alguém pra responder essa questão?
Anexos:
Luz1414:
sim, sim. já me ajudou bastante! consigo desenrolar o desenvolvimento completo.
percebi
tem um método mais rápido e prático para desenvolvê-las?
Sim. Escolhe a primeira linha (a que tem mais zeros) e usa Laplace. Assim tu elimina dois Cofatores.
Olha esse vídeo: https://www.youtube.com/watch?v=pQDLbqNiVz0
tá ok
Aqui deu 747 O.o
o det B
o resultado final deu -795
Mique . Use Laplace. Vai econtrar det A = 233 e det B= 699
Respostas
respondido por:
1
Oi. Segue o Cálculo
Calculando o det de A:
![detA=4.C_{11}+(-3).C_{12}+0.C_{13}+0.C_{14} \\ \\ 4.(-1)^{1+1}. \left[\begin{array}{ccc}2&3&1\\5&2&1\\2&3&-3\end{array}\right] -3(-1)^{1+2}. \left[\begin{array}{ccc}1&3&1\\2&2&1\\4&3&-3\end{array}\right] +0+0 \\ \\ \\ \\detA= 4(-1)^2.(44)-3(-1)^3.(19) \\ \\detA= 4.(44)+3(19) \\ \\ detA=176+57 \\ \\ \boxed{detA=233}](https://tex.z-dn.net/?f=detA%3D4.C_%7B11%7D%2B%28-3%29.C_%7B12%7D%2B0.C_%7B13%7D%2B0.C_%7B14%7D+%5C%5C++%5C%5C+4.%28-1%29%5E%7B1%2B1%7D.++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26amp%3B3%26amp%3B1%5C%5C5%26amp%3B2%26amp%3B1%5C%5C2%26amp%3B3%26amp%3B-3%5Cend%7Barray%7D%5Cright%5D+-3%28-1%29%5E%7B1%2B2%7D.++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26amp%3B3%26amp%3B1%5C%5C2%26amp%3B2%26amp%3B1%5C%5C4%26amp%3B3%26amp%3B-3%5Cend%7Barray%7D%5Cright%5D+%2B0%2B0+%5C%5C++%5C%5C++%5C%5C++%5C%5CdetA%3D+4%28-1%29%5E2.%2844%29-3%28-1%29%5E3.%2819%29+%5C%5C++%5C%5CdetA%3D+4.%2844%29%2B3%2819%29+%5C%5C++%5C%5C+detA%3D176%2B57+%5C%5C++%5C%5C+%5Cboxed%7BdetA%3D233%7D "detA=4.C_{11}+(-3).C_{12}+0.C_{13}+0.C_{14} \\ \\ 4.(-1)^{1+1}. \left[\begin{array}{ccc}2&3&1\\5&2&1\\2&3&-3\end{array}\right] -3(-1)^{1+2}. \left[\begin{array}{ccc}1&3&1\\2&2&1\\4&3&-3\end{array}\right] +0+0 \\ \\ \\ \\detA= 4(-1)^2.(44)-3(-1)^3.(19) \\ \\detA= 4.(44)+3(19) \\ \\ detA=176+57 \\ \\ \boxed{detA=233}")
Calculando o det de B:
![detB=4.C_{11}+(-3).C_{12}+0.C_{13}+0.C_{14} \\ \\ detB=4.(-1)^{1+1}. \left[\begin{array}{ccc}2&3&3\\5&2&3\\2&3&-9\end{array}\right] -3(-1)^{1+2}. \left[\begin{array}{ccc}1&3&3\\2&2&3\\4&3&-9\end{array}\right] +0+0 \\ \\ \\ \\detB= 4(-1)^2.(132)-3(-1)^3.(57) \\ \\detB= 4.(132)+3(57) \\ \\ detB=528+171 \\ \\ \boxed{detB=699}](https://tex.z-dn.net/?f=detB%3D4.C_%7B11%7D%2B%28-3%29.C_%7B12%7D%2B0.C_%7B13%7D%2B0.C_%7B14%7D+%5C%5C++%5C%5C+detB%3D4.%28-1%29%5E%7B1%2B1%7D.++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26amp%3B3%26amp%3B3%5C%5C5%26amp%3B2%26amp%3B3%5C%5C2%26amp%3B3%26amp%3B-9%5Cend%7Barray%7D%5Cright%5D+-3%28-1%29%5E%7B1%2B2%7D.++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26amp%3B3%26amp%3B3%5C%5C2%26amp%3B2%26amp%3B3%5C%5C4%26amp%3B3%26amp%3B-9%5Cend%7Barray%7D%5Cright%5D+%2B0%2B0+%5C%5C++%5C%5C++%5C%5C++%5C%5CdetB%3D+4%28-1%29%5E2.%28132%29-3%28-1%29%5E3.%2857%29+%5C%5C++%5C%5CdetB%3D+4.%28132%29%2B3%2857%29+%5C%5C++%5C%5C+detB%3D528%2B171+%5C%5C++%5C%5C+%5Cboxed%7BdetB%3D699%7D "detB=4.C_{11}+(-3).C_{12}+0.C_{13}+0.C_{14} \\ \\ detB=4.(-1)^{1+1}. \left[\begin{array}{ccc}2&3&3\\5&2&3\\2&3&-9\end{array}\right] -3(-1)^{1+2}. \left[\begin{array}{ccc}1&3&3\\2&2&3\\4&3&-9\end{array}\right] +0+0 \\ \\ \\ \\detB= 4(-1)^2.(132)-3(-1)^3.(57) \\ \\detB= 4.(132)+3(57) \\ \\ detB=528+171 \\ \\ \boxed{detB=699}")
Calculando:
Calculando o det de A:
Calculando o det de B:
Calculando:
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