basic rule in matrix
Just want to know the reason behind the highlighted step ..
Like if the squaring is there so why isn't it
P^-2A^2P^2 ?
![No description](https://cdn.answeroverflow.com/1260968195678928946/Screenshot_20240711-2004002.jpg)
![No description](https://cdn.answeroverflow.com/1260968196005957743/Screenshot_20240711-2003482.jpg)
18 Replies
@Apu
Note for OP
+solved @user1 @user2...
to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.(P^-1AP)^2
= (P^-1AP)(P^-1AP)
=(P^-1A^2P)
The P outside bracket in the first highlight seems like a typo
Oh yes
Could you please elaborate this step like why only ...A^2.. not P^-2A^2P^2 ?
(AB)^2 = ABAB and not A^2B^2
Matrix multiplication is not commutative
![No description](https://cdn.answeroverflow.com/1260974183312003173/rn_image_picker_lib_temp_c4396143-ab81-436c-922e-c4edd3731301.jpg)
yes friend
just remove the brackets now
P will be multiplied with P^-1, i.e. identity matrix
hence A will be multiplied by A, A^2
P^-1 on the left and P on the right remain as is
just like Bs are not multiplied together
P and P-1 wont be
you’re not allowed to move anything
So AA = A^2
But (AB)^2= ABAB
Is this correct ?
![No description](https://cdn.answeroverflow.com/1260975266810040490/1260974183731429376remix-1720710327338.png)
yes correct
Okay 👍
that’s how you multiply
Please check this one@πrate
![No description](https://cdn.answeroverflow.com/1260975895754313819/rn_image_picker_lib_temp_d9bae0ba-d3a6-452d-80bf-8ff19fa044e9.jpg)
![No description](https://cdn.answeroverflow.com/1260990688724521020/1260975896152768604remix-1720714002806.png)
Okay ..
Understood
@πrate
![No description](https://cdn.answeroverflow.com/1261008181413281884/image0.png)
+ solved @πrate
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