Solveeit Logo
Login

Question

Mathematics Question on Invertible Matrices

If A=[2−2 −22]A = \begin{bmatrix}2&-2\\\ -2&2\end{bmatrix} then An=2kA,A^n = 2^k A, where k =

A

2n−12^{n -1}

B

n + 1

C

n - 1

D

2(n -1)

Answer

2(n -1)

Explanation

Solution

A2=[2−2 −22][2−2 −22]=[8−8 −88]=4A=22AA^{2} = \begin{bmatrix}2&-2\\\ -2&2\end{bmatrix}\begin{bmatrix}2&-2\\\ -2&2\end{bmatrix} =\begin{bmatrix}8&-8\\\ -8&8\end{bmatrix} =4A = 2^{2}A
A3=A2.A=4A.A=4(4A)=16A=24AA^{3} =A^{2} .A = 4A .A = 4 \left(4A\right) = 16A = 2^{4} A
A4=A3.A=16A.A=16(4A)=64A=26AA^{4}= A^{3} .A = 16A.A= 16\left(4A\right) =64A=2^{6}A
∴\therefore By inspection k=2(n−1)k = 2(n - 1)