Q. 34 - 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.

$\left\{\begin{array}{l}x-y+z=-4\\ 2x-3y+4z=-15\\ 5x+y-2z=12\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the system of equations $\left\{\begin{array}{l}x-y+z=-4\\ 2x-3y+4z=-15\\ 5x+y-2z=12\end{array}\right.$is  $(1,3,-2)$

Step 1. Given information

The given system of equations is   

$\left\{\begin{array}{l}x-y+z=-4\\ 2x-3y+4z=-15\\ 5x+y-2z=12\end{array}\right.$

Step 2. Determinant

The determinant D of the coefficients of the variables 

$$\begin{gathered} D=\left|\begin{array}{ccc}1& -1& 1\\ 2& -3& 4\\ 5& 1& -2\end{array}\right| \\ D={(-1)}^{1+1}\left(1\right)\left|\begin{array}{cc}-3& 4\\ 1& -2\end{array}\right|+{(-1)}^{1+2}(-1)\left|\begin{array}{cc}2& 4\\ 5& -2\end{array}\right|+{(-1)}^{1+3}\left(1\right)\left|\begin{array}{cc}2& -3\\ 5& 1\end{array}\right| \\ D=1(6-4)+1(-4-20)+1(2+15) \\ D=2-24+17 \\ D=-5 \end{gathered}$$

$D\ne 0,$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 

$$\begin{gathered} D_x=\left|\begin{array}{ccc}-4& -1& 1\\ -15& -3& 4\\ 12& 1& -2\end{array}\right| \\ {D}_x={(-1)}^{1+1}(-4)\left|\begin{array}{cc}-3& 4\\ 1& -2\end{array}\right|+{(-1)}^{1+2}(-1)\left|\begin{array}{cc}-15& 4\\ 12& -2\end{array}\right|+{(-1)}^{1+3}\left(1\right)\left|\begin{array}{cc}-15& -3\\ 12& 1\end{array}\right| \\ {D}_x=-4(6-4)+1(30-48)+1(-15+36) \\ {D}_x=-8-18+21 \\ {D}_x=-5 \end{gathered}$$

Step 4. Value of D y

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

$$\begin{gathered} D_y=\left|\begin{array}{ccc}1& -4& 1\\ 2& -15& 4\\ 5& 12& -2\end{array}\right| \\ {D}_y={(-1)}^{1+1}\left(1\right)\left|\begin{array}{cc}-15& 4\\ 12& -2\end{array}\right|+{(-1)}^{1+2}(-4)\left|\begin{array}{cc}2& 4\\ 5& -2\end{array}\right|+{(-1)}^{1+3}\left(1\right)\left|\begin{array}{cc}2& -15\\ 5& 12\end{array}\right| \\ {D}_y=1(30-48)+4(-4-20)+1(24+75) \\ {D}_y=-18-96+99 \\ {D}_y=-15 \end{gathered}$$

Step 5. Value of D z

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

$$\begin{gathered} D_z=\left|\begin{array}{ccc}1& -1& -4\\ 2& -3& -15\\ 5& 1& 12\end{array}\right| \\ {D}_z={(-1)}^{1+1}\left(1\right)\left|\begin{array}{cc}-3& -15\\ 1& 12\end{array}\right|+{(-1)}^{1+2}(-1)\left|\begin{array}{cc}2& -15\\ 5& 12\end{array}\right|+{(-1)}^{1+3}(-4)\left|\begin{array}{cc}2& -3\\ 5& 1\end{array}\right| \\ {D}_z=1(-36+15)+1(24+75)-4(2+15) \\ {D}_z=-21+99-68 \\ {D}_z=10 \end{gathered}$$

Step 6. Solution of system

Solution for x  

$$\begin{gathered} x=\frac{D_x}{D} \\ x=\frac{-5}{-5} \\ x=1 \end{gathered}$$

Solution for y  

$$\begin{gathered} y=\frac{D_y}{D} \\ y=\frac{-15}{-5} \\ y=3 \end{gathered}$$

Solution for z  

$$\begin{gathered} z=\frac{D_z}{D} \\ z=\frac{10}{-5} \\ z=-2 \end{gathered}$$

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