Matematica discreta Esempi

$7 x - 3 y + 3 z = 6$ , $7 x + 4 y - 3 z = 11$ , $28 x - 5 y + 6 z = 29$

Passaggio 1

Write the system as a matrix.

$[\begin{array}{c} 7 & - 3 & 3 & 6 \\ 7 & 4 & - 3 & 11 \\ 28 & - 5 & 6 & 29 \end{array}]$

Passaggio 2

Trova la forma ridotta a scala per righe di Echelon.

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Passaggio 2.1

Multiply each element of $R_{1}$ by $\frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

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Passaggio 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{7}{7} & \frac{- 3}{7} & \frac{3}{7} & \frac{6}{7} \\ 7 & 4 & - 3 & 11 \\ 28 & - 5 & 6 & 29 \end{array}]$

Passaggio 2.1.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{3}{7} & \frac{3}{7} & \frac{6}{7} \\ 7 & 4 & - 3 & 11 \\ 28 & - 5 & 6 & 29 \end{array}]$

Passaggio 2.2

Perform the row operation $R_{2} = R_{2} - 7 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Passaggio 2.2.1

Perform the row operation $R_{2} = R_{2} - 7 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{3}{7} & \frac{3}{7} & \frac{6}{7} \\ 7 - 7 \cdot 1 & 4 - 7 (- \frac{3}{7}) & - 3 - 7 (\frac{3}{7}) & 11 - 7 (\frac{6}{7}) \\ 28 & - 5 & 6 & 29 \end{array}]$

Passaggio 2.2.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{3}{7} & \frac{3}{7} & \frac{6}{7} \\ 0 & 7 & - 6 & 5 \\ 28 & - 5 & 6 & 29 \end{array}]$

Passaggio 2.3

Perform the row operation $R_{3} = R_{3} - 28 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Passaggio 2.3.1

Perform the row operation $R_{3} = R_{3} - 28 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{3}{7} & \frac{3}{7} & \frac{6}{7} \\ 0 & 7 & - 6 & 5 \\ 28 - 28 \cdot 1 & - 5 - 28 (- \frac{3}{7}) & 6 - 28 (\frac{3}{7}) & 29 - 28 (\frac{6}{7}) \end{array}]$

Passaggio 2.3.2

Semplifica $R_{3}$ .

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

Passaggio 2.4

Multiply each element of $R_{2}$ by $\frac{1}{7}$ to make the entry at $2, 2$ a $1$ .

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Passaggio 2.4.1

Multiply each element of $R_{2}$ by $\frac{1}{7}$ to make the entry at $2, 2$ a $1$ .

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

Passaggio 2.4.2

Semplifica $R_{2}$ .

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

Passaggio 2.5

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

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Passaggio 2.5.1

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

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

Passaggio 2.5.2

Semplifica $R_{3}$ .

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

Passaggio 2.6

Perform the row operation $R_{1} = R_{1} + \frac{3}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Passaggio 2.6.1

Perform the row operation $R_{1} = R_{1} + \frac{3}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

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

Passaggio 2.6.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{49} & \frac{57}{49} \\ 0 & 1 & - \frac{6}{7} & \frac{5}{7} \\ 0 & 0 & 0 & 0 \end{array}]$

Passaggio 3

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

$x + \frac{3}{49} z = \frac{57}{49}$

$y - \frac{6}{7} z = \frac{5}{7}$

$0 = 0$

Passaggio 4

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

$(\frac{57}{49} - \frac{3 z}{49}, \frac{5}{7} + \frac{6 z}{7}, z)$