 4:03




$begingroup$
Thank you so much Guys :)
$endgroup$
– user3767495
Dec 25 '18 at 4:03


















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