You can probably find out what you are looking for if you download my course pack for Linear Algebra (201-NYC-05) from my course web page and check out the problem sheet for chapter 9; the link to my web page is in the description of every video. Let me know. :-)
Since (A+B)^2=A^2 + AB + BA + B^2 in general, the first equality you wrote implies that AB=BA. This is often stated as A & B commute. It is impossible to imply your second equality from this property. I would suggest to revisit the original problem that led you to ask this question.
Your post is incomplete, but I am guessing that it ends with the following: the inverse of I-A. Simply multiply I-A by I+A+A^2 and show that the result is equal to I.
@@slcmathpcanother doubt i have: (A^n)^m=A^(nm) where n,m belong to natural number and n,m may or may not be same........Is it true for every matrix A?
Let Q=(A^n)....then Q(A^-1)^n= (A.A.A....)(n times)A^-1.A^-1.....(n times)...[by definition of Apow_n...]...by associativity first the innermost A.A^-1 becomes 'I' then recursively the whole expression will become 'I' (BY CANDY CRUSH LOGIC)....hence Q inverse is (A^-1)^n...
Thank you so much for the help!!!!
- from A level further maths student
Thank you so much for that tip! Never thought of translating the equations first.
🤣🤣🤣
the video is helpful tbh
Thank man much appreciated 😊, Hope you're having a nice day
Thanks
Thank u.it's realy helpful.
very helpful and to the point
thanku!
Nice explanation sir
Thank you 🙏🏻 😭👏🏻👏🏻👏🏻👏🏻👏🏻💗
Thank you...
Thank u so so much🙏😊
Superb
Pls work 4 us properties of adjoint too😢😢plssss
You can probably find out what you are looking for if you download my course pack for Linear Algebra (201-NYC-05) from my course web page and check out the problem sheet for chapter 9; the link to my web page is in the description of every video. Let me know. :-)
I think it's just a generalization of 6 when all the matrices are the same (A).
It is not a generalization, but a special case. It is best seen by expanding A^n = AAA***A (n-times).
Ah, I see. Thanks for the video, btw. :)
How can I deduce (A+B)^2=A^2+2AB+B^2 to B=A^-1??
Since (A+B)^2=A^2 + AB + BA + B^2 in general, the first equality you wrote implies that AB=BA. This is often stated as A & B commute. It is impossible to imply your second equality from this property. I would suggest to revisit the original problem that led you to ask this question.
If A^3 = 0 then I + A+A^2 is
Your post is incomplete, but I am guessing that it ends with the following: the inverse of I-A.
Simply multiply I-A by I+A+A^2 and show that the result is equal to I.
Can you help me out.....is det(A^n)=(det(A))^n where n is a natural number true for every A?
Yes indeed, the determinant is a multiplicative function.
@@slcmathpcanother doubt i have: (A^n)^m=A^(nm) where n,m belong to natural number and n,m may or may not be same........Is it true for every matrix A?
Expand both sides and you will see that it is true. ;-)
IAdj.AI=IAI`N-1
Proof of 7?
Prove by induction.
Let Q=(A^n)....then Q(A^-1)^n= (A.A.A....)(n times)A^-1.A^-1.....(n times)...[by definition of Apow_n...]...by associativity first the innermost A.A^-1 becomes 'I' then recursively the whole expression will become 'I' (BY CANDY CRUSH LOGIC)....hence Q inverse is (A^-1)^n...