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

**daRitz** · April 27th 2008, 03:23 AM

please someone help!!

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.

**flyingsquirrel** · April 27th 2008, 03:25 AM

Hi

Evaluate $X^2$ and $X^3$ , guess what is $X^n$ and show it using induction.

**Moo** · April 27th 2008, 03:27 AM

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

**flyingsquirrel** · April 27th 2008, 03:29 AM

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

**Isomorphism** · April 27th 2008, 03:30 AM

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, $X^n = \begin{pmatrix} 2^{n-1} & 2^{n-1} \\ 2^{n-1} & 2^{n-1} \end{pmatrix}$

**Moo** · April 27th 2008, 03:31 AM

Originally Posted by flyingsquirrel

Take a look at $X^3$ , Moo

Oh yeah, $2^{n-1}$

**daRitz** · April 27th 2008, 03:32 AM

does the formula X^n= 2^n-1 work?

**Moo** · April 27th 2008, 03:33 AM

No, it's $X^n=\begin{pmatrix} 2^{n-1} & 2^{n-1} \\ 2^{n-1} & 2^{n-1} \end{pmatrix}$

**daRitz** · April 27th 2008, 03:37 AM

now can anyone help me with all the other problems??

**flyingsquirrel** · April 27th 2008, 03:39 AM

For $Y$ the idea is exactly the same : evaluate its first powers, guess $Y^n$ and show it by induction.

**Isomorphism** · April 27th 2008, 03:40 AM

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....

**daRitz** · April 27th 2008, 04:02 AM

you're really helpfull thanks!

but im alittle confused, am i solving for A^n?

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

**Isomorphism** · April 27th 2008, 04:06 AM

Originally Posted by daRitz

you're really helpfull thanks!

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

$a^nX^n=a^n \begin{pmatrix} 2^{n-1} & 2^{n-1} \\ 2^{n-1} & 2^{n-1} \end{pmatrix} = \begin{pmatrix} a^n2^{n-1} & a^n2^{n-1} \\ a^n2^{n-1} & a^n2^{n-1} \end{pmatrix}$

**daRitz** · April 27th 2008, 04:20 AM

thanks!

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?

**Isomorphism** · April 27th 2008, 04:22 AM

Originally Posted by daRitz

thanks!

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

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