Question

a=(aij) 3x3 tal que aij=j² + 2j

Answer 1

• A matriz A de ordem 3x3 dada pela lei aij=j² + 2j é igual a:

$\large\displaystyle\begin{aligned} A_{3\cdot3}\Leftrightarrow \left[\begin{array}{ccc}3&8&15\\3&8&15\\3&8&15\end{array}\right]\end{aligned}$

Dado que cada termo da matriz a₃.₃ = (aij) segue a lei aij=j² + 2j, temos que:

$\large\displaystyle\begin{aligned} A_{3\cdot3}\Leftrightarrow \left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right] \end{aligned}$

• Dada a lei  aij=j² + 2j. Logo:

$\large\displaystyle\begin{aligned} A_{3\cdot3}\Leftrightarrow \left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right] \end{aligned}$

$\large\displaystyle\begin{aligned} A_{3\cdot3}\Leftrightarrow \left[\begin{array}{ccc}(1^2+2\cdot1)&(2^2+2\cdot2)&(3^2+2\cdot3)\\(1^2+2\cdot1)&(2^2+2\cdot2)&(3^2+2\cdot3)\\(1^2+2\cdot1)&(2^2+2\cdot2)&(3^2+2\cdot3)\end{array}\right] \end{aligned}$

• Agora, basta resolvermos normalmente, feito isso teremos enfim encontrado a matriz dada inicialmente crua.

$\large\displaystyle\begin{aligned} A_{3\cdot3}\Leftrightarrow \left[\begin{array}{ccc}(1^2+2\cdot1)&(2^2+2\cdot2)&(3^2+2\cdot3)\\(1^2+2\cdot1)&(2^2+2\cdot2)&(3^2+2\cdot3)\\(1^2+2\cdot1)&(2^2+2\cdot2)&(3^2+2\cdot3)\end{array}\right] \end{aligned}$

$\large\displaystyle\begin{aligned} A_{3\cdot3}\Leftrightarrow \left[\begin{array}{ccc}(1+2)&(4+4)&(9+6)\\(1+2)&(4+4)&(9+6)\\(1+2)&(4+4)&(9+6)\end{array}\right] \end{aligned}$

$\large\displaystyle\text{ \begin{aligned} \therefore \boxed{\boxed{\green{A_{3\cdot3}\Leftrightarrow \left[\begin{array}{ccc}3&8&15\\3&8&15\\3&8&15\end{array}\right] }}}\end{aligned} }$

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Matrizes.

$\blue{\square}$ brainly.com.br/tarefa/156868

Answer 2

Resposta:

$A_3_X_3=\left[\begin{array}{ccc}3&8&15\\3&8&15\\3&8&15\end{array}\right]$

Explicação passo a passo:

$A_3_X_3=\left[\begin{array}{ccc}a_1_1&a_1_2&a_1_3\\a_2_1&a_2_2&a_2_3\\a_3_1&a_3_2&a_3_3\end{array}\right]\\\\ a_1_1=1^2+2.1=1+2=3\\a_1_2=2^2+2.2=4+4=8\\a_1_3=3^2+2.3=9+6=15\\a_2_1=1^2+2.1=1+2=3\\a_2_2=2^2+2.2=4+4=8\\a_2_3=3^2+2.3=9+6=15\\a_3_1=1^2+2.1=1+4=3\\a_3_2=2^2+2.2=4+4=8\\a_3_3=3^2+2.3=9+6=15$