Finite Math Examples

$- x - y - z = - 4$ , $x + x + z = 0$ , $x + y - z = - 6$

Step 1

Add $x$ and $x$ .

$- x - y - z = - 4$

$2 x + z = 0$

$x + y - z = - 6$

Step 2

Write the system as a matrix.

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

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

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

Step 3.1.2

Simplify $R_{1}$ .

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 4 \\ 0 & - 2 & - 1 & - 8 \\ 1 & 1 & - 1 & - 6 \end{array}]$

Step 3.3

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

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Step 3.3.1

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

$[\begin{array}{c} 1 & 1 & 1 & 4 \\ 0 & - 2 & - 1 & - 8 \\ 1 - 1 & 1 - 1 & - 1 - 1 & - 6 - 4 \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 4 \\ 0 & - 2 & - 1 & - 8 \\ 0 & 0 & - 2 & - 10 \end{array}]$

Step 3.4

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

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Step 3.4.1

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

$[\begin{array}{c} 1 & 1 & 1 & 4 \\ - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot - 1 & - \frac{1}{2} \cdot - 8 \\ 0 & 0 & - 2 & - 10 \end{array}]$

Step 3.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 4 \\ 0 & 1 & \frac{1}{2} & 4 \\ 0 & 0 & - 2 & - 10 \end{array}]$

Step 3.5

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

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Step 3.5.1

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

$[\begin{array}{c} 1 & 1 & 1 & 4 \\ 0 & 1 & \frac{1}{2} & 4 \\ - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot - 10 \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

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

Step 3.6

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

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Step 3.6.1

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

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

Step 3.6.2

Simplify $R_{2}$ .

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

Step 3.7

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

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Step 3.7.1

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

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

Step 3.7.2

Simplify $R_{1}$ .

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

Step 3.8

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

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Step 3.8.1

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

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

Step 3.8.2

Simplify $R_{1}$ .

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

Step 4

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

$x = - \frac{5}{2}$

$y = \frac{3}{2}$

$z = 5$

Step 5

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

$(- \frac{5}{2}, \frac{3}{2}, 5)$