Sejam A=\left[a_{i j}\right]_{2 \times 2} e B=\left[b_{i j}\right]_{2 \times 2} matrizes quadradas de ordem 2 tais que:
a_{i j}=\left\{\begin{array}{cc} i^{i}, & \text { se } i=j \\\ \frac{i}{j}, & \text { se } i \neq j \end{array} \quad \text { e } \quad b_{i j}=\left\{\begin{array}{cl} (i+1)^{i}, & \text { se } i=j \\\ \frac{1}{j+1}, & \text { se } i \neq j \end{array} .\right.\right.
Determine 6 A-12 B.

Passo 1

O enunciado define duas matrizes 2\times2\), \[A = [a_{ij}] \quad\text{e}\quad B = [b_{ij}],
com
a_{ij} = \begin{cases} i^i, & i=j,\\[6pt] \dfrac{i}{j}, & i\neq j, \end{cases} \qquad b_{ij} = \begin{cases} (i+1)^i, & i=j,\\[6pt] \dfrac{1}{\,j+1\,}, & i\neq j. \end{cases}
Precisamos calcular 6A - 12B\]. Vamos primeiro obter as matrizes \(A e B.]

Passo 2

Calcular os elementos de A:
a_{11} = 1^1 = 1, \quad a_{12} = \frac{1}{2}, \quad a_{21} = \frac{2}{1} = 2, \quad a_{22} = 2^2 = 4.
Logo,
A = \begin{pmatrix} 1 & \tfrac12 \\[4pt] 2 & 4 \end{pmatrix}.]

Passo 3

Calcular os elementos de B:
b_{11} = (1+1)^1 = 2, \quad b_{12} = \frac{1}{2+1} = \tfrac13, \quad b_{21} = \frac{1}{1+1} = \tfrac12, \quad b_{22} = (2+1)^2 = 9.
Logo,
B = \begin{pmatrix} 2 & \tfrac13 \\[4pt] \tfrac12 & 9 \end{pmatrix}.]

Passo 4

Escalar as matrizes:
6A = 6\begin{pmatrix}1 & \tfrac12 \\[4pt] 2 & 4\end{pmatrix} = \begin{pmatrix}6 & 3 \\[4pt] 12 & 24\end{pmatrix}, \qquad 12B = 12\begin{pmatrix}2 & \tfrac13 \\[4pt] \tfrac12 & 9\end{pmatrix} = \begin{pmatrix}24 & 4 \\[4pt] 6 & 108\end{pmatrix}.]

Passo 5

Subtrair termo a termo:
6A - 12B = \begin{pmatrix}6-24 & 3-4 \\[4pt] 12-6 & 24-108\end{pmatrix} = \begin{pmatrix}-18 & -1 \\[4pt] 6 & -84\end{pmatrix}.]

Resposta

6A - 12B = \begin{pmatrix} -18 & -1 \\[4pt] 6 & -84 \end{pmatrix}.]

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