Question

quem souber ajude .............

Answer 1

A matriz é quadrada de ordem 3, então, é da forma

$A=\left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right]$

Temos que

$a_{ij}=\begin{cases}4i-3j,~i~\textless~j\\3i-2j,~i=j\\2i+4j,~i~\textgreater~j\end{cases}$

Sendo i a i-ésima linha e j a j-ésima da coluna, temos:

i = j quando estamos nos elementos da diagonal principal
i < j quando estamos em elementos acima da diagonal principal
i > j quando estamos em elementos abaixo da diagonal principal

Para simplificar o caso i = j, temos $a_{ij}=3i-2j=3i-2i=i$

Então, no geral, temos os seguintes elementos:

$a_{11}=1~(i=j)\\a_{12}=4(1)-3(2)=-2~(i~\textless~j)\\a_{13}=4(1)-3(3)=-5~(i~\textless~j)\\a_{21}=2(2)+4(1)=~~8~(i~\textgreater~j)\\a_{22}=2~(i=j)\\a_{23}=4(2)-3(3)=-1~(i~\textless~j)\\a_{31}=2(3)+4(1)=~10~(i~\textgreater~j)\\a_{32}=2(3)+4(2)=~14~(i~\textgreater~j)\\a_{33}=3~(i=j)$

Então:

$\boxed{\boxed{A=\left[\begin{array}{ccc}1&-2&-5\\8&~~2&-1\\10&~~14&~~3\end{array}\right]}}$

Encontrando o determinante de A, usando o Teorema de Laplace:

$det(A)=\left|\begin{array}{ccc}1&-2&-5\\8&~~2&-1\\10&~~14&~~3\end{array}\right|\\\\\\det(A)=(-1)^{1+1}\cdot det(A_{1})+(-1)^{1+2}\cdot8\cdot det(A_{2})+(-1)^{1+3}\cdot10\cdot det(A_{3})\\\\det(A)=det(A_{1})-8det(A_{2})+10det(A_{3})$

onde

$det(A_{1})=\left|\begin{array}{cc}2&-1\\14&~~3\end{array}\right|=2\cdot3-(-1)\cdot14=20\\\\\\det(A_{2})=\left|\begin{array}{cc}-2&-5\\14&~~3\end{array}\right|=(-2)\cdot3-(-5)\cdot14=64\\\\\\det(A_{3})=\left|\begin{array}{cc}-2&-5\\~~2&-1\end{array}\right|=(-2)\cdot(-1)-(-5)\cdot2=12$

Daí,

$det(A)=det(A_{1})-8\cdot det(A_{2})+10\cdot det(A_{3})\\\\det(A)=20-8\cdot64+10\cdot12\\\\det(A)=20-512+120\\\\\boxed{\boxed{det(A)=-372}}$