Q. 40 - 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.

$\left\{\begin{array}{l}x+4y-3z=0\\ 3x-y+3z=0\\ x+y+6z=0\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the system of equations$\left\{\begin{array}{l}x+4y-3z=0\\ 3x-y+3z=0\\ x+y+6z=0\end{array}\right.$ is  $(0,0,0)$

Step 1. Given information

The given system of equations is   

$\left\{\begin{array}{l}x+4y-3z=0\\ 3x-y+3z=0\\ x+y+6z=0\end{array}\right.$

Step 2. Determinant

The determinant D of the coefficients of the variables 

$$\begin{gathered} D=\left|\begin{array}{ccc}1& 4& -3\\ 3& -1& 3\\ 1& 1& 6\end{array}\right| \\ D={(-1)}^{1+1}\left(1\right)\left|\begin{array}{cc}-1& 3\\ 1& 6\end{array}\right|+{(-1)}^{1+2}\left(4\right)\left|\begin{array}{cc}3& 3\\ 1& 6\end{array}\right|+{(-1)}^{1+3}(-3)\left|\begin{array}{cc}3& -1\\ 1& 1\end{array}\right| \\ D=1(-6-3)-4(18-3)-3(3+1) \\ D=-9-60-12 \\ D=-81 \end{gathered}$$

$D\ne 0,$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. Solution of system

All constants of the system of the equation are zeros

so

 $$\begin{gathered} D_x=0 \\ {D}_y=0 \\ {D}_z=0 \end{gathered}$$

Determinants $D\ne 0$ and all constants of the system of the equation are zeros so the system of equations has a trivial solution

$$\begin{gathered} x=0 \\ y=0 \\ z=0 \end{gathered}$$

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