Question

Seja uma Matriz A3x3 tal que aij= 3i^2 - 5j, Calcule o determinante de A.​

Answer 1

Vamos começar lembrando que, no elemento geral da matriz (aij), "i" representa a linha e "j", a coluna deste elemento.

Dito isso, vamos ver como fica montada essa matriz:

$A~=~\left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right]\\\\\\\\A~=~\left[\begin{array}{ccc}3.(1)^2-5.(1)~&~3.(1)^2-5.(2)~&~3.(1)^2-5.(3)\\3.(2)^2-5.(1)~&~3.(2)^2-5.(2)~&~3.(2)^2-5.(3)\\3.(3)^2-5.(1)~&~3.(3)^2-5.(2)~&~3.(3)^2-5.(3)\end{array}\right] \\\\\\\\A~=~\left[\begin{array}{ccc}3\,.\,1-5~&~3\,.\,1-10~&~3\,.\,1-15\\3\,.\,4-5~&~3\,.\,4-10~&~3\,.\,4-15\\3\,.\,9-5~&~3\,.\,9-10~&~3\,.\,9-15\end{array}\right]$

$A~=~\left[\begin{array}{ccc}3-5~&~3-10~&~3-15\\12-5~&~12-10~&~12-15\\27-5~&~27-10~&~27-15\end{array}\right]\\\\\\\\A~=~\left[\begin{array}{ccc}-2&-7&-12\\7&2&-3\\22&17&12\end{array}\right]$

Podemos agora calcular o determinante da matriz:

$det(A)~=~\left|\begin{array}{ccc}-2&-7&-12\\7&2&-3\\22&17&12\end{array}\right|\\\\\\\\Utilizando~a~regra~de~Sarrus,~temos:\\\\\\det(A)~=~\left|\begin{array}{ccc}-2&-7&-12\\7&2&-3\\22&17&12\\-2&-7&-12\\7&2&-3\end{array}\right|$

$det(A)=\left(~(-2).2.12+7.17.(-12)+22.(-7).(-3)~\right)-\left(~(-12).2.22+(-3).17.(-2)+12.(-7).7~\right)$

$det(A)~=~\left(~-48-1428+462~\right)~-~~ \left(~-528+102-588~\right)\\\\\\det(A)~=~\left(~-1014~\right)~-~~ \left(~-1014~\right)\\\\\\det(A)~=~-1014~+~1014\\\\\\\boxed{det(A)~=~0}$