Precalculus Examples

$x - y + 3 z = 3$ , $2 x + y - z = 0$ , $2 x + 3 y - 5 z = - 2$

Step 1

Write the system as a matrix.

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

Step 2

Find the reduced row echelon form.

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Step 2.1

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 2.1.1

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

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

Step 2.1.2

Simplify $R_{2}$ .

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & 3 & 3 \\ 0 & 3 & - 7 & - 6 \\ 0 & 5 & - 11 & - 8 \end{array}]$

Step 2.3

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

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Step 2.3.1

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

$[\begin{array}{c} 1 & - 1 & 3 & 3 \\ \frac{0}{3} & \frac{3}{3} & \frac{- 7}{3} & \frac{- 6}{3} \\ 0 & 5 & - 11 & - 8 \end{array}]$

Step 2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & 3 & 3 \\ 0 & 1 & - \frac{7}{3} & - 2 \\ 0 & 5 & - 11 & - 8 \end{array}]$

Step 2.4

Perform the row operation $R_{3} = R_{3} - 5 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 2.4.1

Perform the row operation $R_{3} = R_{3} - 5 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - 1 & 3 & 3 \\ 0 & 1 & - \frac{7}{3} & - 2 \\ 0 - 5 \cdot 0 & 5 - 5 \cdot 1 & - 11 - 5 (- \frac{7}{3}) & - 8 - 5 \cdot - 2 \end{array}]$

Step 2.4.2

Simplify $R_{3}$ .

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

Step 2.5

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

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Step 2.5.1

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

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

Step 2.5.2

Simplify $R_{3}$ .

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

Step 2.6

Perform the row operation $R_{2} = R_{2} + \frac{7}{3} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 2.6.1

Perform the row operation $R_{2} = R_{2} + \frac{7}{3} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - 1 & 3 & 3 \\ 0 + \frac{7}{3} \cdot 0 & 1 + \frac{7}{3} \cdot 0 & - \frac{7}{3} + \frac{7}{3} \cdot 1 & - 2 + \frac{7}{3} \cdot 3 \\ 0 & 0 & 1 & 3 \end{array}]$

Step 2.6.2

Simplify $R_{2}$ .

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

Step 2.7

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

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Step 2.7.1

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

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

Step 2.7.2

Simplify $R_{1}$ .

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

Step 2.8

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

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Step 2.8.1

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

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & 0 + 0 & - 6 + 1 \cdot 5 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 1 & 3 \end{array}]$

Step 2.8.2

Simplify $R_{1}$ .

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

Step 3

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

$x = - 1$

$y = 5$

$z = 3$

Step 4

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

$(- 1, 5, 3)$