let `A=[[1,0],[1,1]]` and `I=[[1,0],[0,1]]` prove that `A^n=nA-(n-1)I, ngeq1`

If A=[[a,b],[0,1]] then prove that A^n=[[a^n,(b(a^n-1))/(a-1)],[0,1]]

If A = [(4,2),(-1,1)] and I = [(1,0),( 0,1)] prove that (A - 2I) (A - 3I) = 0

If A=[[1,1],[0,1]] , prove that A^n=[[1,n],[0,1]] for all n epsilon N

Let A=[(1,1,1),(0,1,1),(0,0,1)] prove that by induction that A^n=[(1,n,n(n+1)/2),(0,1,n),(0,0,1)] for all n in N .

Let A=[[0,1],[0,0]] , prove that: (aI+bA)^n = a^(n-1) bA , where I is the unit matrix of order 2 and n is a positive integer.

If A={:((4,2),(-1,1))and I={:((1,0),(0,1)), prove that, (A-2I)(A-3I)={:((0,0),(0,0)).

Let A=[[0, 1],[ 0, 0]] show that (a I+b A)^n=a^n I+n a^(n-1)b A , where I is the identity matrix of order 2 and n in N .

Let A = [[0,1],[0,0]] , show that (aI + bA)^n = a^nI + na^(n-1) bA , where I is the identity matrix of order 2 and n in N

If A=[(1,0),(1,1)] and I=[(1,0),(0,1)] then which one of the following holds for all nge1 by the principle of mathematica induction? (A) A^n=2^(n-1) A+(n-1)I (B) A^n=nA+(n-1) I (C) A^n=2^(n-1) A-(n-1)I (D) A^n=nA-(n-1) AI

If A=[(1,0),(1,1)] and I=[(1,0),(0,1)] then which one of the following holds for all nge1 by the principle of mathematica induction? (A) A^n=2^(n-1) A+(n-1)I (B) A^n=nA+(n-1) I (C) A^n=2^(n-1) A-(n-1)I (D) A^n=nA-(n-1) AI