Precalcolo Esempi

$x - y - 2 z = - 6$ , $3 x + 2 y = - 25$ , $- 4 x + y - z = 12$

Passaggio 1

Write the system as a matrix.

$[\begin{array}{c} 1 & - 1 & - 2 & - 6 \\ 3 & 2 & 0 & - 25 \\ - 4 & 1 & - 1 & 12 \end{array}]$

Passaggio 2

Trova la forma ridotta a scala per righe di Echelon.

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Passaggio 2.1

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

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Passaggio 2.1.1

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

$[\begin{array}{c} 1 & - 1 & - 2 & - 6 \\ 3 - 3 \cdot 1 & 2 - 3 \cdot - 1 & 0 - 3 \cdot - 2 & - 25 - 3 \cdot - 6 \\ - 4 & 1 & - 1 & 12 \end{array}]$

Passaggio 2.1.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 2 & - 6 \\ 0 & 5 & 6 & - 7 \\ - 4 & 1 & - 1 & 12 \end{array}]$

Passaggio 2.2

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

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Passaggio 2.2.1

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

$[\begin{array}{c} 1 & - 1 & - 2 & - 6 \\ 0 & 5 & 6 & - 7 \\ - 4 + 4 \cdot 1 & 1 + 4 \cdot - 1 & - 1 + 4 \cdot - 2 & 12 + 4 \cdot - 6 \end{array}]$

Passaggio 2.2.2

Semplifica $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - 2 & - 6 \\ 0 & 5 & 6 & - 7 \\ 0 & - 3 & - 9 & - 12 \end{array}]$

Passaggio 2.3

Multiply each element of $R_{2}$ by $\frac{1}{5}$ to make the entry at $2, 2$ a $1$ .

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Passaggio 2.3.1

Multiply each element of $R_{2}$ by $\frac{1}{5}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 1 & - 2 & - 6 \\ \frac{0}{5} & \frac{5}{5} & \frac{6}{5} & \frac{- 7}{5} \\ 0 & - 3 & - 9 & - 12 \end{array}]$

Passaggio 2.3.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 2 & - 6 \\ 0 & 1 & \frac{6}{5} & - \frac{7}{5} \\ 0 & - 3 & - 9 & - 12 \end{array}]$

Passaggio 2.4

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

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Passaggio 2.4.1

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

$[\begin{array}{c} 1 & - 1 & - 2 & - 6 \\ 0 & 1 & \frac{6}{5} & - \frac{7}{5} \\ 0 + 3 \cdot 0 & - 3 + 3 \cdot 1 & - 9 + 3 (\frac{6}{5}) & - 12 + 3 (- \frac{7}{5}) \end{array}]$

Passaggio 2.4.2

Semplifica $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - 2 & - 6 \\ 0 & 1 & \frac{6}{5} & - \frac{7}{5} \\ 0 & 0 & - \frac{27}{5} & - \frac{81}{5} \end{array}]$

Passaggio 2.5

Multiply each element of $R_{3}$ by $- \frac{5}{27}$ to make the entry at $3, 3$ a $1$ .

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Passaggio 2.5.1

Multiply each element of $R_{3}$ by $- \frac{5}{27}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - 1 & - 2 & - 6 \\ 0 & 1 & \frac{6}{5} & - \frac{7}{5} \\ - \frac{5}{27} \cdot 0 & - \frac{5}{27} \cdot 0 & - \frac{5}{27} (- \frac{27}{5}) & - \frac{5}{27} (- \frac{81}{5}) \end{array}]$

Passaggio 2.5.2

Semplifica $R_{3}$ .

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

Passaggio 2.6

Perform the row operation $R_{2} = R_{2} - \frac{6}{5} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Passaggio 2.6.1

Perform the row operation $R_{2} = R_{2} - \frac{6}{5} R_{3}$ to make the entry at $2, 3$ a $0$ .

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

Passaggio 2.6.2

Semplifica $R_{2}$ .

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

Passaggio 2.7

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

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Passaggio 2.7.1

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

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

Passaggio 2.7.2

Semplifica $R_{1}$ .

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

Passaggio 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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Passaggio 2.8.1

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

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

Passaggio 2.8.2

Semplifica $R_{1}$ .

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

Passaggio 3

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

$x = - 5$

$y = - 5$

$z = 3$

Passaggio 4

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

$(- 5, - 5, 3)$