Question

calcule o valor do determinante

Answer 1

Resposta:

$\sf A=\left|\begin{array}{cccc}\sf1&\sf1&\sf3&\sf1\\\sf1&\sf3&\sf3&\sf2\\\sf2&\sf5&\sf3&\sf3\\\sf1&\sf1&\sf1&\sf1\end{array}\right|$

$\sf L_4\rightarrow a_{41}=1,a_{42}=1,a_{43}=1,a_{44}$

Escolhi a quarta linha para aplicar o teorema de Laplace:

$\sf det\,A=1\cdot cof_{41}+1\cdot cof_{42}+1\cdot cof_{43}+1\cdot cof_{44}$

$\sf det\,A=cof_{41}+cof_{42}+cof_{43}+cof_{44}$

Cofator:

$\sf cof_{ij}=(-\,1)^{i+j}\cdot d_{ij}$

$\therefore$

$\sf det\,A=(-\,1)^{4+1}\cdot d_{41}+(-\,1)^{4+2}\cdot d_{42}+(-\,1)^{4+3}\cdot d_{43}+(-\,1)^{4+4}\cdot d_{44}$

$\sf det\,A=(-\,1)^5\cdot d_{41}+(-\,1)^6\cdot d_{42}+(-\,1)^7\cdot d_{43}+(-\,1)^8\cdot d_{44}$

$\sf det\,A=-\,1\cdot d_{41}+1\cdot d_{42}-1\cdot d_{43}+1\cdot d_{44}$

$\sf det\,A=-\,d_{41}+d_{42}-d_{43}+d_{44}$

$\sf det\,A=-\,\left|\begin{array}{ccc}\sf1&\sf3&\sf1\\\sf3&\sf3&\sf2\\\sf5&\sf3&\sf3\end{array}\right|+\left|\begin{array}{ccc}\sf1&\sf3&\sf1\\\sf1&\sf3&\sf2\\\sf2&\sf3&\sf3\end{array}\right|-\left|\begin{array}{ccc}\sf1&\sf1&\sf1\\\sf1&\sf3&\sf2\\\sf2&\sf5&\sf3\end{array}\right|+\left|\begin{array}{ccc}\sf1&\sf1&\sf3\\\sf1&\sf3&\sf3\\\sf2&\sf5&\sf3\end{array}\right|$

$\sf det\,A=-(9+30+9-15-27-6)+(9+3+12-6-9-6)-(9+4+5-6-3-10)+(9+6+15-18-3-15)$

$\sf det\,A=0+3-(-\,1)-6$

$\sf det\,A=3+1-6$

$\sf\red{\sf det\,A=-\,2}\leftarrow valor~do~determinante.$