Question

Calcule está matriz B11 e B14

Answer 1

Explicação passo-a-passo:

a)

$\sf B_{ij}=(-1)^{i+j}\cdot D_{ij}$

$\sf B_{11}=(-1)^{1+1}\cdot D_{11}$

$\sf D_{11}=\Big|\begin{array}{ccc} \sf -1 & \sf 2 & \sf 5 \\ \sf -2 & \sf -4 & \sf 3 \\ \sf -3 & \sf 0 & \sf -1 \end{array}\Big|$

$\sf D_{11}=(-1)\cdot(-4)\cdot(-1)+2\cdot3\cdot(-3)+5\cdot(-2)\cdot0-(-3)\cdot(-4)\cdot5-0\cdot3\cdot(-1)-(-1)\cdot(-2)\cdot2$

$\sf D_{11}=-4-18-0-60+0-4$

$\sf D_{11}=-22-64$

$\sf D_{11}=-86$

Assm:

$\sf B_{11}=(-1)^{1+1}\cdot D_{11}$

$\sf B_{11}=(-1)^{2}\cdot(-86)$

$\sf B_{11}=1\cdot(-86)$

$\sf \red{B_{11}=-86}$

b)

$\sf B_{ij}=(-1)^{i+j}\cdot D_{ij}$

$\sf B_{14}=(-1)^{1+4}\cdot D_{14}$

$\sf D_{14}=\Big|\begin{array}{ccc} \sf 2 & \sf -1 & \sf 2 \\ \sf 0 & \sf -2 & \sf -4 \\ \sf -1 & \sf -3 & \sf 0 \end{array}\Big|$

$\sf D_{14}=2\cdot(-2)\cdot0+(-1)\cdot(-4)\cdot(-1)+2\cdot0\cdot(-3)-(-1)\cdot(-2)\cdot2-(-3)\cdot(-4)\cdot2-0\cdot0\cdot(-1)$

$\sf D_{14}=0-4-0-4-24+0$

$\sf D_{14}=-8-24$

$\sf D_{14}=-32$

Assm:

$\sf B_{14}=(-1)^{1+4}\cdot D_{14}$

$\sf B_{14}=(-1)^{5}\cdot(-32)$

$\sf B_{14}=(-1)\cdot(-32)$

$\sf \red{B_{14}=32}$

Answer 2

Olá!

Calculando B11

Pelo Teorema de Laplace, temos que o cofator  $B_{11}$,   da matriz B será:

Fica assim:

$B_{ij}=(-1)^{i+j} ~.~ detD_{ij}$    ,   onde  i   e  j  representam a linha e coluna do elemento e $D_{ij}$  é a matriz formada depois de eliminarmos a linha e coluna do elemento  $b_{11}$  .

$B_{ij}=(-1)^{1+1} ~\times~ det\left\begin{vmatrix}{-1&~~ 2&~~ 5\\-2&-4&~~3\\-3&~~ 0&-1\end{vmatrix}\right$

O determinante vamos resolver utilizando regra de Sarrus.

$det\left\begin{vmatrix}{-1&~~ 2&~~ 5&-1&~~2\\-2&-4&~~3&-2&-4\\-3&~~ 0&-1&-3&~~0\end{vmatrix}\right\\ \\ \\ \\ \\ \left[(-1).(-4).(-1)+2.3.(-3)+5.(-2).0\right]~-~\left[5.(-4).(-3)+(-1).3.0+2.(-2).(-1)\right]=\\(-4-18)~-~(60+4)\\ \\ -22-64\\ \\\boxed{-86}$

Então:

$B_{11}=(-1)^{2} ~\times(-86)\\ B_{11}=1 ~\times(-86)\\ \\ \boxed{B_{11}=-86}$

==================================================================

Calculando B14

$B_{ij}=(-1)^{1+4} ~\times~ det\left\begin{vmatrix}{~~2&-1&~~2\\~~0&-2&-4\\-1&-3&~~0\end{vmatrix}\right$

O determinante vamos resolver utilizando regra de Sarrus.

$det\left\begin{vmatrix}{~~2&-1&~~2&~~2&-1\\~~0&-2&-4&~~0&-2\\-1&-3&~~0&-1&-3\end{vmatrix}\right\\ \\ \\ \\ \\ \left[2.(-2).0+(-1).(-4).(-1)+(2).0.(-3)\right]~-~\left[(2).(-2).(-1)+2.(-4).(-3)+(-1).0.0\right]=\\ \\-4~-~(4+24)\\ \\ -4-28\\ \\\\ \boxed{-32}$

Então:

$B_{14}=(-1)^{5} ~\times(-32)\\ B_{14}=-1 ~\times(-32)\\ \\ \boxed{B_{14}=32}$

:)