Question

Seja A = (a ij) a matriz quadrada de ordem 3, onde

______| i + 2, se i < j |
____aij | i + j, se i = j |
______| j - 1, se i > j |


O valor do determinante de A é igual a:

a) -40

b) 56

c) 40

d) -56

e) 0
Alguém me ajuda presiso do cálculo todo completo

Answer 1

Desejamos calcular o determinante da matriz A, sendo ele uma matriz cuja a lei de formação é igual a:

i + 2, se i < j

i + j, se i = j

j - 1, se i > j

Sendo a matriz A dada da seguinte forma:

$\Large\displaystyle\begin{gathered}\sf A_{3\times 3}= \left[\begin{array}{ccc}\tt a_{11}&\tt a_{12}&\tt a_{13}\\\tt a_{21}&\tt a_{22}&\tt a_{23}\\\tt a_{31}&\tt a_{32}&\tt a_{33}\end{array}\right] \end{gathered}$

Onde $\large\displaystyle\begin{gathered}\tt a_{ij} \end{gathered}$ sendo $\large\displaystyle\begin{gathered}\tt i=linhas \end{gathered}$ e $\large\displaystyle\begin{gathered}\tt j=colunas \end{gathered}$. Com isso, temos que:

$\Large\displaystyle\begin{gathered}\sf A_{3\times 3}= \left[\begin{array}{ccc}\tt i+j&\tt i+2&\tt i+2\\\tt j-1&\tt i+j&\tt i+2\\\tt j-1&\tt j-1&\tt i+j\end{array}\right] \end{gathered}$

$\Large\displaystyle\begin{gathered}\sf A_{3\times 3}= \left[\begin{array}{ccc}\tt 1+1&\tt 1+2&\tt 1+2\\\tt 1-1&\tt 2+2&\tt 2+2\\\tt 1-1&\tt 2-1&\tt 3+3\end{array}\right] \end{gathered}$

$\Large\displaystyle\begin{gathered}\therefore \sf A_{3\times 3}= \left[\begin{array}{ccc}\tt 2&\tt 3&\tt 3\\\tt 0&\tt 4&\tt 4\\\tt 0&\tt 1&\tt 6\end{array}\right] \end{gathered}$

Agora é só calcular o determinante, e para isso irei utilizar a Regra de Sarrus.

$\Large\displaystyle\begin{gathered}\tt \det\left(\sf A_{3\times 3}\right)= \left|\begin{array}{ccc}\tt 2&\tt 3&\tt 3\\\tt 0&\tt 4&\tt 4\\\tt 0&\tt 1&\tt 6\end{array}\right| \left|\begin{array}{ccc}\tt 2&\tt 3\\\tt 0&\tt 4\\\tt 0&\tt 1\end{array}\right| \end{gathered}$

$\Large\displaystyle\begin{gathered}\tt \det\left(\sf A_{3\times 3}\right)= 48+0+0-\left(0+8+0\right) \end{gathered}$

$\Large\displaystyle\begin{gathered}\tt \det\left(\sf A_{3\times 3}\right)= 48-8 \end{gathered}$

$\Large\displaystyle\text{ \begin{gathered}\boxed{\tt \det\left(\sf A_{3\times 3}\right)= 40}\end{gathered} }$