If A = [2&-2 -2&2] then An = 2k A, where k =

Question

If A = [2&-2 -2&2] then An = 2k A, where k = (A) 2n -1 (B) n + 1 (C) n - 1 (D) 2(n -1)

Tardigrade Admin · Accepted Answer

A2 = [2&-2 -2&2][2&-2 -2&2] =[8&-8 -8&8] =4A = 22A A3 =A2 .A = 4A .A = 4 (4A) = 16A = 24 A A4= A3 .A = 16A.A= 16(4A) =64A=26A ∴ By inspection k = 2(n - 1)

Q. If $A = [2 - 2 - 2 2]$ then $A^{n} = 2^{k} A,$ where k =

$2^{n - 1}$

n + 1

n - 1

2(n -1)

$A^{2} = [2 - 2 - 2 2] [2 - 2 - 2 2] = [8 - 8 - 8 8] = 4 A = 2^{2} A$
$A^{3} = A^{2} . A = 4 A . A = 4 (4 A) = 16 A = 2^{4} A$
$A^{4} = A^{3} . A = 16 A . A = 16 (4 A) = 64 A = 2^{6} A$
$∴$ By inspection $k = 2 (n - 1)$

Q. If A=[2−2​−22​] then An=2kA, where k =

2n−1