Question

Fie matricea A =
(1. 1)
(-2. -1)
Găsiți cel mai mic număr natural n€N* pentru care A^n = I2​

Answer 1

$\it A^2=\begin{pmatrix}1&1\\ \\ -2&-1\end{pmatrix}\cdot\begin{pmatrix}1&1\\ \\ -2&-1\end{pmatrix}=\begin{pmatrix}-1&0\\ \\ 0&-1\end{pmatrix}=-I_2\\ \\ \\ A^4=(A^2)^2=(-I_2)^2=I_2$

Așadar,  pentru n = 4,   A⁴ =  I₂

Answer 2

Răspuns:

n = 4

Explicație pas cu pas:

$A = \left[\begin{array}{ccc}1&1\\-2&-1\\\end{array}\right]$

$A^{2} = \left[\begin{array}{ccc}1&1\\-2&-1\\\end{array}\right] *\left[\begin{array}{ccc}1&1\\-2&-1\\\end{array}\right] = \left[\begin{array}{ccc}1-2&1-1\\-2+2&-2+1\\\end{array}\right] = \left[\begin{array}{ccc}-1&0\\0&-1\\\end{array}\right]$

$A^{3} = \left[\begin{array}{ccc}-1&0\\0&-1\\\end{array}\right] *\left[\begin{array}{ccc}1&1\\-2&-1\\\end{array}\right] = \left[\begin{array}{ccc}-1&-1\\2&1\\\end{array}\right]$

$A^{4} = \left[\begin{array}{ccc}-1&-1\\2&1\\\end{array}\right] * \left[\begin{array}{ccc}1&1\\-2&-1\\\end{array}\right] = \left[\begin{array}{ccc}-1+2&-1+1\\2-2&2-1\\\end{array}\right] = \left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right] = I_{2}$

Așadar, pentru n = 4, A² = I₂