Question

Ache a transposta da matriz A= (aij)2x2 tal que (aij) = sen (i pi/3) + cos (j pi/6)

Answer 1

$A=(a_{i,j})2x2$

 Para facilitar vamos montar a base da nossa matriz, lembrando que i representa as linhas e j as colunas.

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

 Agora vamos calcular $a_{1,1},a_{1,2}, a_{2,1}$ e$a_{2,2}$.

$a_{1,1}=Sen(\frac{\pi}{3})+Cos(\frac{\pi}{6})\\a_{1,1}=Sen(60^o)+Cos(30^o)\\a_{1,1}=\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\\a_{1,1}=\sqrt{3}\\\\a_{1,2}=Sen(\frac{\pi}{3})+Cos(\frac{2.\pi}{6})\\a_{1,2}=Sen(60^o)+Cos(\frac{\pi}{3})\\a_{1,2}=\frac{\sqrt{3}}{2}+Cos(60^o)\\a_{1,2}=\frac{\sqrt{3}}{2}+\frac{1}{2}\\a_{1,2} = \frac{1+\sqrt{3}}{2}\\\\a_{2,1}=Sen(\frac{2.\pi}{3})+Cos(\frac{\pi}{6})\\a_{2,1}=Sen(60^o)+Cos(30^o)\\a_{2,1}=\sqrt{3}$

$a_{2,2}=Sen(\frac{2.\pi}{3})+Cos(\frac{2.\pi}{6})\\a_{2,2}=Sen(60^o)+Cos(60^o)\\a_{2,2}=\frac{1+\sqrt{3}}{2}$

 Agora vamos aplicar isso na matriz:

$A=\left[\begin{array}{ccc}\sqrt{3}&\frac{1+\sqrt{3}}{2}\\\sqrt{3}&\frac{1+\sqrt{3}}{2}\end{array}\right]$

 Para encontrar sua transposta basta trocar sua linha pela coluna e sua coluna pela linha:

$A=\left[\begin{array}{ccc}\sqrt{3}&\sqrt{3}\\\frac{1+\sqrt{3}}{2}&\frac{1+\sqrt{3}}{2}\end{array}\right]$

Dúvidas só perguntar!