Finite Math Examples

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

Step 1

Write the system as a matrix.

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

Step 2

Find the reduced row echelon form.

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Step 2.1

Swap $R_{2}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

Simplify $R_{1}$ .

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & - 1 & 3 \\ 0 & 0 & - 3 & 6 \\ 0 & 0 & - 6 & 12 \end{array}]$

Step 2.4

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

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Step 2.4.1

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

$[\begin{array}{c} 1 & \frac{1}{2} & - 1 & 3 \\ - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot 6 \\ 0 & 0 & - 6 & 12 \end{array}]$

Step 2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & - 1 & 3 \\ 0 & 0 & 1 & - 2 \\ 0 & 0 & - 6 & 12 \end{array}]$

Step 2.5

Perform the row operation $R_{3} = R_{3} + 6 R_{2}$ to make the entry at $3, 3$ a $0$ .

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Step 2.5.1

Perform the row operation $R_{3} = R_{3} + 6 R_{2}$ to make the entry at $3, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{2} & - 1 & 3 \\ 0 & 0 & 1 & - 2 \\ 0 + 6 \cdot 0 & 0 + 6 \cdot 0 & - 6 + 6 \cdot 1 & 12 + 6 \cdot - 2 \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

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

Step 2.6

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

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Step 2.6.1

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

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

Step 2.6.2

Simplify $R_{1}$ .

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

Step 3

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

$x + \frac{1}{2} y = 1$

$z = - 2$

$0 = 0$

Step 4

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

$(1 - \frac{y}{2}, y, - 2)$