Chapter 7, Problem 193RE

a)

To determine

To evaluate: using Gaussian elimination to solve the system of equations.

Answer to Problem 193RE

The value of $x=\frac{53}{33},y=-\frac{17}{33},z=\frac{61}{66}$ .

Explanation of Solution

Given information :

System of equations are given as

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

Writing the given system of equations in the form of $AX=B$ ,

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

Applying Gaussian elimination rule and finding the determinant,

Finding the values of x , y and z ,

  $\begin{array}{c}\frac{66}{7}z=\frac{61}{7}\\ 66z=61\\ z=\frac{61}{66}\\ \text{Further,}\\ 7y-8z=-11\\ 7y-8\left(\frac{61}{66}\right)=-11\\ 7y=-11+\frac{488}{66}\\ y=-\frac{238}{66\times 7}\\ y=-\frac{17}{33}\\ \text{Further},\\ x-3y+2z=5\\ x-3\left(-\frac{17}{33}\right)+2\left(\frac{61}{66}\right)=5\\ x+\frac{102}{66}+\frac{122}{66}=5\\ x=5-\frac{102}{66}-\frac{122}{66}\\ x=5-\frac{224}{66}\\ x=\frac{330-224}{66}\\ x=\frac{53}{33}\end{array}$

The value of $x=\frac{53}{33},y=-\frac{17}{33},z=\frac{61}{66}$ .

b)

To determine

To evaluate: using Cramer’s rule to solve the system of equations.

Answer to Problem 193RE

The value of $x=\frac{53}{33},y=-\frac{17}{33},z=\frac{61}{66}$ .

Explanation of Solution

Given information :

System of equations are given as

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

Writing the given system of equations in the form of $AX=B$ ,

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

Applying Cramer’s rule and finding the determinant,

  $\begin{array}{c}\Delta =\left|\begin{array}{ccc}1& -3& 2\\ 2& 1& -4\\ 2& 4& 2\end{array}\right|\\ =1\left|\begin{array}{cc}1& -4\\ 4& 2\end{array}\right|-2\left|\begin{array}{cc}-3& 2\\ 4& 2\end{array}\right|+2\left|\begin{array}{cc}-3& 2\\ 1& -4\end{array}\right|\\ =1\left(2+16\right)-2\left(-6-8\right)+2\left(12-2\right)\\ =18+28+20\\ =66\end{array}$

Finding ,

  $\begin{array}{c}{\Delta }_x=\left|\begin{array}{ccc}5& -3& 2\\ -1& 1& -4\\ 3& 4& 2\end{array}\right|\\ =5\left|\begin{array}{cc}1& -4\\ 4& 2\end{array}\right|+1\left|\begin{array}{cc}-3& 2\\ 4& 2\end{array}\right|+3\left|\begin{array}{cc}-3& 2\\ 1& -4\end{array}\right|\\ =5\left(2+16\right)+1\left(-6-8\right)+3\left(12-2\right)\\ =90-14+30\\ =106\end{array}$

Finding ,

  $\begin{array}{c}{\Delta }_y=\left|\begin{array}{ccc}1& 5& 2\\ 2& -1& -4\\ 2& 3& 2\end{array}\right|\\ =1\left|\begin{array}{cc}-1& -4\\ 3& 2\end{array}\right|-2\left|\begin{array}{cc}5& 2\\ 3& 2\end{array}\right|+2\left|\begin{array}{cc}5& 2\\ -1& -4\end{array}\right|\\ =1\left(-2+12\right)-2\left(10-6\right)+2\left(-20+2\right)\\ =10-8-36\\ =-34\end{array}$

Finding ,

  $\begin{array}{c}{\Delta }_z=\left|\begin{array}{ccc}1& -3& 5\\ 2& 1& -1\\ 2& 4& 3\end{array}\right|\\ =1\left|\begin{array}{cc}1& -1\\ 4& 3\end{array}\right|-2\left|\begin{array}{cc}-3& 5\\ 4& 3\end{array}\right|+2\left|\begin{array}{cc}-3& 5\\ 1& -1\end{array}\right|\\ =1\left(3+4\right)-2\left(-9-20\right)+2\left(3-5\right)\\ =7+58-4\\ =61\end{array}$

  $\begin{array}{c}x=\frac{{\Delta }_x}{\Delta }\\ =\frac{106}{66}\\ =\frac{53}{33}\\ y=\frac{{\Delta }_y}{\Delta }\\ =-\frac{34}{66}\\ =-\frac{17}{33}\\ z=\frac{{\Delta }_z}{\Delta }\\ =\frac{61}{66}\end{array}$

The value of $x=\frac{53}{33},y=-\frac{17}{33},z=\frac{61}{66}$ .

Both the methods are preferable but if the questions needs to be solve quickly then Gaussian elimination will be more preferable as it gives the result fast.