Answer to Question #223067 in Linear Algebra for Haroun

Question

A is a 2x2 matrix

Let A = {(2) (0) (0) (-1)}. Find A³²

Answer 1

$A=\begin{pmatrix} 2 & 0 \\ 0 & -1 \end{pmatrix}$

$A^2=\begin{pmatrix} 2 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 0 & -1 \end{pmatrix}=\begin{pmatrix} 2^2 & 0 \\ 0 & (-1)^2 \end{pmatrix}$

$=\begin{pmatrix} 4 & 0 \\ 0 & 1 \end{pmatrix}$

$A^4=(A^2)^2=\begin{pmatrix} 4 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 4 & 0 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 2^4 & 0 \\ 0 & (-1)^4 \end{pmatrix}$

$=\begin{pmatrix} 16 & 0 \\ 0 & 1 \end{pmatrix}$

$A^8=(A^4)^2=\begin{pmatrix} 16 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 16 & 0 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 2^8 & 0 \\ 0 & (-1)^8 \end{pmatrix}$

$=\begin{pmatrix} 256 & 0 \\ 0 & 1 \end{pmatrix}$

$A^{16}=(A^8)^2=\begin{pmatrix} 256 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 256 & 0 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 2^{16} & 0 \\ 0 & (-1)^{16} \end{pmatrix}$

$=\begin{pmatrix} 65536 & 0 \\ 0 & 1 \end{pmatrix}$

$A^{32}=(A^{16})^2=\begin{pmatrix} 65536 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 65536 & 0 \\ 0 & 1 \end{pmatrix}$

$=\begin{pmatrix} 2^{32} & 0 \\ 0 & (-1)^{32} \end{pmatrix}=\begin{pmatrix} 4294967296 & 0 \\ 0 & 1 \end{pmatrix}$