Chapter 8.5, Problem 11E

To determine

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

  $\{\begin{array}{c}4x-y+z=-5\\ 2x+2y+3z=10\\ 5z-2y+6z=1\end{array}$

Answer to Problem 11E

  $\left(x,y,z\right)=\left(-1,3,2\right)$

Explanation of Solution

First find the coefficient matrix of this system,

  $A=\left[\begin{array}{ccc}4& -1& 1\\ 2& 2& 3\\ 5& -2& 6\end{array}\right]$

Determinant of the coefficient matrix A is:

  $\begin{array}{c}D=\left|\begin{array}{ccc}4& -1& 1\\ 2& 2& 3\\ 5& -2& 6\end{array}\right|\\ =4\left|\begin{array}{cc}2& 3\\ -2& 6\end{array}\right|-\left(-1\right)\left|\begin{array}{cc}2& 3\\ 5& 6\end{array}\right|+1\left|\begin{array}{cc}2& 2\\ 5& -2\end{array}\right|\\ =4\left(12+6\right)+1\left(12-15\right)+1\left(-4-10\right)\end{array}$

  $\begin{array}{c}=72-3-14\\ =55\end{array}$

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

Consider,

  $\begin{array}{c}{D}_x=\left|\begin{array}{ccc}-5& -1& 1\\ 10& 2& 3\\ 1& -2& 6\end{array}\right|\\ =-5\left|\begin{array}{cc}2& 3\\ -2& 6\end{array}\right|-10\left|\begin{array}{cc}-1& 1\\ -2& 6\end{array}\right|+1\left|\begin{array}{cc}-1& 1\\ 2& 3\end{array}\right|\\ =-5\left(12+6\right)-10\left(-6+2\right)+1\left(-3-2\right)\end{array}$

  $\begin{array}{c}=-90+40-5\\ =-55\end{array}$

  $\begin{array}{c}{D}_y=\left|\begin{array}{ccc}4& -5& 1\\ 2& 10& 3\\ 5& 1& 6\end{array}\right|\\ =4\left|\begin{array}{cc}10& 3\\ 1& 6\end{array}\right|-2\left|\begin{array}{cc}-5& 1\\ 1& 6\end{array}\right|+5\left|\begin{array}{cc}-5& 1\\ 10& 3\end{array}\right|\\ =4\left(60-3\right)-2\left(-30-1\right)+5\left(-15-10\right)\end{array}$

  $\begin{array}{c}=228+62-125\\ =165\end{array}$

  $\begin{array}{c}{D}_z=\left|\begin{array}{ccc}4& -1& -5\\ 2& 2& 10\\ 5& -2& 1\end{array}\right|\\ =4\left|\begin{array}{cc}2& 10\\ -2& 1\end{array}\right|-2\left|\begin{array}{cc}-1& -5\\ -2& 1\end{array}\right|+5\left|\begin{array}{cc}-1& -5\\ 2& 10\end{array}\right|\\ =4\left(2+20\right)-2\left(-1-10\right)+5\left(-10+10\right)\end{array}$

  $\begin{array}{c}=88+22+0\\ =110\end{array}$

Now determine x ,

  $\begin{array}{c}x=\frac{D_x}{D}\\ =\frac{-55}{55}\\ =-1\end{array}$

Now determine y ,

  $\begin{array}{c}y=\frac{D_y}{D}\\ =\frac{165}{55}\\ =3\end{array}$

Now determine z ,

  $\begin{array}{c}z=\frac{D_z}{D}\\ =\frac{110}{55}\\ =2\end{array}$

So, the solution of the given system is,

  $\left(x,y,z\right)=\left(-1,3,2\right)$