Q. 20 - Precalculus Enhanced with Graphing Utilities

Question

In Problems $19$ and $20$, use Cramer’s Rule, if possible, to solve each system.

$\left\{\begin{array}{rr}4x-3y+2z=& 15\\ -2x+y-3z=& -15\\ 5x-5y+2z=& 18\end{array}\right.$

Step-by-Step Solution

Answer

The solution is $(1,1,4)$

Step 1. Given information

System of equations is given as 

$\left\{\begin{array}{rr}4x-3y+2z=& 15\\ -2x+y-3z=& -15\\ 5x-5y+2z=& 18\end{array}\right.$

Step 2. Check Cramer's rule

First find determinant of coefficients,

$$\begin{gathered} D=\left|\begin{array}{ccc}4& -3& 2\\ -2& 1& -3\\ 5& -5& 2\end{array}\right| \\ =4(1\times 2-(-3)\times (-5\left)\right)-(-3)\left(\right(-2)\times 2-(-3)\times 5)+2\left(\right(-2)\times (-5)-1\times 5) \\ =4(-13)+3\left(11\right)+2\left(5\right) \\ =-9 \end{gathered}$$

Since $D\ne 0$, We can use Cramer's rule

Step 3. Finding determinants

For D_x, replace the first column of D with constants given on the other side, For D_y replace second column and for D_z replace third column.

$$\begin{gathered} D_x=\left|\begin{array}{ccc}15& -3& 2\\ -15& 1& -3\\ 18& -5& 2\end{array}\right| \\ =15(1\times 2-(-3)\times (-5\left)\right)-(-3)\left(\right(-15)\times 2-(-3)\times 18)+2\left(\right(-15)\times (-5)-1\times 18) \\ =15(-13)+3\left(24\right)+2\left(57\right) \\ =9 \end{gathered}$$

$$\begin{gathered} D_y=\left|\begin{array}{ccc}4& 15& 2\\ -2& -15& -3\\ 5& 18& 2\end{array}\right| \\ =4\left(\right(-15)\times 2-(-3)\times 18)-15\left(\right(-2)\times 2-(-3)\times 5)+2\left(\right(-2)\times 18-(-15)\times 5) \\ =4\left(24\right)-15\left(11\right)+2\left(39\right) \\ =9 \end{gathered}$$

$$\begin{gathered} D_z=\left|\begin{array}{ccc}4& -3& 15\\ -2& 1& -15\\ 5& -5& 18\end{array}\right| \\ =4(1\times 18-(-15)\times (-5\left)\right)-(-3)\left(\right(-2)\times 18-(-15)\times 5)+15\left(\right(-2)\times (-5)-1\times 5) \\ =4(-57)+3\left(39\right)+15\left(5\right) \\ =-36 \end{gathered}$$

Step 4. Finding values

Now substitute the values of determinants as,

$$\begin{gathered} x=\frac{D_x}{D} \\ =\frac{-9}{-9} \\ =1 \end{gathered}$$

$$\begin{gathered} y=\frac{D_y}{D} \\ =\frac{-9}{-9} \\ =1 \end{gathered}$$

$$\begin{gathered} z=\frac{D_z}{D} \\ =\frac{-36}{-9} \\ =4 \end{gathered}$$