Question

O determinante é igual a ???

Answer 1

Resposta:

D

Explicação passo-a-passo:

Pelo Teorema de Laplace:

$A=\left(\begin{array}{cccc} 2&3&0&5 \\ 1&2&3&0 \\ 1&0&4&1 \\ 4&0&2&1 \end{array}\right)$

$\text{Det}~A=a_{12}\cdot A_{12}+a_{22}\cdot A_{22}+a_{32}\cdot A_{32}+a_{42}\cdot A_{42}$

Temos que:

$A_{12}=(-1)^{1+2}\cdot D_{12}$

$D_{12}=\left(\begin{array}{ccc} 1&3&0 \\ 1&4&1 \\ 4&2&1 \end{array} \right)$

$\text{Det}~D_{12}=1\cdot4\cdot1+3\cdot1\cdot4+0\cdot1\cdot2-4\cdot4\cdot0-2\cdot1\cdot1-1\cdot1\cdot3$

$\text{Det}~D_{12}=4+12+0-0-2-3$

$\text{Det}~D_{12}=11$

$A_{12}=(-1)^3\cdot11$

$A_{12}=(-1)\cdot11$

$A_{12}=-11$

$A_{22}=(-1)^{2+2}\cdot D_{22}$

$D_{22}=\left(\begin{array}{ccc} 2&0&5 \\ 1&4&1 \\ 4&2&1 \end{array} \right)$

$\text{Det}~D_{12}=2\cdot4\cdot1+0\cdot1\cdot4+5\cdot1\cdot2-4\cdot4\cdot5-2\cdot1\cdot2-1\cdot1\cdot0$

$\text{Det}~D_{22}=8+0+10-80-4-0$

$\text{Det}~D_{22}=-66$

$A_{12}=(-1)^4\cdot(-66)$

$A_{12}=1\cdot(-66)$

$A_{12}=-66$

Logo:

$\text{Det}~A=3\cdot(-11)+2\cdot(-66)+0\cdot A_{32}+0\cdot A_{42}$

$\text{Det}~A=-33-132+0+0$

$\text{Det}~A=-165$

Letra D

Answer 2

Vamos desenvolver o determinante da matriz segundo os elementos da 2ª linha.

$\mathsf{A}=\begin{bmatrix}2&3&0&5\\1&2&3&0\\1&0&4&1\\4&0&2&1\end{bmatrix}$

$\mathsf{det~A=a_{21}.A_{21}+a_{22}.A_{22}+a_{23}.A_{23}}$

$\mathsf{A_{21}}=- 1.\begin{vmatrix}3&0&5\\0&4&1\\0&2&1\end{vmatrix}</p><p>$

$\mathsf{A_{21}=-1[3.2]=-6}$

$\mathsf{A_{22}}=\begin{vmatrix}2&0&5\\1&4&1\\4&2&1\end{vmatrix}$

$\mathsf{A_{22}=2(4-2)+5(2-16)=4-70=-66}$

$\mathsf{A_{23}}=-1.\begin{vmatrix}2&3&5\\1&0&1\\4&0&1\end{vmatrix}$

$\mathsf{A_{23}=-1[-3(1-4)]=-1.(9)=-9}$

$\mathsf{det~A=a_{21}.A_{21}+a_{22}.A_{22}+a_{23}.A_{23}}\\\mathsf{det~A=1.(-6)+2.(-66)+3.(-9)}\\\mathsf{det~A=-6-132-27=-165}$