Question

Se considera matricea A=$\left[\begin{array}{cc}3&0\\0&1\\\end{array}\right]$
Sa se determine transpusa matricei B=A+$A^{2} +A^{3} +...+A^{2016}$

Answer 1

Răspuns:

$B^T=\begin{bmatrix} 3+3^2+...+3^{2016}&0\\0&2016\end{bmatrix}$

Explicație pas cu pas:

Este un exercițiu ușor de a arăta că $diag(\lambda_1,...,\lambda_n)^k=diag(\lambda_1^k,...,\lambda_n^k)$

Answer 2

Explicație pas cu pas:

$A=\left(\begin{array}{ccc}3&0\\0&1\end{array}\right)$

Calculam A².

$A^2=\left(\begin{array}{ccc}3&0\\0&1\end{array}\right) *\left(\begin{array}{ccc}3&0\\0&1\end{array}\right) =\left(\begin{array}{ccc}3^2&0\\0&1\end{array}\right)$

Calculam A³.

$A^3=A^2*A=\left(\begin{array}{ccc}3^2&0\\0&1\end{array}\right) *\left(\begin{array}{ccc}3^3&0\\0&1\end{array}\right)$

Observam ca Aⁿ are forma:

$A^n=\left(\begin{array}{ccc}3^n&0\\0&1\end{array}\right)$

Demonstram prin inductie ca:

$P(n): A^n=\left(\begin{array}{ccc}3^n&0\\0&1\end{array}\right) ~este~adevarata.$

1) Etapa de verificare:

$P(1): A=\left(\begin{array}{ccc}3^1&0\\0&1\end{array}\right) =\left(\begin{array}{ccc}3&0\\0&1\end{array}\right)$

$P(2): A^2=\left(\begin{array}{ccc}3^2&0\\0&1\end{array}\right) ~verificata~anterior$

2) Etapa inductiva:

$Presupunem~P(k): A^k=\left(\begin{array}{ccc}3^k&0\\0&1\end{array}\right) ~adevarata$

$Demonstram~P(k+1): A^{k+1}=\left(\begin{array}{ccc}3^{k+1}&0\\0&1\end{array}\right) ~adevarata.$

$A^{k+1}=A^k*A=\left(\begin{array}{ccc}3^k&0\\0&1\end{array}\right) *\left(\begin{array}{ccc}3&0\\0&1\end{array}\right) =\left(\begin{array}{ccc}3^k*3&0\\0&1\end{array}\right) =\left(\begin{array}{ccc}3^{k+1}&0\\0&1\end{array}\right)$

Deci, avem ca P(k+1) este adevarata.

Din 1) si 2), conform principiului I al inductiei matematice, P(n) este adevarata.

Determinam B.

$B=A+A^2+....+A^{2016}=\left(\begin{array}{ccc}3&0\\0&1\end{array}\right) +\left(\begin{array}{ccc}3^2&0\\0&1\end{array}\right) +...+\left(\begin{array}{ccc}3^{2016}&0\\0&1\end{array}\right) =\left(\begin{array}{ccc}3+3^2+...+3^{2016}&0\\0&1+1+...+1\end{array}\right) =\left(\begin{array}{ccc}3*\frac{3^{2016}-1}{3-1} &0\\0&2016\end{array}\right) =\left(\begin{array}{ccc}\frac{3^{2017}-3}{2} &0\\0&1\end{array}\right)$

Determinam transpusa matricii B.

$B^t=\left(\begin{array}{ccc}\frac{3^{2017}-3}{2} &0\\0&1\end{array}\right)$