Matematica discreta Esempi

$10 x + 20 y + 30 z = 14040$ , $5 y + 7 z = 3096$

Passaggio 1

Write the system as a matrix.

$[\begin{array}{c} 10 & 20 & 30 & 14040 \\ 0 & 5 & 7 & 3096 \end{array}]$

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 $\frac{1}{10}$ to make the entry at $1, 1$ a $1$ .

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Passaggio 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{10}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{10}{10} & \frac{20}{10} & \frac{30}{10} & \frac{14040}{10} \\ 0 & 5 & 7 & 3096 \end{array}]$

Passaggio 2.1.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & 2 & 3 & 1404 \\ 0 & 5 & 7 & 3096 \end{array}]$

Passaggio 2.2

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

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Passaggio 2.2.1

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

$[\begin{array}{c} 1 & 2 & 3 & 1404 \\ \frac{0}{5} & \frac{5}{5} & \frac{7}{5} & \frac{3096}{5} \end{array}]$

Passaggio 2.2.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 3 & 1404 \\ 0 & 1 & \frac{7}{5} & \frac{3096}{5} \end{array}]$

Passaggio 2.3

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

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Passaggio 2.3.1

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

$[\begin{array}{c} 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & 3 - 2 (\frac{7}{5}) & 1404 - 2 (\frac{3096}{5}) \\ 0 & 1 & \frac{7}{5} & \frac{3096}{5} \end{array}]$

Passaggio 2.3.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{1}{5} & \frac{828}{5} \\ 0 & 1 & \frac{7}{5} & \frac{3096}{5} \end{array}]$

Passaggio 3

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

$x + \frac{1}{5} z = \frac{828}{5}$

$y + \frac{7}{5} z = \frac{3096}{5}$

Passaggio 4

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

$(\frac{828}{5} - \frac{z}{5}, \frac{3096}{5} - \frac{7 z}{5}, z)$