Finite Math Examples

$2 x + 5 y - 2 z = 14$ , $5 x - 6 y + 2 z = 0$ , $4 x - y + 3 z = - 7$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 2 & 5 & - 2 & 14 \\ 5 & - 6 & 2 & 0 \\ 4 & - 1 & 3 & - 7 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} \frac{2}{2} & \frac{5}{2} & \frac{- 2}{2} & \frac{14}{2} \\ 5 & - 6 & 2 & 0 \\ 4 & - 1 & 3 & - 7 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{5}{2} & - 1 & 7 \\ 5 & - 6 & 2 & 0 \\ 4 & - 1 & 3 & - 7 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & \frac{5}{2} & - 1 & 7 \\ 5 - 5 \cdot 1 & - 6 - 5 (\frac{5}{2}) & 2 - 5 \cdot - 1 & 0 - 5 \cdot 7 \\ 4 & - 1 & 3 & - 7 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{5}{2} & - 1 & 7 \\ 0 & - \frac{37}{2} & 7 & - 35 \\ 4 & - 1 & 3 & - 7 \end{array}]$

Step 2.3

Perform the row operation $R_{3} = R_{3} - 4 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 2.3.1

Perform the row operation $R_{3} = R_{3} - 4 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{5}{2} & - 1 & 7 \\ 0 & - \frac{37}{2} & 7 & - 35 \\ 4 - 4 \cdot 1 & - 1 - 4 (\frac{5}{2}) & 3 - 4 \cdot - 1 & - 7 - 4 \cdot 7 \end{array}]$

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{5}{2} & - 1 & 7 \\ 0 & - \frac{37}{2} & 7 & - 35 \\ 0 & - 11 & 7 & - 35 \end{array}]$

Step 2.4

Multiply each element of $R_{2}$ by $- \frac{2}{37}$ to make the entry at $2, 2$ a $1$ .

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Step 2.4.1

Multiply each element of $R_{2}$ by $- \frac{2}{37}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{5}{2} & - 1 & 7 \\ - \frac{2}{37} \cdot 0 & - \frac{2}{37} (- \frac{37}{2}) & - \frac{2}{37} \cdot 7 & - \frac{2}{37} \cdot - 35 \\ 0 & - 11 & 7 & - 35 \end{array}]$

Step 2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{5}{2} & - 1 & 7 \\ 0 & 1 & - \frac{14}{37} & \frac{70}{37} \\ 0 & - 11 & 7 & - 35 \end{array}]$

Step 2.5

Perform the row operation $R_{3} = R_{3} + 11 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 2.5.1

Perform the row operation $R_{3} = R_{3} + 11 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{5}{2} & - 1 & 7 \\ 0 & 1 & - \frac{14}{37} & \frac{70}{37} \\ 0 + 11 \cdot 0 & - 11 + 11 \cdot 1 & 7 + 11 (- \frac{14}{37}) & - 35 + 11 (\frac{70}{37}) \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{5}{2} & - 1 & 7 \\ 0 & 1 & - \frac{14}{37} & \frac{70}{37} \\ 0 & 0 & \frac{105}{37} & - \frac{525}{37} \end{array}]$

Step 2.6

Multiply each element of $R_{3}$ by $\frac{37}{105}$ to make the entry at $3, 3$ a $1$ .

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Step 2.6.1

Multiply each element of $R_{3}$ by $\frac{37}{105}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{5}{2} & - 1 & 7 \\ 0 & 1 & - \frac{14}{37} & \frac{70}{37} \\ \frac{37}{105} \cdot 0 & \frac{37}{105} \cdot 0 & \frac{37}{105} \cdot \frac{105}{37} & \frac{37}{105} (- \frac{525}{37}) \end{array}]$

Step 2.6.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{5}{2} & - 1 & 7 \\ 0 & 1 & - \frac{14}{37} & \frac{70}{37} \\ 0 & 0 & 1 & - 5 \end{array}]$

Step 2.7

Perform the row operation $R_{2} = R_{2} + \frac{14}{37} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 2.7.1

Perform the row operation $R_{2} = R_{2} + \frac{14}{37} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{5}{2} & - 1 & 7 \\ 0 + \frac{14}{37} \cdot 0 & 1 + \frac{14}{37} \cdot 0 & - \frac{14}{37} + \frac{14}{37} \cdot 1 & \frac{70}{37} + \frac{14}{37} \cdot - 5 \\ 0 & 0 & 1 & - 5 \end{array}]$

Step 2.7.2

Simplify $R_{2}$ .

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

Step 2.8

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

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Step 2.8.1

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

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

Step 2.8.2

Simplify $R_{1}$ .

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

Step 2.9

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

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Step 2.9.1

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

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

Step 2.9.2

Simplify $R_{1}$ .

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

Step 3

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

$x = 2$

$y = 0$

$z = - 5$

Step 4

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

$(2, 0, - 5)$