Question

qual o valor do determinante da matriz. obs: aplique teorema de Laplace
$1 \: \: \: \: \: \: \: 0 \: \ \: \: \: \: \: \: 2 \: \: \: \: 0 \\ \: 3 \: \: - 2 \: \: \: \: \: \: \: \:1 \: \: \: 5 \\ 6 \: \: \: \: \: \: \: \: 0 \: \: - 1 \: \: \: \: 4 \\ - 5 \: \: \: \: 0 \: \: \: \: \: \: \: \: 3 \: \: \: \: 2$
a)12
b)-60
c)156
d)-78​

Answer 1

$\mathtt{A=\begin{vmatrix}1&0&2&0\\3&-2&1&5\\6&0&-1&4\\-5&0&3&2\end{vmatrix}}$

$\mathtt{det~A=a_{11}.A_{11}+a_{12}.A_{12}+a_{13}.A_{13}+a_{14}.A_{14}}\\\mathtt{a_{11}A_{11}+0.A_{12}+a_{13}.A_{13}+0.A_{14}}\\\mathtt{a_{11}.A_{11}+a_{13}.A_{13}}$

$\mathtt{A_{11}={(-1)}^{1+1}.\begin{vmatrix}-2&1&5\\0&-1&4\\0&3&2\end{vmatrix}} \\\mathtt{A_{11} = - 2( - 2 - 12) = 28}$

$\mathtt{A_{13}={(-1)}^{1+3}.\begin{vmatrix}3& - 2&5\\6&0&4\\ - 5&0&2\end{vmatrix}} \\\mathtt{A_{13} = 2(12 + 20) = 64 }$

$\mathtt{det~A=1.28+2.64=28+128=156}$

$\huge\boxed{\boxed{\mathtt{\maltese~~Alternativa~c}}}$