o determinante

[tex](800) \\ - 1 2 - 3 \\ (416)[/tex]
e igual a:

a) 105
b) 120
c) 61
d) -105
e) -12

nessa segunda coluna de números e pra ser -1 2 -3

me ajudem por favor

[tex]\Large\displaystyle\text{$\begin{gathered} \tt \det A_{3\times 3}=\left|\begin{array}{ccc}8&0&0\\-1&2&-3\\4&1&6\end{array}\right| \end{gathered}$}[/tex]

Aplicando então a Regra de Sarrus, temos que:

[tex]\Large\displaystyle\text{$\begin{gathered} \tt \det A_{3\times 3}=\left|\begin{array}{ccc}8&0&0\\-1&2&-3\\4&1&6\end{array}\right| \left|\begin{array}{ccc}8&0\\-1&2\\4&1\end{array}\right| \end{gathered}$}[/tex]

[tex]\Large\displaystyle\text{$\begin{gathered} \tt \det A_{3\times 3}=96+24 \end{gathered}$}[/tex]

[tex]\Large\displaystyle\text{$\begin{gathered} \therefore \green{\underline{\boxed{\tt \det A_{3\times 3}=120}}}\ \ (\checkmark). \end{gathered}$}[/tex]

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NIELSONSOUZA36ON NIELSONSOUZA36ON · Answer 2

NIELSONSOUZA36ON NIELSONSOUZA36ON

[tex] \large\boxed{ \begin{array}{l} \begin{bmatrix}8&0&0 \\ - 1&2& - 3 \\ 4&1&6 \end{bmatrix} \\ \\ \rm \: det = \begin{bmatrix}8&0&0 \\ - 1&2& - 3 \\ 4&1&6\end{bmatrix} \begin{bmatrix}8 &0 \\ - 1&2 \\ 4&1 \end{bmatrix} \\ \\ \rm \: det = 96 + 24 \\ \\ \boxed{ \boxed{ \rm det = 120}}\end{array}}[/tex]

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