Question

Resolva o sistema, aplicando a regra de Cramer e assinale a alternativa correta:

$\left \{ { {x-y+2z=-2} \atop {3x-2y+4z=-5}}\atop{{y-3z=2} \atop } \right.$

a V = {(3, 2, 1)} 
 b V = {(-1, -1, -1 )} 
 c V = {(2, -1, 3 )} 
 d V = {( 2, -1, 4)} 
 e V = {( -2, 3, -1)}

Answer 1

Oi Wesley.

$3x-2y+4z=-5\\ 0x+y-3z=2\\ \\ \Delta =\begin{matrix} 1 & -1 & 2 \\ 3 & -2 & 4 \\ 0 & 1 & -3 \end{matrix}\begin{matrix} 1 & -1 \\ 3 & -2 \\ 0 & 1 \end{matrix}\\ \\ \Delta =6+0+6+0-4-9\\ \Delta =12-13\\ \Delta =-1$

$\Delta x=\begin{matrix} -2 & -1 & 2 \\ -5 & -2 & 4 \\ 2 & 1 & -3 \end{matrix}\begin{matrix} -2 & -1 \\ -5 & -2 \\ 2 & 1 \end{matrix}\\ \\ \Delta x=-12-8-10+8+8+15\\ \Delta x=-30+31\\ \Delta x=1$

$\Delta y=\begin{matrix} 1 & -2 & 2 \\ 3 & -5 & 4 \\ 0 & 2 & -3 \end{matrix}\begin{matrix} 1 & -2 \\ 3 & -5 \\ 0 & 2 \end{matrix}\\ \\ \Delta y=15+0+12+0-8-18\\ \Delta y=27-26\\ \Delta y=1$

$\Delta z=\begin{matrix} 1 & -1 & -2 \\ 3 & -2 & -5 \\ 0 & 1 & 2 \end{matrix}\begin{matrix} 1 & -1 \\ 3 & -2 \\ 0 & 1 \end{matrix}\\ \\ \Delta z=-4+0-6+0+5+6\\ \Delta z=-10+11\\ \Delta z=1$

$x=\frac { \Delta x }{ \Delta } \leftrightarrow \frac { 1 }{ -1 } \Leftrightarrow -1\\ \\ y=\frac { \Delta y }{ \Delta } \leftrightarrow \frac { 1 }{ -1 } \Leftrightarrow -1\\ \\ z=\frac { \Delta z }{ \Delta } \leftrightarrow \frac { 1 }{ -1 } \Leftrightarrow -1$


R:B