Chapter 7, Problem 41RE

To determine

Solve the given system of equations using Gaussian elimination method.

Answer to Problem 41RE

The solution of the above system of equations is $–\frac{3}{4},0,–\frac{5}{4}$ .

Explanation of Solution

Given:

The system of equations

  $\begin{array}{l}2x+6z=–9\hfill \\ 3x–2y+11z=–16\hfill \\ 3x–y+7z=–11\hfill \end{array}$

Consider the system of equations

  $\begin{array}{l}2x+6z=–\mathrm{9.......}\left(1\right)\hfill \\ 3x–2y+11z=–\mathrm{16......}\left(2\right)\hfill \\ 3x–y+7z=–\mathrm{11.......}\left(3\right)\hfill \end{array}$

Use method of Gaussian elimination

Multiply equation (3) by 2 to get

  $6x–2y+14z=–\mathrm{22......}\left(4\right)$

Subtract equation (2) from equation (4)

  $\begin{array}{l}\left(6x–2y+14z\right)–\left(3x–2y+11z\right)=–22–\left(–16\right)\hfill \\ 6x–\cancel{2y}+14z–3x+\cancel{2y}–11z=–6\hfill \\ 3x+3z=–\mathrm{6.....}\left(5\right)\hfill \end{array}$

Multiply equation (5) by 2 to get

  $6x+6z=–\mathrm{12.......}\left(6\right)$

Subtract equation (6) from equation (5)

  $\begin{array}{l}\left(6x+6z\right)–\left(2x+6z\right)=–12–\left(–9\right)\hfill \\ 6x+\cancel{6z}–2x–\cancel{6z}=–12+9\hfill \end{array}$

  $4x=–3$ Divide both sides by 4

  $x=–\frac{3}{4}$

Multiply equation (7) by 4 and then subtract equation (5) from the resultant to get

  $\begin{array}{l}4\left(13y–2z\right)–\left(y–8z\right)=4\times 38–32\hfill \\ 52y–\cancel{8z}–y+\cancel{8z}=152–32\hfill \\ 51y=120\hfill \end{array}$

Divide both sides by 51 to get

  $y=\frac{40}{17}$

Substitute $x=–\frac{3}{4}$ in equation (1)

  $\begin{array}{l}2\left(-\frac{3}{4}\right)+6z=–9\hfill \\ –\frac{6}{4}+6z=–9\hfill \end{array}$

Add $\frac{6}{4}$ to both sides to get

  $6z=–\frac{15}{2}$

Divide both sides by 6 to get

  $z=–\frac{5}{4}$

Substitute $x=–\frac{3}{4}$ and $z=–\frac{5}{4}$ in equation (1)

  $\begin{array}{l}3\left(-\frac{3}{4}\right)–y+7\left(-\frac{5}{4}\right)=–11\hfill \\ –\frac{9}{4}–y–\frac{35}{4}=–11\hfill \\ –11–y=–11\hfill \end{array}$

Add 11 to both sides of equation

  $y=0$ Hence, the solution of the above system of equations is $–\frac{3}{4},0,–\frac{5}{4}$ .