Finite Math Examples

$3 x - 4 y + 4 z = 10$ , $3 x + 5 y - z = 15$ , $12 x - 7 y + 11 z = 45$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 3 & - 4 & 4 & 10 \\ 3 & 5 & - 1 & 15 \\ 12 & - 7 & 11 & 45 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} \frac{3}{3} & \frac{- 4}{3} & \frac{4}{3} & \frac{10}{3} \\ 3 & 5 & - 1 & 15 \\ 12 & - 7 & 11 & 45 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{4}{3} & \frac{10}{3} \\ 3 & 5 & - 1 & 15 \\ 12 & - 7 & 11 & 45 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{4}{3} & \frac{10}{3} \\ 3 - 3 \cdot 1 & 5 - 3 (- \frac{4}{3}) & - 1 - 3 (\frac{4}{3}) & 15 - 3 (\frac{10}{3}) \\ 12 & - 7 & 11 & 45 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{4}{3} & \frac{10}{3} \\ 0 & 9 & - 5 & 5 \\ 12 & - 7 & 11 & 45 \end{array}]$

Step 2.3

Perform the row operation $R_{3} = R_{3} - 12 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 2.3.1

Perform the row operation $R_{3} = R_{3} - 12 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{4}{3} & \frac{10}{3} \\ 0 & 9 & - 5 & 5 \\ 12 - 12 \cdot 1 & - 7 - 12 (- \frac{4}{3}) & 11 - 12 (\frac{4}{3}) & 45 - 12 (\frac{10}{3}) \end{array}]$

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{4}{3} & \frac{10}{3} \\ 0 & 9 & - 5 & 5 \\ 0 & 9 & - 5 & 5 \end{array}]$

Step 2.4

Multiply each element of $R_{2}$ by $\frac{1}{9}$ to make the entry at $2, 2$ a $1$ .

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Step 2.4.1

Multiply each element of $R_{2}$ by $\frac{1}{9}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{4}{3} & \frac{10}{3} \\ \frac{0}{9} & \frac{9}{9} & \frac{- 5}{9} & \frac{5}{9} \\ 0 & 9 & - 5 & 5 \end{array}]$

Step 2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{4}{3} & \frac{10}{3} \\ 0 & 1 & - \frac{5}{9} & \frac{5}{9} \\ 0 & 9 & - 5 & 5 \end{array}]$

Step 2.5

Perform the row operation $R_{3} = R_{3} - 9 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 2.5.1

Perform the row operation $R_{3} = R_{3} - 9 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{4}{3} & \frac{10}{3} \\ 0 & 1 & - \frac{5}{9} & \frac{5}{9} \\ 0 - 9 \cdot 0 & 9 - 9 \cdot 1 & - 5 - 9 (- \frac{5}{9}) & 5 - 9 (\frac{5}{9}) \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{4}{3} & \frac{4}{3} & \frac{10}{3} \\ 0 & 1 & - \frac{5}{9} & \frac{5}{9} \\ 0 & 0 & 0 & 0 \end{array}]$

Step 2.6

Perform the row operation $R_{1} = R_{1} + \frac{4}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 2.6.1

Perform the row operation $R_{1} = R_{1} + \frac{4}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{4}{3} \cdot 0 & - \frac{4}{3} + \frac{4}{3} \cdot 1 & \frac{4}{3} + \frac{4}{3} (- \frac{5}{9}) & \frac{10}{3} + \frac{4}{3} \cdot \frac{5}{9} \\ 0 & 1 & - \frac{5}{9} & \frac{5}{9} \\ 0 & 0 & 0 & 0 \end{array}]$

Step 2.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{16}{27} & \frac{110}{27} \\ 0 & 1 & - \frac{5}{9} & \frac{5}{9} \\ 0 & 0 & 0 & 0 \end{array}]$

Step 3

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

$x + \frac{16}{27} z = \frac{110}{27}$

$y - \frac{5}{9} z = \frac{5}{9}$

$0 = 0$

Step 4

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

$(\frac{110}{27} - \frac{16 z}{27}, \frac{5}{9} + \frac{5 z}{9}, z)$