Find formulas for the entries of Mn

Question

Let M=[1 -2

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Find formulas for the entries of Mⁿ, where n is a positive interger

Answer 1

Dang, this one brought me back. When you try to brute force this one, you'll get an infinite series of polynomials... not the best way to express a general answer, right?

Okay, here's how to think of this one - there's this tool called "diagonalizing" a matrix. You have to find matrixes P, A and P-1 such that A is diagonal, and M = PAP^-1.

Now, this leaves a really cool result.  If M = PAP^-1, then M^2 = (PAP^-1)(PAP^-1), right? 

But the associative property holds for matrixes, so M^2 = (PAP^-1)(PAP^-1) = PA(P^-1P)AP^-1

Now, since P and P-1 are inverses, then P^-1P = I, the identity matrix, and we can effectively drop I out of the equation

PA(P^-1P)AP^-1  = PA(I)AP^-1 = P(A^2)P^-1

And you can do that drop-out-P^-1P trick for LITERALLY any power of M, getting a general answer of

 

M^N = PA^NP^-1.

So, the trick is to find P, A, and P^-1.  then you only have to express your answer in terms of the entries in the diagonal matrix, and P and P inverse.

Okay, I'm way over time on this one, so I'm going to hand you the three matrixes, and show you how to find the polynomial.  I'll leave it for someone else to show you how to diagonalize in the first place.

I'm going to use the following matrix notation for the rest of this - in person I would be doing this in a more visual layout, but this interface doesn't make it easy.  You may find it more useful to copy each line down on unlined paper in a visual notation such as you used to post the problem.

M ={{1,-2},{1,4}}   
P= {{-2,-1},{1,1}}
A= {{2,0},{0,3}}
P^-1={{-1,-1},{1,2}}

so, for any N on the diagonal matrix A,

A^N = {{2^N,0},{0,3^N}}

Now we multiple on the left by P

PA^N = {{-2,-1},{1,1}}.{{2^N,0},{0,3^N}} = {{-2(2^N), -(3^N)},{(2^N),(3^N)}}

then on the right by P^-1

{{-2(2^N),-(3^N)},{(2^N),(3^N)}}.{{-1,-1},{1,2}} = {{(2*2^N-3^N),(2*(2^N)-2*(3^N))},{((3^N)-(2^N)),(2*(3^N)-(2^N))}}

leaving this as our final answer

M^N = {{(2*(2^N)-(3^N)),(2*(2^N)-2*(3^N))},{((3^N)-(2^N)),(2*(3^N)-(2^N))}}

And verify for N=1 and N=2

M^1 = {{(4-3),(4-6)},{(3-2),(6-2)}} = {{1,-2},{1,4}} = M check

M^2 = {{(8-9),(8-18)},{(9-4),(18-4)}} = {{-1,-10},{5,14}} = M^2 check

(I multiplied out M^2 the old fashioned way as a test but haven't typed that work in here.)

Were you able to follow all that?