Matematica discreta Esempi

$6 x + y = 59$ , $41 x + 7 y = 404$

Passaggio 1

Write the system as a matrix.

$[\begin{array}{c} 6 & 1 & 59 \\ 41 & 7 & 404 \end{array}]$

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 $\frac{1}{6}$ to make the entry at $1, 1$ a $1$ .

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Passaggio 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{6}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{6}{6} & \frac{1}{6} & \frac{59}{6} \\ 41 & 7 & 404 \end{array}]$

Passaggio 2.1.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{6} & \frac{59}{6} \\ 41 & 7 & 404 \end{array}]$

Passaggio 2.2

Perform the row operation $R_{2} = R_{2} - 41 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Passaggio 2.2.1

Perform the row operation $R_{2} = R_{2} - 41 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{6} & \frac{59}{6} \\ 41 - 41 \cdot 1 & 7 - 41 (\frac{1}{6}) & 404 - 41 (\frac{59}{6}) \end{array}]$

Passaggio 2.2.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{6} & \frac{59}{6} \\ 0 & \frac{1}{6} & \frac{5}{6} \end{array}]$

Passaggio 2.3

Multiply each element of $R_{2}$ by $6$ to make the entry at $2, 2$ a $1$ .

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Passaggio 2.3.1

Multiply each element of $R_{2}$ by $6$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{6} & \frac{59}{6} \\ 6 \cdot 0 & 6 (\frac{1}{6}) & 6 (\frac{5}{6}) \end{array}]$

Passaggio 2.3.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{6} & \frac{59}{6} \\ 0 & 1 & 5 \end{array}]$

Passaggio 2.4

Perform the row operation $R_{1} = R_{1} - \frac{1}{6} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Passaggio 2.4.1

Perform the row operation $R_{1} = R_{1} - \frac{1}{6} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{1}{6} \cdot 0 & \frac{1}{6} - \frac{1}{6} \cdot 1 & \frac{59}{6} - \frac{1}{6} \cdot 5 \\ 0 & 1 & 5 \end{array}]$

Passaggio 2.4.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 9 \\ 0 & 1 & 5 \end{array}]$

Passaggio 3

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

$x = 9$

$y = 5$

Passaggio 4

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

$(9, 5)$