Chapter 8, Problem 3CT

To determine

Find solution of the given system of linear equations

Answer to Problem 3CT

The solution of the system of linear equations is:

  $x=1,y=3andz=-\frac{1}{2}$

Explanation of Solution

Given:

System of equations

  $\begin{array}{l}4x+3y–2z=14\hfill \\ –x–y+2z=–5\hfill \\ 3x+y–4z=8\hfill \end{array}$

Let consider the following system of equations

  $\begin{array}{l}4x+3y–2z=14\hfill \\ –x–y+2z=–5\hfill \\ 3x+y–4z=8\hfill \end{array}$

Here, the variables are already aligned.

Now, using the coefficients and constant terms as the matrix entries, the augmented matrix is:

  $\left[\begin{array}{ccccc}4& 3& -2& ⋮& 14\\ -1& -1& 2& ⋮& -5\\ 3& 1& -4& ⋮& 8\end{array}\right]$

The system can be solved by the Gaussian elimination method.

First, convert the given matrix into its row-echelon form:

  $\left[\begin{array}{ccccc}4& 3& -2& ⋮& 14\\ 0& -\frac{1}{4}& \frac{3}{2}& ⋮& -\frac{3}{2}\\ 0& -\frac{5}{4}& -\frac{5}{2}& ⋮& -\frac{5}{2}\end{array}\right]$

  $\begin{array}{l}{R}_2\to R_2+\frac{1}{4}{R}_1\\ R_3\to R_3-\frac{3}{4}{R}_1\end{array}$

  $\left[\begin{array}{ccccc}4& 3& -2& ⋮& 14\\ 0& -\frac{1}{4}& -\frac{3}{2}& ⋮& -\frac{3}{2}\\ 0& 0& -10& ⋮& 5\end{array}\right]$

  $R_3\to R_3-5R_2$

  $\left[\begin{array}{ccccc}4& 3& -2& ⋮& 14\\ 0& 1& -6& ⋮& 6\\ 0& 0& 1& ⋮& -\frac{1}{2}\end{array}\right]$

  $\begin{array}{l}{R}_2\to -4R_2\\ R_3\to -\frac{1}{10}{R}_3\end{array}$

So, the linear equations become:

  $\begin{array}{l}4x+3y–2z=14\hfill \\ y–6z=6\hfill \\ z=–\frac{1}{2}\hfill \end{array}$

Now, solve these equations by back-substitution.

The third equation gives $z=–\frac{1}{2}$

Putting this value in secondequation,

  $\begin{array}{l}y–6\left(-\frac{1}{2}\right)=6\hfill \\ y=3\hfill \end{array}$

Putting these values in third equation,

  $\begin{array}{l}4x+3\left(3\right)–2\left(-\frac{1}{2}\right)=14\hfill \\ x=1\hfill \end{array}$

Hence, the solution of the system of linear equations is:

  $x=1,y=3andz=-\frac{1}{2}$