Matematica discreta Esempi

$x - y = 1$ , $2 x - 8 z = 8$ , $8 y + z = 6$

Passaggio 1

Write the system as a matrix.

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

Passaggio 2

Trova la forma ridotta a scala per righe di Echelon.

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Passaggio 2.1

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

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Passaggio 2.1.1

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

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

Passaggio 2.1.2

Semplifica $R_{2}$ .

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

Passaggio 2.2

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

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Passaggio 2.2.1

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

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

Passaggio 2.2.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & 0 & 1 \\ 0 & 1 & - 4 & 3 \\ 0 & 8 & 1 & 6 \end{array}]$

Passaggio 2.3

Perform the row operation $R_{3} = R_{3} - 8 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Passaggio 2.3.1

Perform the row operation $R_{3} = R_{3} - 8 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - 1 & 0 & 1 \\ 0 & 1 & - 4 & 3 \\ 0 - 8 \cdot 0 & 8 - 8 \cdot 1 & 1 - 8 \cdot - 4 & 6 - 8 \cdot 3 \end{array}]$

Passaggio 2.3.2

Semplifica $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & 0 & 1 \\ 0 & 1 & - 4 & 3 \\ 0 & 0 & 33 & - 18 \end{array}]$

Passaggio 2.4

Multiply each element of $R_{3}$ by $\frac{1}{33}$ to make the entry at $3, 3$ a $1$ .

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Passaggio 2.4.1

Multiply each element of $R_{3}$ by $\frac{1}{33}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - 1 & 0 & 1 \\ 0 & 1 & - 4 & 3 \\ \frac{0}{33} & \frac{0}{33} & \frac{33}{33} & \frac{- 18}{33} \end{array}]$

Passaggio 2.4.2

Semplifica $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & 0 & 1 \\ 0 & 1 & - 4 & 3 \\ 0 & 0 & 1 & - \frac{6}{11} \end{array}]$

Passaggio 2.5

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

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Passaggio 2.5.1

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

$[\begin{array}{c} 1 & - 1 & 0 & 1 \\ 0 + 4 \cdot 0 & 1 + 4 \cdot 0 & - 4 + 4 \cdot 1 & 3 + 4 (- \frac{6}{11}) \\ 0 & 0 & 1 & - \frac{6}{11} \end{array}]$

Passaggio 2.5.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & 0 & 1 \\ 0 & 1 & 0 & \frac{9}{11} \\ 0 & 0 & 1 & - \frac{6}{11} \end{array}]$

Passaggio 2.6

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

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Passaggio 2.6.1

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

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & 0 + 0 & 1 + \frac{9}{11} \\ 0 & 1 & 0 & \frac{9}{11} \\ 0 & 0 & 1 & - \frac{6}{11} \end{array}]$

Passaggio 2.6.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{20}{11} \\ 0 & 1 & 0 & \frac{9}{11} \\ 0 & 0 & 1 & - \frac{6}{11} \end{array}]$

Passaggio 3

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

$x = \frac{20}{11}$

$y = \frac{9}{11}$

$z = - \frac{6}{11}$

Passaggio 4

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

$(\frac{20}{11}, \frac{9}{11}, - \frac{6}{11})$