Chapter 8.2, Problem 99E

To determine

To find: the solution to the given system of linear equations

Answer to Problem 99E

The solutions to the given system of equation are $x=1,y=5\text{and}z=-1$ .

Explanation of Solution

Given information: The system of equation is $\{\begin{array}{l}2x-y+7z=-10\\ 3x+2y-4z=17\\ 6x-5y+z=-20\end{array}$

Concept used:

Co-efficient matrix of a system of linear equation of the form $\{\begin{array}{l}{a}_{1}x+b_{1}y=c_1\\ a_{2}x+b_{2}y=c_2\end{array}$ is given by

  $\left[\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right]$

The augmented matrix of the system of linear equation is given by

  $\left[\begin{array}{cccc}{a}_1& b_1& ⋮& c_1\\ a_2& b_2& ⋮& c_2\end{array}\right]$

A matrix is in row-echelon form is

1. If any row containing all elements as zero is in the bottom of the matrix.
2. Each row that does not consist entirely of zeros has the first nonzero entry as 1.
3. For successive nonzero rows the leading 1 in the higher row is further than the leading in the lower row.

And a matrix is in reduced row echelon form if every column that has a leading 1 has zeros in every position above and below its leading 1.

To solve a system of linear equation, the corresponding augmented matrix need to reduce in row echelon form using Gaussian elimination method.

Calculation:

Write the system of equation $\{\begin{array}{l}2x-y+7z=-10\\ 3x+2y-4z=17\\ 6x-5y+z=-20\end{array}$ in augmented form

  $\left[\begin{array}{cccc}2& -1& 7& ⋮-10\\ 3& 2& -4& ⋮17\\ 6& -5& 1& ⋮-20\end{array}\right]$

Perform $R_2\to 2R_2-3R_1,R_3\to R_3-3R_1$

  $\left[\begin{array}{cccc}2& -1& 7& ⋮-10\\ 0& 7& -29& ⋮64\\ 0& -2& -20& ⋮10\end{array}\right]$

Perform $R_3\to \left(-\frac{1}{2}\right)R_3$

  $\left[\begin{array}{cccc}2& -1& 7& ⋮-10\\ 0& 7& -29& ⋮64\\ 0& 1& 10& ⋮-5\end{array}\right]$

Perform $R_1\to 7R_1+R_2,R_3\to 7R_3-R_2$

  $\left[\begin{array}{cccc}14& 0& 20& ⋮-3\\ 0& 7& -29& ⋮64\\ 0& 0& 99& ⋮-99\end{array}\right]$

Perform $R_3\to \left(\frac{1}{99}\right)R_3$ , $R_1\to \left(\frac{1}{2}\right)R_1$

  $\left[\begin{array}{cccc}7& 0& 10& ⋮-3\\ 0& 7& -29& ⋮64\\ 0& 0& 1& ⋮-1\end{array}\right]$

Perform $R_1\to R_1-10R_3,R_2\to R_2+29R_3$

  $\left[\begin{array}{cccc}7& 0& 0& ⋮7\\ 0& 7& 0& ⋮35\\ 0& 0& 1& ⋮-1\end{array}\right]$

Perform $R_1\to \left(\frac{1}{7}\right)R_1$ , $R_2\to \left(\frac{1}{7}\right)R_2$

  $\left[\begin{array}{cccc}1& 0& 0& ⋮1\\ 0& 1& 0& ⋮5\\ 0& 0& 1& ⋮-1\end{array}\right]$

The matrix is now in reduced row-echelon form. Converting back to a system of linearequations, you have

  $\{\begin{array}{l}x=1\\ y=5\\ z=-1\end{array}$