Precalculus Examples

$4 x + 3 y + z = 7$ , $x - 3 y + 2 z = 5$ , $11 x - 2 y + 3 z = 10$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 4 & 3 & 1 & 7 \\ 1 & - 3 & 2 & 5 \\ 11 & - 2 & 3 & 10 \end{array}]$

Step 2

Find the reduced row echelon form.

Tap for more steps...

Step 2.1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

Tap for more steps...

Step 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{4}{4} & \frac{3}{4} & \frac{1}{4} & \frac{7}{4} \\ 1 & - 3 & 2 & 5 \\ 11 & - 2 & 3 & 10 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & \frac{7}{4} \\ 1 & - 3 & 2 & 5 \\ 11 & - 2 & 3 & 10 \end{array}]$

Step 2.2

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

Tap for more steps...

Step 2.2.1

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

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & \frac{7}{4} \\ 1 - 1 & - 3 - \frac{3}{4} & 2 - \frac{1}{4} & 5 - \frac{7}{4} \\ 11 & - 2 & 3 & 10 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & \frac{7}{4} \\ 0 & - \frac{15}{4} & \frac{7}{4} & \frac{13}{4} \\ 11 & - 2 & 3 & 10 \end{array}]$

Step 2.3

Perform the row operation $R_{3} = R_{3} - 11 R_{1}$ to make the entry at $3, 1$ a $0$ .

Tap for more steps...

Step 2.3.1

Perform the row operation $R_{3} = R_{3} - 11 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & \frac{7}{4} \\ 0 & - \frac{15}{4} & \frac{7}{4} & \frac{13}{4} \\ 11 - 11 \cdot 1 & - 2 - 11 (\frac{3}{4}) & 3 - 11 (\frac{1}{4}) & 10 - 11 (\frac{7}{4}) \end{array}]$

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & \frac{7}{4} \\ 0 & - \frac{15}{4} & \frac{7}{4} & \frac{13}{4} \\ 0 & - \frac{41}{4} & \frac{1}{4} & - \frac{37}{4} \end{array}]$

Step 2.4

Multiply each element of $R_{2}$ by $- \frac{4}{15}$ to make the entry at $2, 2$ a $1$ .

Tap for more steps...

Step 2.4.1

Multiply each element of $R_{2}$ by $- \frac{4}{15}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & \frac{7}{4} \\ - \frac{4}{15} \cdot 0 & - \frac{4}{15} (- \frac{15}{4}) & - \frac{4}{15} \cdot \frac{7}{4} & - \frac{4}{15} \cdot \frac{13}{4} \\ 0 & - \frac{41}{4} & \frac{1}{4} & - \frac{37}{4} \end{array}]$

Step 2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & \frac{7}{4} \\ 0 & 1 & - \frac{7}{15} & - \frac{13}{15} \\ 0 & - \frac{41}{4} & \frac{1}{4} & - \frac{37}{4} \end{array}]$

Step 2.5

Perform the row operation $R_{3} = R_{3} + \frac{41}{4} R_{2}$ to make the entry at $3, 2$ a $0$ .

Tap for more steps...

Step 2.5.1

Perform the row operation $R_{3} = R_{3} + \frac{41}{4} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & \frac{7}{4} \\ 0 & 1 & - \frac{7}{15} & - \frac{13}{15} \\ 0 + \frac{41}{4} \cdot 0 & - \frac{41}{4} + \frac{41}{4} \cdot 1 & \frac{1}{4} + \frac{41}{4} (- \frac{7}{15}) & - \frac{37}{4} + \frac{41}{4} (- \frac{13}{15}) \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & \frac{7}{4} \\ 0 & 1 & - \frac{7}{15} & - \frac{13}{15} \\ 0 & 0 & - \frac{68}{15} & - \frac{272}{15} \end{array}]$

Step 2.6

Multiply each element of $R_{3}$ by $- \frac{15}{68}$ to make the entry at $3, 3$ a $1$ .

Tap for more steps...

Step 2.6.1

Multiply each element of $R_{3}$ by $- \frac{15}{68}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & \frac{7}{4} \\ 0 & 1 & - \frac{7}{15} & - \frac{13}{15} \\ - \frac{15}{68} \cdot 0 & - \frac{15}{68} \cdot 0 & - \frac{15}{68} (- \frac{68}{15}) & - \frac{15}{68} (- \frac{272}{15}) \end{array}]$

Step 2.6.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & \frac{7}{4} \\ 0 & 1 & - \frac{7}{15} & - \frac{13}{15} \\ 0 & 0 & 1 & 4 \end{array}]$

Step 2.7

Perform the row operation $R_{2} = R_{2} + \frac{7}{15} R_{3}$ to make the entry at $2, 3$ a $0$ .

Tap for more steps...

Step 2.7.1

Perform the row operation $R_{2} = R_{2} + \frac{7}{15} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & \frac{7}{4} \\ 0 + \frac{7}{15} \cdot 0 & 1 + \frac{7}{15} \cdot 0 & - \frac{7}{15} + \frac{7}{15} \cdot 1 & - \frac{13}{15} + \frac{7}{15} \cdot 4 \\ 0 & 0 & 1 & 4 \end{array}]$

Step 2.7.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{4} & \frac{1}{4} & \frac{7}{4} \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 4 \end{array}]$

Step 2.8

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

Tap for more steps...

Step 2.8.1

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

$[\begin{array}{c} 1 - \frac{1}{4} \cdot 0 & \frac{3}{4} - \frac{1}{4} \cdot 0 & \frac{1}{4} - \frac{1}{4} \cdot 1 & \frac{7}{4} - \frac{1}{4} \cdot 4 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 4 \end{array}]$

Step 2.8.2

Simplify $R_{1}$ .

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

Step 2.9

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

Tap for more steps...

Step 2.9.1

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

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

Step 2.9.2

Simplify $R_{1}$ .

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

Step 3

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

$x = 0$

$y = 1$

$z = 4$

Step 4

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

$(0, 1, 4)$