Chapter 7, Problem 40RE

To determine

Solve the given system of equations using Gaussian elimination method.

Answer to Problem 40RE

The solution forms the ordered triple $\left(3a+4,5+2a,a\right)$

Explanation of Solution

Given:

The system of equations

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

Let consider the following system of equations

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

Multiplying equation (1) with−2 and adding with equation (2).

  $\begin{array}{l}–2x+4y–2z=12\hfill \\ 2x–3y=–7\hfill \\ y–2z=5\hfill \end{array}$

Adding equation (1) with equation (3).

  $\begin{array}{l}x–2y+z=–6\hfill \\ –x+3y–3z=11\hfill \\ y–2z=5\hfill \end{array}$

The revised equivalent system is:

  $\begin{array}{l}\begin{array}{l}x–2y+z=–6\hfill \\ y–2z=5\hfill \end{array} \\ y–2z=5\end{array}$

Adding −1 times equation (2) with equation three gives the system as:

  $\begin{array}{l}\begin{array}{l}x–2y+z=–6\hfill \\ y–2z=5\hfill \end{array} \\ 0=0\end{array}$

This shows that equation (3) depends on equation (2) and (1). Moreover, 0=0 is a true statement, and the system will be having infinitely many solutions.

Now the original system of equation has become,

  $\begin{array}{l}x–2y+z=–6\hfill \\ y–2z=5\hfill \end{array}$

Now, $y=5+2z$ . Substitute y in terms of z in equation (1).

  $x–2y+z=–6$ write the equation.

  $x–2\left(5+2x\right)+z=–6$ Distributive property

  $x–3z=10–6$ Combining like terms

  $x=3z+4$ solve for $x$

Put $z=a$ , being a real value. Then $y=5+2a$ and $x=3a+4$ . Sothe solution forms the ordered triple $\left(3a+4,5+2a,a\right)$