Question

Qual o valor de $D=\begin{bmatrix}1 &2 & 3 & 4\\ 5 & 6 & 7 &8 \\ -6 & 0&-2 &2 \\ -1 & -1 & -7 &10 \end{bmatrix}$?
Obs: utilizar Teorema de Laplace :)

Answer 1

$D = \left[\begin{array}{cccc}1&2&3&4\\5&6&7&8\\-6&0&-2&2\\-1&-1&-7&10\end{array}\right]$

Usando a expansão na coluna 2

$D=-2*\left[\begin{array}{ccc}5&7&8\\-6&-2&2\\-1&-7&10\end{array}\right] +6*\left[\begin{array}{ccc}1&3&4\\-6&-2&2\\-1&-7&10\end{array}\right] +(-1)*\left[\begin{array}{ccc}1&3&4\\5&7&8\\-6&-2&2\end{array}\right]$

Resolvendo pela Regra de Sarrus

$D = -2*696+6*323+0+(-1)*(-16)$

Multiplicando

$D = -1392+1968 + 16$

Somando

$D = 592$

Valor

Dúvidas? Comente.

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Answer 2

Resposta:

$\textsf{Segue a resposta abaixo}$

Explicação passo-a-passo:

$\sf D=\left[\begin{array}{cccc}\sf1&\sf2&\sf3&\sf4\\\sf5&\sf6&\sf7&\sf8\\\sf-6&\sf0&\sf-2&\sf2\\\sf-1&\sf-1&\sf-7&\sf10\end{array}\right]$

$\displaystyle{\mathsf{D=\sum\mathsf{a_{ij}\cdot A_{ij}}}}$

$\mathsf{ D=a_{11}\cdot A_{11}+a_{12}\cdot A_{12}+a_{13}\cdot A_{13}+a_{14}\cdot A_{14}}$

$\mathsf{D=1\cdot A_{11}+2\cdot A_{12}+3\cdot A_{13}+4\cdot A_{14} }$

$\mathsf{A_{11}=(-1)^{1+1}\cdot\left[\begin{array}{ccc}\sf 6&\sf7&\sf8\\\sf0&\sf-2&\sf2\\\sf-1&\sf-7&\sf10\end{array}\right]=1\cdot(-66)=-66 }$

$\mathsf{ A_{12}=(-1)^{1+2}\cdot\left[\begin{array}{ccc}\sf 5&\sf7&\sf8\\\sf-6&\sf-2&\sf2\\\sf-1&\sf-7&\sf10\end{array}\right]=-1\cdot696=-696}$

$\mathsf{ A_{13}=(-1)^{1+3}\cdot\left[\begin{array}{ccc}\sf 5&\sf6&\sf8\\\sf-6&\sf0&\sf2\\\sf-1&\sf-1&\sf10\end{array}\right]=1\cdot406=406}$

$\mathsf{ A_{14}=(-1)^{1+4}\cdot\left[\begin{array}{ccc}\sf 5&\sf6&\sf7\\\sf-6&\sf0&\sf-2\\\sf-1&\sf-1&\sf-7\end{array}\right]=-1\cdot(-208)=208}$

$\mathsf{D=1\cdot(-66)+2\cdot(-696)+3\cdot406+4\cdot208 }$

$\mathsf{D=-66+(-1392)+1218+832}$

$\mathsf{ D=-1458+1218+832}$

$\mathsf{D=-240+ 832}$

$\boxed{\boxed{ \mathsf{ D=592}}}$