Question

1) Aplicando o teorema de Laplace , calcule os determinantes:

a) det A= $\left[\begin{array}{ccc}3&2&-1\\6&0&0\\2&-3&5\end{array}\right]$

b) Det A= $\left[\begin{array}{ccc}x&y&1\\3&1&1\\2&-3&1\end{array}\right]$

2) RESOLVA AS EQUAÇOES

a) $\left[\begin{array}{ccc}x-2&6\\3&5\end{array}\right] = 2$

b) $\left[\begin{array}{ccc}x+3&5\\1&x-1\end{array}\right] = 0$

Answer 1

Explicação passo-a-passo:

1)

Pelo Teorema de Laplace:

$\sf det~(A)=a_{21}\cdot A_{21}+a_{22}\cdot A_{22}+a_{23}\cdot A_{23}$

$\sf det~(A)=6\cdot A_{21}+0\cdot A_{22}+0\cdot A_{23}$

$\sf det~(A)=a_{21}\cdot A_{21}+0+0$

$\sf det~(A)=a_{21}\cdot A_{21}$

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

Temos que:

$\sf D_{12}=\left|\begin{array}{cc} \sf 2 & \sf -1 \\ \sf -3 & \sf 5 \end{array}\right|$

$\sf D_{12}=2\cdot5-(-3)\cdot(-1)$

$\sf D_{12}=10-3$

$\sf D_{12}=7$

Assim:

$\sf A_{12}=(-1)^{2+1}\cdot D_{21}$

$\sf A_{12}=(-1)^{3}\cdot7$

$\sf A_{12}=(-1)\cdot7$

$\sf A_{12}=-7$

Logo:

$\sf det~(A)=a_{21}\cdot A_{21}$

$\sf det~(A)=6\cdot(-7)$

$\sf \red{det~(A)=-42}$

b)

Pelo Teorema de Laplace:

$\sf det~(A)=a_{13}\cdot A_{13}+a_{23}\cdot A_{23}+a_{33}\cdot A_{33}$

$\sf det~(A)=1\cdot A_{13}+1\cdot A_{23}+1\cdot A_{33}$

$\sf det~(A)=A_{13}+A_{23}+A_{33}$

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

Temos que:

$\sf D_{13}=\left|\begin{array}{cc} \sf 3 & \sf 1 \\ \sf 2 & \sf -3 \end{array}\right|$

$\sf D_{13}=3\cdot(-3)-2\cdot1$

$\sf D_{13}=-9-2$

$\sf D_{13}=-11$

$\sf D_{23}=\left|\begin{array}{cc} \sf x & \sf y \\ \sf 2 & \sf -3 \end{array}\right|$

$\sf D_{23}=x\cdot(-3)-2\cdot y$

$\sf D_{23}=-3x-2y$

$\sf D_{33}=\left|\begin{array}{cc} \sf x & \sf y \\ \sf 3 & \sf 1 \end{array}\right|$

$\sf D_{33}=x\cdot1-3\cdot y$

$\sf D_{33}=x-3y$

Assim:

$\sf A_{13}=(-1)^{1+3}\cdot D_{13}$

$\sf A_{13}=(-1)^{4}\cdot(-11)$

$\sf A_{13}=1\cdot(-11)$

$\sf A_{13}=-11$

$\sf A_{23}=(-1)^{2+3}\cdot D_{23}$

$\sf A_{23}=(-1)^{5}\cdot(-3x-2y)$

$\sf A_{13}=(-1)\cdot(-3x-2y)$

$\sf A_{13}=3x+2y$

$\sf A_{33}=(-1)^{3+3}\cdot D_{33}$

$\sf A_{33}=(-1)^{6}\cdot(x-3y)$

$\sf A_{33}=1\cdot(x-3y)$

$\sf A_{33}=x-3y$

Logo:

$\sf det~(A)=A_{13}+A_{23}+A_{33}$

$\sf det~(A)=-11+3x+2y+x-3y$

$\sf \red{det~(A)=4x-y-11}$

2)

a)

$\sf 5\cdot(x-2)-3\cdot6=2$

$\sf 5x-10-18=2$

$\sf 5x-28=2$

$\sf 5x=2+28$

$\sf 5x=30$

$\sf x=\dfrac{30}{5}$

$\sf \red{x=6}$

b)

$\sf (x+3)\cdot(x-1)-1\cdot5=0$

$\sf x^2-x+3x-3-5=0$

$\sf x^2+2x-8=0$

$\sf \Delta=2^2-4\cdot1\cdot(-8)$

$\sf \Delta=4+32$

$\sf \Delta=36$

$\sf x=\dfrac{-2\pm\sqrt{36}}{2\cdot1}=\dfrac{-2\pm6}{2}$

$\sf x'=\dfrac{-2+6}{2}~\Rightarrow~x'=\dfrac{4}{2}~\Rightarrow~\red{x'=2}$

$\sf x"=\dfrac{-2-6}{2}~\Rightarrow~x"=\dfrac{-8}{2}~\Rightarrow~\red{x"=-4}$