Question

Resolver os sistemas 3 x 3 utilizando a regra de Cramer.

Answer 1

Explicação passo-a-passo:

a) $\sf \begin{cases} \sf 2x+2y+3z=1 \\ \sf 2x+y+z=0 \\ \sf 5x+2y+z=0 \end{cases}$

$\sf D=\left(\begin{array}{ccc} \sf 2& \sf 2& \sf 3 \\ \sf 2& \sf 1& \sf 1 \\ \sf 5& \sf 2& \sf 1\end{array}\right)$

$\sf det~(D)=2+10+12-15-4-4$

$\sf det~(D)=1$

$\sf D_x=\left(\begin{array}{ccc} \sf 1& \sf 2& \sf 3 \\ \sf 0& \sf 1& \sf 1 \\ \sf 0& \sf 2& \sf 1\end{array}\right)$

$\sf det~(D_x)=1+0+0-0-2-0$

$\sf det~(D_x)=-1$

$\sf D_y=\left(\begin{array}{ccc} \sf 2& \sf 1& \sf 3 \\ \sf 2& \sf 0& \sf 1 \\ \sf 5& \sf 0& \sf 1\end{array}\right)$

$\sf det~(D_y)=0+5+0-0-0-2$

$\sf det~(D_y)=3$

$\sf D_z=\left(\begin{array}{ccc} \sf 2& \sf 2& \sf 1 \\ \sf 2& \sf 1& \sf 0 \\ \sf 5& \sf 2& \sf 0\end{array}\right)$

$\sf det~(D_z)=0+0+4-5-0-0$

$\sf det~(D_z)=-1$

Assim:

• $\sf x=\dfrac{det~(D_x)}{det~(D)}~\rightarrow~x=\dfrac{-1}{1}~\rightarrow~x=-1$

• $\sf y=\dfrac{det~(D_y)}{det~(D)}~\rightarrow~y=\dfrac{3}{1}~\rightarrow~y=3$

• $\sf z=\dfrac{det~(D_z)}{det~(D)}~\rightarrow~z=\dfrac{-1}{1}~\rightarrow~z=-1$

$\sf S=\{(-1,3,-1)\}$

b) $\sf \begin{cases} \sf 2x+2y+3z=10 \\ \sf 2x+y+z=0 \\ \sf 5x+2y+z=5 \end{cases}$

$\sf D=\left(\begin{array}{ccc} \sf 2& \sf 2& \sf 3 \\ \sf 2& \sf 1& \sf 1 \\ \sf 5& \sf 2& \sf 1\end{array}\right)$

$\sf det~(D)=2+10+12-15-4-4$

$\sf det~(D)=1$

$\sf D_x=\left(\begin{array}{ccc} \sf 10& \sf 2& \sf 3 \\ \sf 0& \sf 1& \sf 1 \\ \sf 5& \sf 2& \sf 1\end{array}\right)$

$\sf det~(D_x)=10+10+0-15-20-0$

$\sf det~(D_x)=-15$

$\sf D_y=\left(\begin{array}{ccc} \sf 2& \sf 10& \sf 3 \\ \sf 2& \sf 0& \sf 1 \\ \sf 5& \sf 5& \sf 1\end{array}\right)$

$\sf det~(D_y)=0+50+30-0-10-20$

$\sf det~(D_y)=50$

$\sf D_z=\left(\begin{array}{ccc} \sf 2& \sf 2& \sf 10 \\ \sf 2& \sf 1& \sf 0 \\ \sf 5& \sf 2& \sf 5\end{array}\right)$

$\sf det~(D_z)=10+0+40-50-0-20$

$\sf det~(D_z)=-20$

Assim:

• $\sf x=\dfrac{det~(D_x)}{det~(D)}~\rightarrow~x=\dfrac{-15}{1}~\rightarrow~x=-15$

• $\sf y=\dfrac{det~(D_y)}{det~(D)}~\rightarrow~y=\dfrac{50}{1}~\rightarrow~y=50$

• $\sf z=\dfrac{det~(D_z)}{det~(D)}~\rightarrow~z=\dfrac{-20}{1}~\rightarrow~z=-20$

$\sf S=\{(-15,50,-20)\}$

c) $\sf \begin{cases} \sf 2x+2y+3z=10 \\ \sf 2x+y+z=5 \\ \sf 5x+2y+z=5 \end{cases}$

$\sf D=\left(\begin{array}{ccc} \sf 2& \sf 2& \sf 3 \\ \sf 2& \sf 1& \sf 1 \\ \sf 5& \sf 2& \sf 1\end{array}\right)$

$\sf det~(D)=2+10+12-15-4-4$

$\sf det~(D)=1$

$\sf D_x=\left(\begin{array}{ccc} \sf 10& \sf 2& \sf 3 \\ \sf 5& \sf 1& \sf 1 \\ \sf 5& \sf 2& \sf 1\end{array}\right)$

$\sf det~(D_x)=10+10+30-15-20-10$

$\sf det~(D_x)=5$

$\sf D_y=\left(\begin{array}{ccc} \sf 2& \sf 10& \sf 3 \\ \sf 2& \sf 5& \sf 1 \\ \sf 5& \sf 5& \sf 1\end{array}\right)$

$\sf det~(D_y)=10+50+30-75-10-20$

$\sf det~(D_y)=-15$

$\sf D_z=\left(\begin{array}{ccc} \sf 2& \sf 2& \sf 10 \\ \sf 2& \sf 1& \sf 5 \\ \sf 5& \sf 2& \sf 5\end{array}\right)$

$\sf det~(D_z)=10+50+40-50-20-20$

$\sf det~(D_z)=10$

Assim:

• $\sf x=\dfrac{det~(D_x)}{det~(D)}~\rightarrow~x=\dfrac{5}{1}~\rightarrow~x=5$

• $\sf y=\dfrac{det~(D_y)}{det~(D)}~\rightarrow~y=\dfrac{-15}{1}~\rightarrow~y=-15$

• $\sf z=\dfrac{det~(D_z)}{det~(D)}~\rightarrow~z=\dfrac{10}{1}~\rightarrow~z=10$

$\sf S=\{(5,-15,10)\}$

d) $\sf \begin{cases} \sf 4x+2y+3z=10 \\ \sf 2x+y+2z=7 \\ \sf 5x+2y+2z=5 \end{cases}$

$\sf D=\left(\begin{array}{ccc} \sf 4& \sf 2& \sf 3 \\ \sf 2& \sf 1& \sf 2 \\ \sf 5& \sf 2& \sf 2\end{array}\right)$

$\sf det~(D)=8+20+12-15-16-8$

$\sf det~(D)=1$

$\sf D_x=\left(\begin{array}{ccc} \sf 10& \sf 2& \sf 3 \\ \sf 7& \sf 1& \sf 2 \\ \sf 5& \sf 2& \sf 2\end{array}\right)$

$\sf det~(D_x)=20+20+42-15-40-28$

$\sf det~(D_x)=-1$

$\sf D_y=\left(\begin{array}{ccc} \sf 4& \sf 10& \sf 3 \\ \sf 2& \sf 7& \sf 2 \\ \sf 5& \sf 5& \sf 2\end{array}\right)$

$\sf det~(D_y)=56+100+30-105-40-40$

$\sf det~(D_y)=1$

$\sf D_z=\left(\begin{array}{ccc} \sf 4& \sf 2& \sf 10 \\ \sf 2& \sf 1& \sf 7 \\ \sf 5& \sf 2& \sf 5\end{array}\right)$

$\sf det~(D_z)=20+70+40-50-56-20$

$\sf det~(D_z)=4$

Assim:

• $\sf x=\dfrac{det~(D_x)}{det~(D)}~\rightarrow~x=\dfrac{-1}{1}~\rightarrow~x=-1$

• $\sf y=\dfrac{det~(D_y)}{det~(D)}~\rightarrow~y=\dfrac{1}{1}~\rightarrow~y=1$

• $\sf z=\dfrac{det~(D_z)}{det~(D)}~\rightarrow~z=\dfrac{4}{1}~\rightarrow~z=4$

$\sf S=\{(-1,1,4)\}$