Finite Math Examples

$5 x - y - 6 z = 2$ , $9 x + y - z = 5$ , $9 x + y - 5 z = 7$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 5 & - 1 & - 6 & 2 \\ 9 & 1 & - 1 & 5 \\ 9 & 1 & - 5 & 7 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

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Step 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{5}{5} & \frac{- 1}{5} & \frac{- 6}{5} & \frac{2}{5} \\ 9 & 1 & - 1 & 5 \\ 9 & 1 & - 5 & 7 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{6}{5} & \frac{2}{5} \\ 9 & 1 & - 1 & 5 \\ 9 & 1 & - 5 & 7 \end{array}]$

Step 2.2

Perform the row operation $R_{2} = R_{2} - 9 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 2.2.1

Perform the row operation $R_{2} = R_{2} - 9 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{6}{5} & \frac{2}{5} \\ 9 - 9 \cdot 1 & 1 - 9 (- \frac{1}{5}) & - 1 - 9 (- \frac{6}{5}) & 5 - 9 (\frac{2}{5}) \\ 9 & 1 & - 5 & 7 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{6}{5} & \frac{2}{5} \\ 0 & \frac{14}{5} & \frac{49}{5} & \frac{7}{5} \\ 9 & 1 & - 5 & 7 \end{array}]$

Step 2.3

Perform the row operation $R_{3} = R_{3} - 9 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 2.3.1

Perform the row operation $R_{3} = R_{3} - 9 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{6}{5} & \frac{2}{5} \\ 0 & \frac{14}{5} & \frac{49}{5} & \frac{7}{5} \\ 9 - 9 \cdot 1 & 1 - 9 (- \frac{1}{5}) & - 5 - 9 (- \frac{6}{5}) & 7 - 9 (\frac{2}{5}) \end{array}]$

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{6}{5} & \frac{2}{5} \\ 0 & \frac{14}{5} & \frac{49}{5} & \frac{7}{5} \\ 0 & \frac{14}{5} & \frac{29}{5} & \frac{17}{5} \end{array}]$

Step 2.4

Multiply each element of $R_{2}$ by $\frac{5}{14}$ to make the entry at $2, 2$ a $1$ .

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Step 2.4.1

Multiply each element of $R_{2}$ by $\frac{5}{14}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{6}{5} & \frac{2}{5} \\ \frac{5}{14} \cdot 0 & \frac{5}{14} \cdot \frac{14}{5} & \frac{5}{14} \cdot \frac{49}{5} & \frac{5}{14} \cdot \frac{7}{5} \\ 0 & \frac{14}{5} & \frac{29}{5} & \frac{17}{5} \end{array}]$

Step 2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{6}{5} & \frac{2}{5} \\ 0 & 1 & \frac{7}{2} & \frac{1}{2} \\ 0 & \frac{14}{5} & \frac{29}{5} & \frac{17}{5} \end{array}]$

Step 2.5

Perform the row operation $R_{3} = R_{3} - \frac{14}{5} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 2.5.1

Perform the row operation $R_{3} = R_{3} - \frac{14}{5} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{6}{5} & \frac{2}{5} \\ 0 & 1 & \frac{7}{2} & \frac{1}{2} \\ 0 - \frac{14}{5} \cdot 0 & \frac{14}{5} - \frac{14}{5} \cdot 1 & \frac{29}{5} - \frac{14}{5} \cdot \frac{7}{2} & \frac{17}{5} - \frac{14}{5} \cdot \frac{1}{2} \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{6}{5} & \frac{2}{5} \\ 0 & 1 & \frac{7}{2} & \frac{1}{2} \\ 0 & 0 & - 4 & 2 \end{array}]$

Step 2.6

Multiply each element of $R_{3}$ by $- \frac{1}{4}$ to make the entry at $3, 3$ a $1$ .

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Step 2.6.1

Multiply each element of $R_{3}$ by $- \frac{1}{4}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{6}{5} & \frac{2}{5} \\ 0 & 1 & \frac{7}{2} & \frac{1}{2} \\ - \frac{1}{4} \cdot 0 & - \frac{1}{4} \cdot 0 & - \frac{1}{4} \cdot - 4 & - \frac{1}{4} \cdot 2 \end{array}]$

Step 2.6.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{6}{5} & \frac{2}{5} \\ 0 & 1 & \frac{7}{2} & \frac{1}{2} \\ 0 & 0 & 1 & - \frac{1}{2} \end{array}]$

Step 2.7

Perform the row operation $R_{2} = R_{2} - \frac{7}{2} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 2.7.1

Perform the row operation $R_{2} = R_{2} - \frac{7}{2} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{6}{5} & \frac{2}{5} \\ 0 - \frac{7}{2} \cdot 0 & 1 - \frac{7}{2} \cdot 0 & \frac{7}{2} - \frac{7}{2} \cdot 1 & \frac{1}{2} - \frac{7}{2} (- \frac{1}{2}) \\ 0 & 0 & 1 & - \frac{1}{2} \end{array}]$

Step 2.7.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & - \frac{6}{5} & \frac{2}{5} \\ 0 & 1 & 0 & \frac{9}{4} \\ 0 & 0 & 1 & - \frac{1}{2} \end{array}]$

Step 2.8

Perform the row operation $R_{1} = R_{1} + \frac{6}{5} R_{3}$ to make the entry at $1, 3$ a $0$ .

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Step 2.8.1

Perform the row operation $R_{1} = R_{1} + \frac{6}{5} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 + \frac{6}{5} \cdot 0 & - \frac{1}{5} + \frac{6}{5} \cdot 0 & - \frac{6}{5} + \frac{6}{5} \cdot 1 & \frac{2}{5} + \frac{6}{5} (- \frac{1}{2}) \\ 0 & 1 & 0 & \frac{9}{4} \\ 0 & 0 & 1 & - \frac{1}{2} \end{array}]$

Step 2.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & 0 & - \frac{1}{5} \\ 0 & 1 & 0 & \frac{9}{4} \\ 0 & 0 & 1 & - \frac{1}{2} \end{array}]$

Step 2.9

Perform the row operation $R_{1} = R_{1} + \frac{1}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 2.9.1

Perform the row operation $R_{1} = R_{1} + \frac{1}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{1}{5} \cdot 0 & - \frac{1}{5} + \frac{1}{5} \cdot 1 & 0 + \frac{1}{5} \cdot 0 & - \frac{1}{5} + \frac{1}{5} \cdot \frac{9}{4} \\ 0 & 1 & 0 & \frac{9}{4} \\ 0 & 0 & 1 & - \frac{1}{2} \end{array}]$

Step 2.9.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{1}{4} \\ 0 & 1 & 0 & \frac{9}{4} \\ 0 & 0 & 1 & - \frac{1}{2} \end{array}]$

Step 3

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

$x = \frac{1}{4}$

$y = \frac{9}{4}$

$z = - \frac{1}{2}$

Step 4

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

$(\frac{1}{4}, \frac{9}{4}, - \frac{1}{2})$