Q. 36 - 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.

$\left\{\begin{array}{l}x+4y-3z=-8\\ 3x-y+3z=12\\ x+y+6z=1\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the system of equations$\left\{\begin{array}{l}x+4y-3z=-8\\ 3x-y+3z=12\\ x+y+6z=1\end{array}\right.$ is$\left(3,\frac{-8}{3},\frac{1}{9}\right)$

Step 1. Given information

The given system of equations is   

$\left\{\begin{array}{l}x+4y-3z=-8\\ 3x-y+3z=12\\ x+y+6z=1\end{array}\right.$

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. Value of D x

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

$$\begin{gathered} D_x=\left|\begin{array}{ccc}-8& 4& -3\\ 12& -1& 3\\ 1& 1& 6\end{array}\right| \\ {D}_x={(-1)}^{1+1}(-8)\left|\begin{array}{cc}-1& 3\\ 1& 6\end{array}\right|+{(-1)}^{1+2}\left(4\right)\left|\begin{array}{cc}12& 3\\ 1& 6\end{array}\right|+{(-1)}^{1+3}(-3)\left|\begin{array}{cc}12& -1\\ 1& 1\end{array}\right| \\ {D}_x=-243 \end{gathered}$$

Step 4. Value of D y

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

$$\begin{gathered} D_y=\left|\begin{array}{ccc}1& -8& -3\\ 3& 12& 3\\ 1& 1& 6\end{array}\right| \\ {D}_y={(-1)}^{1+1}\left(1\right)\left|\begin{array}{cc}12& 3\\ 1& 6\end{array}\right|+{(-1)}^{1+2}(-8)\left|\begin{array}{cc}3& 3\\ 1& 6\end{array}\right|+{(-1)}^{1+3}(-3)\left|\begin{array}{cc}3& 12\\ 1& 1\end{array}\right| \\ {D}_y=216 \end{gathered}$$

Step 5. Value of D z

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

$$\begin{gathered} D_z=\left|\begin{array}{ccc}1& 4& -8\\ 3& -1& 12\\ 1& 1& 1\end{array}\right| \\ {D}_z={(-1)}^{1+1}\left(1\right)\left|\begin{array}{cc}-1& 12\\ 1& 1\end{array}\right|+{(-1)}^{1+2}\left(4\right)\left|\begin{array}{cc}3& 12\\ 1& 1\end{array}\right|+{(-1)}^{1+3}(-3)\left|\begin{array}{cc}3& -1\\ 1& 1\end{array}\right| \\ {D}_z=-9 \end{gathered}$$

Step 6. Solution of system

Solution for x  

$$\begin{gathered} x=\frac{D_x}{D} \\ x=\frac{-243}{-81} \\ x=3 \end{gathered}$$

Solution for y  

$$\begin{gathered} y=\frac{D_y}{D} \\ y=\frac{216}{-81} \\ y=\frac{-8}{3} \end{gathered}$$

Solution for z  

$$\begin{gathered} z=\frac{D_z}{D} \\ z=\frac{-9}{-81} \\ z=\frac{1}{9} \end{gathered}$$

So the solution of the system of equations is $\left(3,\frac{-8}{3},\frac{1}{9}\right)$