Finite Math Examples

$7 x - y - z = 0$ , $x + y + z = 8$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 7 & - 1 & - 1 & 0 \\ 1 & 1 & 1 & 8 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Multiply each element of $R_{1}$ by $\frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

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Step 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{7}{7} & \frac{- 1}{7} & \frac{- 1}{7} & \frac{0}{7} \\ 1 & 1 & 1 & 8 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{7} & - \frac{1}{7} & 0 \\ 1 & 1 & 1 & 8 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & - \frac{1}{7} & - \frac{1}{7} & 0 \\ 1 - 1 & 1 + \frac{1}{7} & 1 + \frac{1}{7} & 8 - 0 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{7} & - \frac{1}{7} & 0 \\ 0 & \frac{8}{7} & \frac{8}{7} & 8 \end{array}]$

Step 2.3

Multiply each element of $R_{2}$ by $\frac{7}{8}$ to make the entry at $2, 2$ a $1$ .

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Step 2.3.1

Multiply each element of $R_{2}$ by $\frac{7}{8}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{7} & - \frac{1}{7} & 0 \\ \frac{7}{8} \cdot 0 & \frac{7}{8} \cdot \frac{8}{7} & \frac{7}{8} \cdot \frac{8}{7} & \frac{7}{8} \cdot 8 \end{array}]$

Step 2.3.2

Simplify $R_{2}$ .

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

Step 2.4

Perform the row operation $R_{1} = R_{1} + \frac{1}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 2.4.1

Perform the row operation $R_{1} = R_{1} + \frac{1}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

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

Step 2.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 7 \end{array}]$

Step 3

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

$x = 1$

$y + z = 7$

Step 4

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

$(1, 7 - z, z)$