Chapter 4.6, Problem 6E

To determine

To evaluate: an augmented matrix for system of equations and solve by graphing calculator.

Answer to Problem 6E

The solution is $\left(1.25,-0.5,0.25\right)$

Explanation of Solution

Given information:

  $\begin{array}{l}5x-5y+5z=10\\ 5x-5z=5\\ 5y+10z=0\end{array}$ .

Formula used:

An augmented matrix contains the coefficient matrix of an extra column containing the constant terms.

Calculation:

Step 1.

Using system of equations the augmented matrix is ,

  $\begin{array}{l}5x-5y+5z=10\\ 5x-5z=5\\ 5y+10z=0\end{array}$

  $A=\left[\begin{array}{l}\begin{array}{cccc}5& -5& 5& 10\end{array}\\ \begin{array}{cccc}5& 0& -5& 5\end{array}\\ \begin{array}{cccc}0& 5& 10& 0\end{array}\end{array}\right]$

Keystrokes:

  $\begin{array}{l}\boxed{2nd}[MATRIX]\boxed{⊳}\boxed{⊳}\boxed{ENTER}3\boxed{ENTER}4\boxed{ENTER}\\ 5\boxed{ENTER}(-)5\boxed{ENTER}5\boxed{ENTER}10\boxed{ENTER}\\ 5\boxed{ENTER}0\boxed{ENTER}(-)5\boxed{ENTER}5\boxed{ENTER}\\ 0\boxed{ENTER}5\boxed{ENTER}10\boxed{ENTER}0\boxed{ENTER}\end{array}$

Step 2.

The reduced row echelon form (rref) using the graphing calculator.

Keystrokes:

  $\boxed{2nd}[QUIT]\boxed{2nd}[MATRIX]\boxed{⊳}\boxed{ALPHA}[A][MATRIX]\boxed{ENTER}\boxed{)}\boxed{ENTER}$

The first three columns are same as $3\times 3$ identity matrix. The first three row represents

  $x=1.25$ , the second row represents $y=-0.5$ , the third row represents $z=0.25$

The solution is $\left(1.25,-0.5,0.25\right)$ .

  

Hence, the solution is $\left(1.25,-0.5,0.25\right)$ .