Question

Resolva essa questão de Matriz!

Answer 1

Explicação passo-a-passo:

a)

$\sf A_{ij}=(-1)^{i+j}\cdot D_{ij}$

$\sf A_{22}=(-1)^{2+2}\cdot D_{22}$

• $\sf D_{22}$

$\sf D_{22}=\Big|\begin{array}{cc} \sf 2 & \sf 3 \\ \sf -2 & \sf 3 \end{array}\Big|$

$\sf D_{22}=2\cdot3-(-2)\cdot3$

$\sf D_{22}=6+6$

$\sf D_{22}=12$

Assim:

$\sf A_{22}=(-1)^{2+2}\cdot D_{22}$

$\sf A_{22}=(-1)^{4}\cdot12$

$\sf A_{22}=1\cdot12$

$\sf \red{A_{22}=12}$

b)

$\sf A_{ij}=(-1)^{i+j}\cdot D_{ij}$

$\sf A_{23}=(-1)^{2+3}\cdot D_{23}$

• $\sf D_{23}$

$\sf D_{23}=\Big|\begin{array}{cc} \sf 2 & \sf -1 \\ \sf -2 & \sf 1 \end{array}\Big|$

$\sf D_{23}=2\cdot1-(-2)\cdot(-1)$

$\sf D_{23}=2-2$

$\sf D_{23}=0$

Assim:

$\sf A_{23}=(-1)^{2+3}\cdot D_{23}$

$\sf A_{23}=(-1)^{5}\cdot0$

$\sf A_{23}=(-1)\cdot0$

$\sf \red{A_{23}=0}$

Answer 2

Olá!

Calculando A22

Pelo Teorema de Laplace, temos que o cofator  $A_{22}$,   da matriz A será:

Fica assim:

$A_{ij}=(-1)^{i+j} ~.~ detD_{ij}$    ,   onde  i   e  j  representam a linha e coluna do elemento e $D_{ij}$  é a matriz formada depois de eliminarmos a linha e coluna do elemento  $a_{22}$  .

$A_{ij}=(-1)^{2+2} ~\times~ det\left\begin{vmatrix}{~~ 2&~~3\\-2&~~3\\\end{vmatrix}\right$

O determinante vamos resolver utilizando regra de Sarrus.

$det=\left\begin{vmatrix}{~~ 2&~~3\\-2&~~3\\\end{vmatrix}\right\\ \\ \\ \left[2~.~3\right]~-~\left[3.(-2)\right]\\ \\ 6-(-6)\\ \\ \boxed{12}$

Então:

$A_{22}=(-1)^{4} ~\times(12)\\ A_{22}=1 ~\times(12)\\ \\ \boxed{A_{22}=12}$

==================================================================

Calculando A23

$A_{ij}=(-1)^{2+3} ~\times~ det\left\begin{vmatrix}{~~ 2&-1\\-2&~~1\\\end{vmatrix}\right$

O determinante vamos resolver utilizando regra de Sarrus.

$det=\left\begin{vmatrix}{~~ 2&-1\\-2&~~1\\\end{vmatrix}\right\\ \\ \\ \left[2~.~1\right]~-~\left[(-1).(-2)\right]\\ \\ 2-2\\ \\ \boxed{0}$

$A_{23}=(-1)^{5} ~\times(0)\\ \\ \boxed{A_{23}=0}$

:)