Question

Seja M =$\left[\begin{array}{cccc}1&0&-1&3\\2&3&4&2\\0&2&5&1\\4&1&0&0\end{array}\right]$ Qual o Determinante de M?

Answer 1

Olá!

 Resolvendo por triangularização!!

Escalonemos a matriz considerando as propriedades de determinantes. Segue,

$\begin{vmatrix} 1 & 0 & - 1 & 3 \\ 2 & 3 & 4 & 2 \\ 0 & 2 & 5 & 1 \\ 4 & 1 & 0 & 0 \end{vmatrix} = \\\\ L_2 \rightarrow L_2 - 2 L_1 \\ L_4 \rightarrow L_4 - 4L_1\\\\\\ \begin{vmatrix} 1 & 0 & - 1 & 3 \\ 0 & 3 & 6 & - 4 \\ 0 & 2 & 5 & 1 \\ 0 & 1 & 4 & - 12 \end{vmatrix} = \\\\ L_2 \rightarrow \frac{L_2}{3} \\\\\\ 3 \cdot \begin{vmatrix} 1 & 0 & - 1 & 3 \\ 0 & 1 & 2 & \frac{- 4}{3} \\ 0 & 2 & 5 & 1 \\ 0 & 1 & 4 & - 12 \end{vmatrix} = \\\\ L_3 \rightarrow L_3 - 2 L_2 \\ L_4 \rightarrow L_4 - L_2 \\\\\\ 3 \cdot \begin{vmatrix} 1 & 0 & - 1 & 3 \\ 0 & 1 & 2 & \frac{- 4}{3} \\ 0 & 0 & 1 & \frac{11}{3} \\ 0 & 0 & 2 & \frac{- 32}{3} \end{vmatrix} = \\\\ L_4 \rightarrow L_4 - 2L_3 \\\\\\ 3 \cdot \begin{vmatrix} 1 & 0 & - 1 & 3 \\ 0 & 1 & 2 & \frac{- 4}{3} \\ 0 & 0 & 1 & \frac{11}{3} \\ 0 & 0 & 0 & - 18 \end{vmatrix} =$

 Encontramos o determinante multiplicando os elementos da diagonal principal.

 Daí,

$\det = 3 \cdot 1 \cdot 1 \cdot 1 \cdot (- 18) \\\\ \boxed{\det = - 54}$