# Thread: Mathematical Induction question due in 5hrs could some1 help?

1. ## 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.

2. you know what $A^n$ is, now take that times $A$ and show that equals your desired $A^{n+1}$.

3. so you mean A^(n+1) = A^n * A ? or A * A^n ?

4. 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) ]

5. when i took $A^{n} * A$, i got the desired matrix.