construir a matriz a=[aij]3x2 tal que aij=(i+j)elevado a 2
Soluções para a tarefa
Respondido por
4
A matriz A será do tipo: ![\left[\begin{array}{ccc} a_{11} &a_{12}\\a_{21}&a_{22}\\a_{31}&a_{32}\end{array}\right] \left[\begin{array}{ccc} a_{11} &a_{12}\\a_{21}&a_{22}\\a_{31}&a_{32}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D+a_%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)
em cada elemento, há a identificação da linha "i" e da coluna "j". Assim, temos calcular cada elemento individualmente:
![a_{ij} = (i+j)^{2} \\ a_{11} = (1+1)^{2}=4 \\ a_{12} = (1+2)^{2} = 9 \\ a_{21} = (2+1)^{2} = 9 \\ a_{22} = (2+2)^{2}=16 \\ a_{31} = (3+1)^{2}=16 \\ a_{32} = (3+2)^{2}=25 a_{ij} = (i+j)^{2} \\ a_{11} = (1+1)^{2}=4 \\ a_{12} = (1+2)^{2} = 9 \\ a_{21} = (2+1)^{2} = 9 \\ a_{22} = (2+2)^{2}=16 \\ a_{31} = (3+1)^{2}=16 \\ a_{32} = (3+2)^{2}=25](https://tex.z-dn.net/?f=+a_%7Bij%7D+%3D++%28i%2Bj%29%5E%7B2%7D+++%5C%5C++a_%7B11%7D+%3D++%281%2B1%29%5E%7B2%7D%3D4+%5C%5C+a_%7B12%7D+%3D++%281%2B2%29%5E%7B2%7D+%3D+9++++%5C%5C++a_%7B21%7D+%3D++%282%2B1%29%5E%7B2%7D+%3D+9++%5C%5C+a_%7B22%7D+%3D++%282%2B2%29%5E%7B2%7D%3D16++%5C%5C+a_%7B31%7D+%3D++%283%2B1%29%5E%7B2%7D%3D16++%5C%5C+a_%7B32%7D+%3D++%283%2B2%29%5E%7B2%7D%3D25+)
Logo A =
em cada elemento, há a identificação da linha "i" e da coluna "j". Assim, temos calcular cada elemento individualmente:
Logo A =
Danielsann:
muito obrigado cara vc e fera!
Perguntas interessantes