Q. 39 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 15 – 42, solve each system of equations using Cramer’s Rule if it is applicable. If Cramer’s Rule is not applicable, say so.

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

Step-by-Step Solution

Answer

The solution of the system of equations$\left\{\begin{array}{l}x+2y-z=0\\ 2x-4y+z=0\\ -2x+2y-3z=0\end{array}\right.$ is $(0,0,0)$

Step 1. Given information

The given system of equations is   

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

Step 2. Determinant

The determinant D of the coefficients of the variables 

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

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

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

Step 3. Value of D x

Determine $D_x$by replacing the coefficients of x in D with the constants 

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

Step 4. Value of D y

Determine $D_y$by replacing the coefficients of y in D with the constants 

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

Step 5. Value of D z

Determine $D_z$by replacing the coefficients of z in D with the constants 

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

Step 6. Solution of system

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 $(0,0,0)$