Mathematical Induction question due in 5hrs could some1 help?

• Oct 27th 2008, 11:28 AM
chogz
Mathematical Induction question due in 5hrs could some1 help?
This question is in matrices.

A^n = [ 1 3^n-1 ; 0 3^n ] for all integers n>=1 where A = [ 1 2 ; 0 3 ]

I got the base step

(B) let P(1) equal A^n = [ 1 3^n-1 ; 0 3^n ].

A^1 = [ 1 3^1-1 ; 0 3^1 ] gives A^1 = [ 1 2 ; 0 3 ] which is the same as the given matrix A^1. Thus P(1) is true.

(R) inductive step
Assume P(n) is true for some (n) E A.
I figured that A^(n+1) = [ 1 3^(n+1)-1 ; 0 3^(n+1) ].

However this is the part that i got stuck because i've never actually done any questions regarding matrices with mathematical induction so im lost.

Could someone help me on this question? or even just give me the starting line for it.
• Oct 27th 2008, 11:49 AM
chiph588@
you know what $A^n$ is, now take that times $A$ and show that equals your desired $A^{n+1}$.
• Oct 27th 2008, 11:56 AM
chogz
so you mean A^(n+1) = A^n * A ? or A * A^n ?
• Oct 27th 2008, 12:10 PM
chogz
Given :
A^n = [1 3^(n) -1 ; 0 3^(n) ]
A = [1 2 ; 0 3 ]

when i multiplied A^n by A i got A^n+1 = [ 1x1 + (3^n - 1)x0 , 1x2 + 3^nx(3^n -1) ; 0x1 + 3^nx0 , 0x2 + 3^nx3 ]

which gives me [1 2+3^(n+1)-1 ; 0 3^(n+1) ]

how do i get rid of the +2? because if i add it with 3^(n+1) i wud get 3^(n+1) + 1 which would disagree with the equation A^(n+1) = [ 1 3^(n+1)-1 ; 0 3^(n+1) ]
• Oct 27th 2008, 02:57 PM
chiph588@
when i took $A^{n} * A$, i got the desired matrix.