Solve Using an Augmented Matrix 2x+2y+3z=3 , 6x+3y+3z=3 , 2x+5y+6z=6

$2 x + 2 y + 3 z = 3$ , $6 x + 3 y + 3 z = 3$ , $2 x + 5 y + 6 z = 6$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 2 & 2 & 3 & 3 \\ 6 & 3 & 3 & 3 \\ 2 & 5 & 6 & 6 \end{array}]$

Step 2

Find the reduced row echelon form.

Multiply each element of $R_{1}$ by $\frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{2}{2} & \frac{2}{2} & \frac{3}{2} & \frac{3}{2} \\ 6 & 3 & 3 & 3 \\ 2 & 5 & 6 & 6 \end{array}]$

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 1 & \frac{3}{2} & \frac{3}{2} \\ 6 & 3 & 3 & 3 \\ 2 & 5 & 6 & 6 \end{array}]$

Perform the row operation $R_{2} = R_{2} - 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & \frac{3}{2} & \frac{3}{2} \\ 6 - 6 \cdot 1 & 3 - 6 \cdot 1 & 3 - 6 (\frac{3}{2}) & 3 - 6 (\frac{3}{2}) \\ 2 & 5 & 6 & 6 \end{array}]$

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & \frac{3}{2} & \frac{3}{2} \\ 0 & - 3 & - 6 & - 6 \\ 2 & 5 & 6 & 6 \end{array}]$

Perform the row operation $R_{3} = R_{3} - 2 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & \frac{3}{2} & \frac{3}{2} \\ 0 & - 3 & - 6 & - 6 \\ 2 - 2 \cdot 1 & 5 - 2 \cdot 1 & 6 - 2 (\frac{3}{2}) & 6 - 2 (\frac{3}{2}) \end{array}]$

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & \frac{3}{2} & \frac{3}{2} \\ 0 & - 3 & - 6 & - 6 \\ 0 & 3 & 3 & 3 \end{array}]$

Multiply each element of $R_{2}$ by $- \frac{1}{3}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 1 & \frac{3}{2} & \frac{3}{2} \\ - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot - 6 & - \frac{1}{3} \cdot - 6 \\ 0 & 3 & 3 & 3 \end{array}]$

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & \frac{3}{2} & \frac{3}{2} \\ 0 & 1 & 2 & 2 \\ 0 & 3 & 3 & 3 \end{array}]$

Perform the row operation $R_{3} = R_{3} - 3 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 1 & \frac{3}{2} & \frac{3}{2} \\ 0 & 1 & 2 & 2 \\ 0 - 3 \cdot 0 & 3 - 3 \cdot 1 & 3 - 3 \cdot 2 & 3 - 3 \cdot 2 \end{array}]$

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & \frac{3}{2} & \frac{3}{2} \\ 0 & 1 & 2 & 2 \\ 0 & 0 & - 3 & - 3 \end{array}]$

Multiply each element of $R_{3}$ by $- \frac{1}{3}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 1 & \frac{3}{2} & \frac{3}{2} \\ 0 & 1 & 2 & 2 \\ - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot - 3 \end{array}]$

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & \frac{3}{2} & \frac{3}{2} \\ 0 & 1 & 2 & 2 \\ 0 & 0 & 1 & 1 \end{array}]$

Perform the row operation $R_{2} = R_{2} - 2 R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & 1 & \frac{3}{2} & \frac{3}{2} \\ 0 - 2 \cdot 0 & 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & 2 - 2 \cdot 1 \\ 0 & 0 & 1 & 1 \end{array}]$

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & \frac{3}{2} & \frac{3}{2} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{array}]$

Perform the row operation $R_{1} = R_{1} - \frac{3}{2} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - \frac{3}{2} \cdot 0 & 1 - \frac{3}{2} \cdot 0 & \frac{3}{2} - \frac{3}{2} \cdot 1 & \frac{3}{2} - \frac{3}{2} \cdot 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{array}]$

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{array}]$

Perform the row operation $R_{1} = R_{1} - R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - 0 & 1 - 1 & 0 - 0 & 0 - 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{array}]$

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{array}]$

Step 3

Use the result matrix to declare the final solution to the system of equations.

$x = 0 y = 0 z = 1$

Step 4

The solution is the set of ordered pairs that make the system true.

$(0, 0, 1)$

Solve Using an Augmented Matrix 2x+2y+3z=3 , 6x+3y+3z=3 , 2x+5y+6z=6

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