Chapter 8.5, Problem 16E

To determine

To solve: system of linear equations using Cramer’s rule

  $\{\begin{array}{c}5x-4y+z=-14\\ -x+2y-2z=10\\ 3x+y+z=1\end{array}$

Answer to Problem 16E

  $\left(x,y,z\right)=\left(0,3,-2\right)$

Explanation of Solution

First find the coefficient matrix of this system,

  $A=\left[\begin{array}{ccc}5& -4& 1\\ -1& 2& -2\\ 3& 1& 1\end{array}\right]$

Determinant of the coefficient matrix A is:

  $\begin{array}{c}D=\left|\begin{array}{ccc}5& -4& 1\\ -1& 2& -2\\ 3& 1& 1\end{array}\right|\\ =5\left|\begin{array}{cc}2& -2\\ 1& 1\end{array}\right|-\left(-4\right)\left|\begin{array}{cc}-1& -2\\ 3& 1\end{array}\right|+1\left|\begin{array}{cc}-1& 2\\ 3& 1\end{array}\right|\\ =5\left(2+2\right)+4\left(-1+6\right)+1\left(-1-6\right)\end{array}$

  $\begin{array}{c}=20+20-7\\ =33\end{array}$

Since $D\ne 0$ , Cramer’s rule can be applied.

Consider,

  $\begin{array}{c}{D}_x=\left|\begin{array}{ccc}-14& -4& 1\\ 10& 2& -2\\ 1& 1& 1\end{array}\right|\\ =-14\left|\begin{array}{cc}2& -2\\ 1& 1\end{array}\right|-\left(-4\right)\left|\begin{array}{cc}10& -2\\ 1& 1\end{array}\right|+1\left|\begin{array}{cc}10& 2\\ 1& 1\end{array}\right|\\ =14\left(2+2\right)+4\left(10+2\right)+1\left(10-2\right)\end{array}$

  $\begin{array}{c}=-56+48+8\\ =0\end{array}$

  $\begin{array}{c}{D}_y=\left|\begin{array}{ccc}5& -14& 1\\ -1& 10& -2\\ 3& 1& 1\end{array}\right|\\ =5\left|\begin{array}{cc}10& -2\\ 1& 1\end{array}\right|-\left(-14\right)\left|\begin{array}{cc}-1& -2\\ 3& 1\end{array}\right|+1\left|\begin{array}{cc}-1& 10\\ 3& 1\end{array}\right|\\ =5\left(20+2\right)+14\left(-1+6\right)+1\left(-1-30\right)\end{array}$

  $\begin{array}{c}=60+70-31\\ =99\end{array}$

  $\begin{array}{c}{D}_z=\left|\begin{array}{ccc}5& -4& -14\\ -1& 2& 10\\ 3& 1& 1\end{array}\right|\\ =5\left|\begin{array}{cc}2& 10\\ 1& 1\end{array}\right|-\left(-4\right)\left|\begin{array}{cc}-1& 10\\ 3& 1\end{array}\right|+\left(-14\right)\left|\begin{array}{cc}-1& 2\\ 3& 1\end{array}\right|\\ =5\left(2-10\right)+4\left(-1-30\right)-14\left(-1-6\right)\end{array}$

  $\begin{array}{c}=-40-124+98\\ =-66\end{array}$

Now determine x ,

  $\begin{array}{c}x=\frac{D_x}{D}\\ =\frac{0}{33}\\ =0\end{array}$

Now determine y ,

  $\begin{array}{c}y=\frac{D_y}{D}\\ =\frac{99}{33}\\ =3\end{array}$

Now determine z ,

  $\begin{array}{c}z=\frac{D_z}{D}\\ =\frac{-66}{33}\\ =-2\end{array}$

So, the solution of the given system is,

  $\left(x,y,z\right)=\left(0,3,-2\right)$