Precalculus Examples

$4 x - y + 3 z = 12$ , $x + 4 y + 6 z = - 32$ , $5 x + 3 y + 9 z = 20$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 4 & - 1 & 3 & 12 \\ 1 & 4 & 6 & - 32 \\ 5 & 3 & 9 & 20 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} \frac{4}{4} & \frac{- 1}{4} & \frac{3}{4} & \frac{12}{4} \\ 1 & 4 & 6 & - 32 \\ 5 & 3 & 9 & 20 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\ 1 & 4 & 6 & - 32 \\ 5 & 3 & 9 & 20 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\ 1 - 1 & 4 + \frac{1}{4} & 6 - \frac{3}{4} & - 32 - 3 \\ 5 & 3 & 9 & 20 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\ 0 & \frac{17}{4} & \frac{21}{4} & - 35 \\ 5 & 3 & 9 & 20 \end{array}]$

Step 2.3

Perform the row operation $R_{3} = R_{3} - 5 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 2.3.1

Perform the row operation $R_{3} = R_{3} - 5 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\ 0 & \frac{17}{4} & \frac{21}{4} & - 35 \\ 5 - 5 \cdot 1 & 3 - 5 (- \frac{1}{4}) & 9 - 5 (\frac{3}{4}) & 20 - 5 \cdot 3 \end{array}]$

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\ 0 & \frac{17}{4} & \frac{21}{4} & - 35 \\ 0 & \frac{17}{4} & \frac{21}{4} & 5 \end{array}]$

Step 2.4

Multiply each element of $R_{2}$ by $\frac{4}{17}$ to make the entry at $2, 2$ a $1$ .

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Step 2.4.1

Multiply each element of $R_{2}$ by $\frac{4}{17}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\ \frac{4}{17} \cdot 0 & \frac{4}{17} \cdot \frac{17}{4} & \frac{4}{17} \cdot \frac{21}{4} & \frac{4}{17} \cdot - 35 \\ 0 & \frac{17}{4} & \frac{21}{4} & 5 \end{array}]$

Step 2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\ 0 & 1 & \frac{21}{17} & - \frac{140}{17} \\ 0 & \frac{17}{4} & \frac{21}{4} & 5 \end{array}]$

Step 2.5

Perform the row operation $R_{3} = R_{3} - \frac{17}{4} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 2.5.1

Perform the row operation $R_{3} = R_{3} - \frac{17}{4} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\ 0 & 1 & \frac{21}{17} & - \frac{140}{17} \\ 0 - \frac{17}{4} \cdot 0 & \frac{17}{4} - \frac{17}{4} \cdot 1 & \frac{21}{4} - \frac{17}{4} \cdot \frac{21}{17} & 5 - \frac{17}{4} (- \frac{140}{17}) \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\ 0 & 1 & \frac{21}{17} & - \frac{140}{17} \\ 0 & 0 & 0 & 40 \end{array}]$

Step 2.6

Multiply each element of $R_{3}$ by $\frac{1}{40}$ to make the entry at $3, 4$ a $1$ .

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Step 2.6.1

Multiply each element of $R_{3}$ by $\frac{1}{40}$ to make the entry at $3, 4$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\ 0 & 1 & \frac{21}{17} & - \frac{140}{17} \\ \frac{0}{40} & \frac{0}{40} & \frac{0}{40} & \frac{40}{40} \end{array}]$

Step 2.6.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\ 0 & 1 & \frac{21}{17} & - \frac{140}{17} \\ 0 & 0 & 0 & 1 \end{array}]$

Step 2.7

Perform the row operation $R_{2} = R_{2} + \frac{140}{17} R_{3}$ to make the entry at $2, 4$ a $0$ .

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Step 2.7.1

Perform the row operation $R_{2} = R_{2} + \frac{140}{17} R_{3}$ to make the entry at $2, 4$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\ 0 + \frac{140}{17} \cdot 0 & 1 + \frac{140}{17} \cdot 0 & \frac{21}{17} + \frac{140}{17} \cdot 0 & - \frac{140}{17} + \frac{140}{17} \cdot 1 \\ 0 & 0 & 0 & 1 \end{array}]$

Step 2.7.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\ 0 & 1 & \frac{21}{17} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

Step 2.8

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

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Step 2.8.1

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

$[\begin{array}{c} 1 - 3 \cdot 0 & - \frac{1}{4} - 3 \cdot 0 & \frac{3}{4} - 3 \cdot 0 & 3 - 3 \cdot 1 \\ 0 & 1 & \frac{21}{17} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

Step 2.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & \frac{3}{4} & 0 \\ 0 & 1 & \frac{21}{17} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

Step 2.9

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

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Step 2.9.1

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

$[\begin{array}{c} 1 + \frac{1}{4} \cdot 0 & - \frac{1}{4} + \frac{1}{4} \cdot 1 & \frac{3}{4} + \frac{1}{4} \cdot \frac{21}{17} & 0 + \frac{1}{4} \cdot 0 \\ 0 & 1 & \frac{21}{17} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

Step 2.9.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{18}{17} & 0 \\ 0 & 1 & \frac{21}{17} & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

Step 3

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

$x + \frac{18}{17} z = 0$

$y + \frac{21}{17} z = 0$

$0 = 1$

Step 4

The system is inconsistent so there is no solution.

$No solution$