Question

DETERMINANTES>> Utilize o calculo de determinantes de ordem 1 e 2, as propriedades dos determinantes e pelo menos uma vez o Teorema de Laplace, regra de Sarrus para resolver as questoes a seguir..>>>>>>>>>> SÓ RESPONDA SE SOUBER..SE NAO VOU DENUNCIAR

Answer 1

Explicação passo-a-passo:

a) $\sf A=\left[\begin{array}{c} \sf 24 \end{array}\right]$

$\sf \red{det~(A)=24}$

b) $\sf B=\left[\begin{array}{cc} \sf 10 & \sf -1 \\ \sf 8 & \sf 9 \end{array}\right]$

$\sf det~(B)=10\cdot9-8\cdot(-1)$

$\sf det~(B)=90+8$

$\sf \red{det~(B)=98}$

c) $\sf C=\left[\begin{array}{ccc} \sf 1 & \sf 2 & \sf 0 \\ \sf -2 & \sf 3 &\sf 4 \\ \sf 2 & \sf 5 & \sf 1 \end{array}\right]$

Pela regra de Sarrus:

$\sf det~(C)=1\cdot3\cdot1+2\cdot4\cdot2+0\cdot(-2)\cdot5-2\cdot3\cdot0-5\cdot4\cdot1-1\cdot(-2)\cdot2$

$\sf det~(C)=3+16-0-0-20+4$

$\sf det~(C)=23-20$

$\sf \red{det~(C)=3}$

d) $\sf D=\left[\begin{array}{cccc} \sf 4 & \sf 0 & \sf 2 & \sf -1 \\ \sf 2 & \sf 0 &\sf 3 & \sf 0 \\ \sf 1 & \sf 5 & \sf 12 & \sf 0 \\ \sf -2 & \sf 0 & \sf 3 & \sf 2 \end{array}\right]$

Pela Teorema de Laplace:

$\sf det~(D)=a_{12}\cdot A_{12}+a_{22}\cdot A_{22}+a_{32}\cdot A_{32}+a_{42}\cdot A_{42}$

$\sf det~(D)=0\cdot A_{12}+0\cdot A_{22}+5\cdot A_{32}+0\cdot A_{42}$

$\sf det~(D)=0+0+5\cdot A_{32}+0$

$\sf det~(D)=5\cdot A_{32}$

• $\sf A_{ij}=(-1)^{i+j}\cdot D_{ij}$

$\sf D_{32}=\left|\begin{array}{ccc} \sf 4 & \sf 2 & \sf -1 \\ \sf 2 & \sf 3 &\sf 0 \\ \sf -2 & \sf 3 & \sf 2 \end{array}\right|$

$\sf D_{32}=4\cdot3\cdot2+2\cdot0\cdot(-2)+(-1)\cdot2\cdot3-(-2)\cdot3\cdot(-1)-3\cdot0\cdot4-2\cdot2\cdot2$

$\sf D_{32}=24-0-6-6-0-8$

$\sf D_{32}=24-20$

$\sf D_{32}=4$

Assim:

$\sf A_{32}=(-1)^{3+2}\cdot D_{32}$

$\sf A_{32}=(-1)^{5}\cdot4$

$\sf A_{32}=(-1)\cdot4$

$\sf A_{32}=-4$

Logo:

$\sf det~(D)=5\cdot A_{32}$

$\sf det~(D)=5\cdot(-4)$

$\sf \red{det~(D)=-20}$

Answer 2

sei naoResposta:

Explicação passo-a-passo: