If A = [1&1 1&1] and A100 = 2kA, then the value of k is

Question

If A = [1&1 1&1] and A100 = 2kA, then the value of k is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Let A = [1&1 1&1] A2 = [1&1 1&1][1&1 1&1] = 2[1&1 1&1] = 2A A3 = 22[1&1 1&1], A4 = 23 [1&1 1&1] A3 = 22 A, A4 = 23A ∴ An = 2n-1[1&1 1&1] ⇒ A100 = 2100-1 A ⇒ A100 = 299A ⇒ 2k A = 299A ⇒ k = 99

Q. If $A = [1111]$ and $A^{100} = 2^{k} A,$ then the value of $k$ is

Let $A = [1111]$
$A^{2} = [1111] [1111] = 2 [1111] = 2 A$
$A^{3} = 2^{2} [1111], A^{4} = 2^{3} [1111]$
$A^{3} = 2^{2} A, A^{4} = 2^{3} A$
$∴ A^{n} = 2^{n - 1} [1111]$
$\Rightarrow A^{100} = 2^{100 - 1} A$
$\Rightarrow A^{100} = 2^{99} A$
$\Rightarrow 2^{k} A = 2^{99} A$
$\Rightarrow k = 99$

Q. If A=[11​11​] and A100=2kA, then the value of k is

Let A=[11​11​] A2=[11​11​][11​11​]=2[11​11​]=2A A3=22[11​11​],A4=23[11​11​] A3=22A,A4=23A ∴An=2n−1[11​11​] ⇒A100=2100−1A ⇒A100=299A ⇒2kA=299A ⇒k=99

Q. If $A = [1111]$ and $A^{100} = 2^{k} A,$ then the value of $k$ is

Let $A = [1111]$
$A^{2} = [1111] [1111] = 2 [1111] = 2 A$
$A^{3} = 2^{2} [1111], A^{4} = 2^{3} [1111]$
$A^{3} = 2^{2} A, A^{4} = 2^{3} A$
$∴ A^{n} = 2^{n - 1} [1111]$
$\Rightarrow A^{100} = 2^{100 - 1} A$
$\Rightarrow A^{100} = 2^{99} A$
$\Rightarrow 2^{k} A = 2^{99} A$
$\Rightarrow k = 99$