Chapter 4.6, Problem 5E

To determine

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

Answer to Problem 5E

The solution is $\left(-2,-1,2\right)$

Explanation of Solution

Given information:

  $\begin{array}{l}6x-4y+2z=-4\\ 2x-2y+6z=10\\ 2x+2y+2z=-2\end{array}$ .

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 ,

  $\begin{array}{l}6x-4y+2z=-4\\ 2x-2y+6z=10\\ 2x+2y+2z=-2\end{array}$

  $A=\left[\begin{array}{l}\begin{array}{cccc}6& -4& 2& -4\end{array}\\ \begin{array}{cccc}2& -2& 6& 10\end{array}\\ \begin{array}{cccc}2& 2& 2& -2\end{array}\end{array}\right]$

Keystrokes:

  $\begin{array}{l}\boxed{2nd}[MATRIX]\boxed{⊳}\boxed{⊳}\boxed{ENTER}3\boxed{ENTER}4\boxed{ENTER}\\ 6\boxed{ENTER}(-)4\boxed{ENTER}2\boxed{ENTER}(-)4\boxed{ENTER}\\ 2\boxed{ENTER}(-)2\boxed{ENTER}6\boxed{ENTER}10\boxed{ENTER}\\ 2\boxed{ENTER}2\boxed{ENTER}2\boxed{ENTER}(-)2\boxed{ENTER}\end{array}$

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

  $x=-2$ , the second row represents $y=-1$ , the third row represents $z=2$

The solution is $\left(-2,-1,2\right)$ .

  

Hence, the solution is $\left(-2,-1,2\right)$ .