Question

Considere a matriz A = (aij )3×3 tal que aij = 2i − j.
Calcule o determinante de A.

Answer 1

Explicação passo-a-passo:

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

Sabemos que $a_{ij}=2i-j$.

$a_{11} =2.1-1=2-1=1\\a_{12} =2.1-2=2-2=0\\a_{13} =2.1-3=2-3=-1\\a_{21} =2.2-1=4-1=3\\a_{22} =2.2-2=4-2=2\\a_{23} =2.2-3=4-3=1\\a_{31} =2.3-1=6-1=5\\a_{32} =2.3-2=6-2=4\\a_{33} =2.3-3=6-3=3$

Temos a matriz A.

$A=\left[\begin{array}{ccc}1&0&-1\\3&2&1\\5&4&3\end{array}\right]$

Calculando o determinante(Regra de Sarrus):

$detA=\left|\begin{array}{ccc}1&0&-1\\3&2&1\\5&4&3\end{array}\right|\left\begin{array}{ccc}1&0\\3&2\\5&4\end{array}\right|$

$detA=1.2.3+0.1.5+(-1).3.4-0.3.3-1.1.4-(-1).2.5\\\\detA=6+0-12-0-4+10\\\\detA=0$