# 1,1,1,1 Matrix general formula!!!

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Apr 27th 2008, 04:23 AM
daRitz
1,1,1,1 Matrix general formula!!!

1) if matrix X=
|1 1 |
|1 1 |

find a general formula for X^n

2) if matrix Y=
|1 -1 |
|-1 1 |

find a general formula for Y^n

3) Now find a formula for (X+Y)^n

4)Let A=aX
and B=bY

-find general expressions for A^n, and B^n, using an algebraic method to explain how you arrived at your statement.
• Apr 27th 2008, 04:25 AM
flyingsquirrel
Hi

Evaluate $X^2$ and $X^3$, guess what is $X^n$ and show it using induction.
• Apr 27th 2008, 04:27 AM
Moo
Hello,

Do it by induction.

$\begin{array}{cccc} X^2 & = & \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} & \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \\ & = & \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} & \end{array}$

Prove by induction that $X^n=\begin{pmatrix} ? & ? \\ ? & ? \end{pmatrix}$

Edited :D
• Apr 27th 2008, 04:29 AM
flyingsquirrel
Quote:

Originally Posted by Moo
Hello,
Prove by induction that $X^n=\begin{pmatrix} n & n \\ n & n \end{pmatrix}$

Take a look at $X^3$, Moo :D
• Apr 27th 2008, 04:30 AM
Isomorphism
Quote:

Originally Posted by daRitz
3) Now find a formula for (X+Y)^n

Hint: X+Y = I :)

For the induction hypothesis in the previous problem, $X^n = \begin{pmatrix} 2^{n-1} & 2^{n-1} \\ 2^{n-1} & 2^{n-1} \end{pmatrix}$
• Apr 27th 2008, 04:31 AM
Moo
Quote:

Originally Posted by flyingsquirrel
Take a look at $X^3$, Moo :D

Oh yeah, $2^{n-1}$ :D
• Apr 27th 2008, 04:32 AM
daRitz
does the formula X^n= 2^n-1 work?
• Apr 27th 2008, 04:33 AM
Moo
No, it's $X^n=\begin{pmatrix} 2^{n-1} & 2^{n-1} \\ 2^{n-1} & 2^{n-1} \end{pmatrix}$
• Apr 27th 2008, 04:37 AM
daRitz
now can anyone help me with all the other problems??(Worried)
• Apr 27th 2008, 04:39 AM
flyingsquirrel
For $Y$ the idea is exactly the same : evaluate its first powers, guess $Y^n$ and show it by induction.
• Apr 27th 2008, 04:40 AM
Isomorphism
Quote:

Originally Posted by daRitz
4)Let A=aX
and B=bY
-find general expressions for A^n, and B^n, using an algebraic method to explain how you arrived at your statement.

$A^n = a^n X^n$
You already know $X^n$, so continue....(Giggle)
• Apr 27th 2008, 05:02 AM
daRitz
but im alittle confused, am i solving for A^n?

could you please give me an example by substituting some numbers in?
• Apr 27th 2008, 05:06 AM
Isomorphism
Quote:

Originally Posted by daRitz
but im alittle confused, am i solving for A^n?

could you please give me an example by substituting some numbers in?

You know to compute X^n right?(Moo and flyingsquirrel helped you with that)

So multiply that with a^n.

You should get

$a^nX^n=a^n \begin{pmatrix} 2^{n-1} & 2^{n-1} \\ 2^{n-1} & 2^{n-1} \end{pmatrix} = \begin{pmatrix} a^n2^{n-1} & a^n2^{n-1} \\ a^n2^{n-1} & a^n2^{n-1} \end{pmatrix}$
• Apr 27th 2008, 05:20 AM
daRitz
thanks!(Clapping)
I'm now on the problem "find a general formula for (X+Y)^n

and like you said, its the identity matrix, and its looking as though it's the same thing as with finding X^n and Y^n, but im not sure how to set up the general formula, any thoughts?
• Apr 27th 2008, 05:22 AM
Isomorphism
Quote:

Originally Posted by daRitz
thanks!(Clapping)
I'm now on the problem "find a general formula for (X+Y)^n

and like you said, its the identity matrix, and its looking as though it's the same thing as with finding X^n and Y^n, but im not sure how to set up the general formula, any thoughts?

You know why its called the identity? ;)
It has a property.... You need not set up a general formula (Wink)
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last