Thread: Matrix multiplication problem

Hi,
I'm trying to solve such a problem that says:


A= [a,b;0,1] "2x2 matrix"

Find A^n

I've tried A,A^2,A^3 to get the formula but I couldn't

Hi ahmedzoro10!

What did you get for A^2 and A^3?

Originally Posted by ILikeSerena

Hi ahmedzoro10!

What did you get for A^2 and A^3?

I got a new matrix for each

Originally Posted by ahmedzoro10

I got a new matrix for each

Which new matrices?

A^2= [a^2,ab+b;0,1]
A^3= [a^3,b(a^2+a+1);0,1]
A^4= [a^4,b(a^3+a^2+a+1);0,1]

Right!

I see a clear pattern. Do you?

Hint: multiply b(a^3+a^2+a+1) with (a-1)/(a-1) to find a simpler expression that does not have an infinite number of terms (if n tends to infinity).

Originally Posted by ILikeSerena

Right!

I see a clear pattern. Do you?

Hint: multiply b(a^3+a^2+a+1) with (a-1)/(a-1) to find a simpler expression that does not have an infinite number of terms (if n tends to infinity).

Woow! I got it,Thanks!!

it's [a^n,b((a^n -1)/a-1),0,1],right?

Right!

Originally Posted by ILikeSerena

Right!

Thanks man!

another question,please

I'm trying to prove that:

A(A+B)^-1 B=B((A+B)^-1 A)= (A^-1 + B^-1)^-1

A,B and A+B are non singular matrices

Originally Posted by ahmedzoro10

another question,please

I'm trying to prove that:

A(A+B)^-1 B=B((A+B)^-1 A)= (A^-1 + B^-1)^-1

A,B and A+B are non singular matrices

Hmm, since matrix multiplication is associative, is not generally true...

Originally Posted by ILikeSerena

Hmm, since matrix multiplication is associative, is not generally true...

Oh ok

Originally Posted by ahmedzoro10

Oh ok

It seems your problem statement is a little off.
Is it perhaps incomplete?
Or is there a typo in it?

Originally Posted by ILikeSerena

It seems your problem statement is a little off.
Is it perhaps incomplete?
Or is there a typo in it?

I've already revised it and it seems cool
The second term could be like that B(A+B)^{-1} A (i.e. with no parenthesis)

Originally Posted by ahmedzoro10

I've already revised it and it seems cool
The second term could be like that B(A+B)^{-1} A (i.e. with no parenthesis)

Hmm, as yet I'm not sure of it.

Either way, as you can see here, your equality with the 3rd part does not hold:
WolframAlpha - counter example