1- Obtenha a matriz A :
a) A=[aij]3x2, onde aij= (-1)elevado a i.(2i-3j)
Celio:
Victor, corrija o trecho i²=3j
Victor, tem certeza que é (-1)¹ ? Isto dá sempre -1. Não é (-1)^i ?
Verdade, é (-1) elevado a i
Edite a questão, então, por favor, ok? :)
Soluções para a tarefa
Respondido por
5
Olá, Victor.
![A=[a_{ij}]_{3\times2},\text{ onde }a_{ij}= (-1)^ i.(2i-3j)\\\\
a_{11}=(-1)^1\cdot(2-3)=(-1)(-1)=1\\\\
a_{12}=(-1)^1\cdot(2-6)=(-1)(-4)=4\\\\
a_{21}=(-1)^2\cdot(4-3)=1\cdot1=1\\\\
a_{22}=(-1)^2\cdot(4-6)=1\cdot(-2)=-2\\\\
a_{31}=(-1)^3\cdot(6-3)=(-1)\cdot3=-3\\\\
a_{32}=(-1)^3\cdot(6-6)=0\\\\
A= \left[\begin{array}{ccc}a_{11}&a_{12}\\a_{21}&a_{22}\\a_{31}&a_{32}\end{array}\right] =\left[\begin{array}{ccc}1&4\\1&-2\\-3&0\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Ba_%7Bij%7D%5D_%7B3%5Ctimes2%7D%2C%5Ctext%7B+onde+%7Da_%7Bij%7D%3D+%28-1%29%5E+i.%282i-3j%29%5C%5C%5C%5C%0Aa_%7B11%7D%3D%28-1%29%5E1%5Ccdot%282-3%29%3D%28-1%29%28-1%29%3D1%5C%5C%5C%5C%0Aa_%7B12%7D%3D%28-1%29%5E1%5Ccdot%282-6%29%3D%28-1%29%28-4%29%3D4%5C%5C%5C%5C%0Aa_%7B21%7D%3D%28-1%29%5E2%5Ccdot%284-3%29%3D1%5Ccdot1%3D1%5C%5C%5C%5C%0Aa_%7B22%7D%3D%28-1%29%5E2%5Ccdot%284-6%29%3D1%5Ccdot%28-2%29%3D-2%5C%5C%5C%5C%0Aa_%7B31%7D%3D%28-1%29%5E3%5Ccdot%286-3%29%3D%28-1%29%5Ccdot3%3D-3%5C%5C%5C%5C%0Aa_%7B32%7D%3D%28-1%29%5E3%5Ccdot%286-6%29%3D0%5C%5C%5C%5C%0AA%3D+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_%7B11%7D%26amp%3Ba_%7B12%7D%5C%5Ca_%7B21%7D%26amp%3Ba_%7B22%7D%5C%5Ca_%7B31%7D%26amp%3Ba_%7B32%7D%5Cend%7Barray%7D%5Cright%5D+%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26amp%3B4%5C%5C1%26amp%3B-2%5C%5C-3%26amp%3B0%5Cend%7Barray%7D%5Cright%5D+ "A=[a_{ij}]_{3\times2},\text{ onde }a_{ij}= (-1)^ i.(2i-3j)\\\\
a_{11}=(-1)^1\cdot(2-3)=(-1)(-1)=1\\\\
a_{12}=(-1)^1\cdot(2-6)=(-1)(-4)=4\\\\
a_{21}=(-1)^2\cdot(4-3)=1\cdot1=1\\\\
a_{22}=(-1)^2\cdot(4-6)=1\cdot(-2)=-2\\\\
a_{31}=(-1)^3\cdot(6-3)=(-1)\cdot3=-3\\\\
a_{32}=(-1)^3\cdot(6-6)=0\\\\
A= \left[\begin{array}{ccc}a_{11}&a_{12}\\a_{21}&a_{22}\\a_{31}&a_{32}\end{array}\right] =\left[\begin{array}{ccc}1&4\\1&-2\\-3&0\end{array}\right]")
Muito obrigado.
De nada. Disponha. :)
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