91Ó°ÊÓ

The given system of equations is

$\left\{\begin{array}{l}x+2y-z=0\\ 2x-4y+z=0\\ -2x+2y-3z=0\end{array}\right.$

The determinant D of the coefficients of the variables

$$\begin{gathered} D=\left|\begin{array}{ccc}1& 2& -1\\ 2& -4& 1\\ -2& 2& -3\end{array}\right| \\ D={(-1)}^{1+1}\left(1\right)\left|\begin{array}{cc}-4& 1\\ 2& -3\end{array}\right|+{(-1)}^{1+2}\left(2\right)\left|\begin{array}{cc}2& 1\\ -2& -3\end{array}\right|+{(-1)}^{1+3}(-1)\left|\begin{array}{cc}2& -4\\ -2& 2\end{array}\right| \\ D=1(12-2)-2(-6+2)-1(4-8) \\ D=10+8+4 \\ D=22 \end{gathered}$$

$D≠0,$so Cramer's rule can be used to determine the solution of the system of equations.

so that$x=\frac{D_x}{D},y=\frac{D_y}{D},z=\frac{D_z}{D}$

Determine $D_x$by replacing the coefficients of x in Dwith the constants

$$\begin{gathered} D_x=\left|\begin{array}{ccc}0& 2& -1\\ 0& -4& 1\\ 0& 2& -3\end{array}\right| \\ {D}_x={(-1)}^{1+1}\left(0\right)\left|\begin{array}{cc}-4& 1\\ 2& -3\end{array}\right|+{(-1)}^{1+2}\left(2\right)\left|\begin{array}{cc}0& 1\\ 0& -3\end{array}\right|+{(-1)}^{1+3}(-1)\left|\begin{array}{cc}0& -4\\ 0& 2\end{array}\right| \\ {D}_x=0 \end{gathered}$$

Determine $D_y$by replacing the coefficients of y in Dwith the constants

$$\begin{gathered} D_y=\left|\begin{array}{ccc}1& 0& -1\\ 2& 0& 1\\ -2& 0& -3\end{array}\right| \\ {D}_y={(-1)}^{1+1}\left(1\right)\left|\begin{array}{cc}0& 1\\ 0& -3\end{array}\right|+{(-1)}^{1+2}\left(0\right)\left|\begin{array}{cc}2& 1\\ -2& -3\end{array}\right|+{(-1)}^{1+3}(-1)\left|\begin{array}{cc}2& 0\\ -2& 0\end{array}\right| \\ {D}_y=0 \end{gathered}$$

Determine $D_z$by replacing the coefficients of z in Dwith the constants

$$\begin{gathered} D_z=\left|\begin{array}{ccc}1& 2& 0\\ 2& -4& 0\\ -2& 2& 0\end{array}\right| \\ {D}_z={(-1)}^{1+1}\left(1\right)\left|\begin{array}{cc}-4& 0\\ 2& 0\end{array}\right|+{(-1)}^{1+2}\left(2\right)\left|\begin{array}{cc}2& 0\\ -2& 0\end{array}\right|+{(-1)}^{1+3}\left(0\right)\left|\begin{array}{cc}2& -4\\ -2& 2\end{array}\right| \\ {D}_z=0 \end{gathered}$$

Solution for x

$$\begin{gathered} x=\frac{D_x}{D} \\ x=\frac{0}{22} \\ x=0 \end{gathered}$$

Solution for y

$$\begin{gathered} y=\frac{D_y}{D} \\ y=\frac{0}{22} \\ y=0 \end{gathered}$$

Solution for z

$$\begin{gathered} z=\frac{D_z}{D} \\ z=\frac{0}{22} \\ z=0 \end{gathered}$$

So the solution of the system of equations is$(0,0,0)$