Q. 33 - 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.

$\left\{\begin{array}{l}x+y-z=6\\ 3x-2y+z=-5\\ x+3y-2z=14\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the system of equations$\left\{\begin{array}{l}x+y-z=6\\ 3x-2y+z=-5\\ x+3y-2z=14\end{array}\right.$ is $(1,3,-2)$

Step 1. Given information

The given system of equations is  

$\left\{\begin{array}{l}x+y-z=6\\ 3x-2y+z=-5\\ x+3y-2z=14\end{array}\right.$

Step 2. Determinant

The determinant D of the coefficients of the variables 

$$\begin{gathered} D=\left|\begin{array}{ccc}1& 1& -1\\ 3& -2& 1\\ 1& 3& -2\end{array}\right| \\ D={(-1)}^{1+1}\left(1\right)\left|\begin{array}{cc}-2& 1\\ 3& -2\end{array}\right|+{(-1)}^{1+2}\left(1\right)\left|\begin{array}{cc}3& 1\\ 1& -2\end{array}\right|+{(-1)}^{1+3}(-1)\left|\begin{array}{cc}3& -2\\ 1& 3\end{array}\right| \\ D=1(4-3)-1(-6-1)-1(9+2) \\ D=1+7-11 \\ D=-3 \end{gathered}$$

$D\ne 0,$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

$$\begin{gathered} D_x=\left|\begin{array}{ccc}6& 1& -1\\ -5& -2& 1\\ 14& 3& -2\end{array}\right| \\ {D}_x={(-1)}^{1+1}\left(6\right)\left|\begin{array}{cc}-2& 1\\ 3& -2\end{array}\right|+{(-1)}^{1+2}\left(1\right)\left|\begin{array}{cc}-5& 1\\ 14& -2\end{array}\right|+{(-1)}^{1+3}(-1)\left|\begin{array}{cc}-5& -2\\ 14& 3\end{array}\right| \\ {D}_x=6(4-3)-1(10-14)-1(-15+28) \\ {D}_x=6+4-13 \\ {D}_x=-3 \end{gathered}$$

Step 4. Value of D y

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

$$\begin{gathered} D_y=\left|\begin{array}{ccc}1& 6& -1\\ 3& -5& 1\\ 1& 14& -2\end{array}\right| \\ {D}_y={(-1)}^{1+1}\left(1\right)\left|\begin{array}{cc}-5& 1\\ 14& -2\end{array}\right|+{(-1)}^{1+2}\left(6\right)\left|\begin{array}{cc}3& 1\\ 1& -2\end{array}\right|+{(-1)}^{1+3}(-1)\left|\begin{array}{cc}3& -5\\ 1& 14\end{array}\right| \\ {D}_y=1(10-14)-6(-6-1)-1(42+5) \\ {D}_y=-4+42-47 \\ {D}_y=-9 \end{gathered}$$

Step 5. Value of D z

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

$$\begin{gathered} D_z=\left|\begin{array}{ccc}1& 1& 6\\ 3& -2& -5\\ 1& 3& 14\end{array}\right| \\ {D}_z={(-1)}^{1+1}\left(1\right)\left|\begin{array}{cc}-2& -5\\ 3& 14\end{array}\right|+{(-1)}^{1+2}\left(1\right)\left|\begin{array}{cc}3& -5\\ 1& 14\end{array}\right|+{(-1)}^{1+3}\left(6\right)\left|\begin{array}{cc}3& -2\\ 1& 3\end{array}\right| \\ {D}_z=1(-28+15)-1(42+5)+6(9+2) \\ {D}_z=-13-47+66 \\ {D}_z=6 \end{gathered}$$

Step 6. Solution of system

Solution for x  

$$\begin{gathered} x=\frac{D_x}{D} \\ x=\frac{-3}{-3} \\ x=1 \end{gathered}$$

Solution for y  

$$\begin{gathered} y=\frac{D_y}{D} \\ y=\frac{-9}{-3} \\ y=3 \end{gathered}$$

Solution for z  

$$\begin{gathered} z=\frac{D_z}{D} \\ z=\frac{6}{-3} \\ z=-2 \end{gathered}$$

So the solution of the system of equations is $(1,3,-2)$