Question

considere a seguir a matriz A= (aij) 3x3
2 1 log (2 em baixo e 6 em cima)
1 -2 4
3 log (2 em baixo e 4 em cima) 1
pela regra de sarrus, o determinante dessa matriz é:
a)8
b)9
c)15
d)24

Answer 1

$A=\begin{bmatrix}2&1&\log_26\\1&-2&4\\3&\log_24&1\end{bmatrix}\\\\\\\text{Regra de Sarrus:}\\\\\det(A)=\\(a_{1,1}\cdot{a}_{2,2}\cdot{a}_{3,3}+a_{1,2}\cdot{a}_{2,3}\cdot{a}_{3,1}+a_{1,3}\cdot{a}_{2,1}\cdot{a}_{3,2})-\\(a_{1,3}\cdot{a}_{2,2}\cdot{a}_{3,1}+a_{1,1}\cdot{a}_{2,3}\cdot{a}_{3,2}+a_{1,2}\cdot{a}_{2,1}\cdot{a}_{3,3})\\\\\\\det(A)=\\(2\cdot-2\cdot1+1\cdot4\cdot3+\log_26\cdot1\cdot\log_24)-\\(\log_26\cdot-2\cdot3+2\cdot4\cdot\log_24+1\cdot1\cdot1)\\\\\det(A)=(-4+12+2\log_26)-(-6\log_26+16+1)$

$\det(A)=8+2\log_26-(-6\log_26+17)\\\\\det(A)=8+2\log_26+6\log_26-17\\\\\boxed{\det(A)=-9+8\log_26}$