Matematica discreta Esempi

$5 x - 1 y = 10$ , $x - 2 y = 8$

Passaggio 1

Riscrivi $- 1 y$ come $- y$ .

$5 x - y = 10$

$x - 2 y = 8$

Passaggio 2

Write the system as a matrix.

$[\begin{array}{c} 5 & - 1 & 10 \\ 1 & - 2 & 8 \end{array}]$

Passaggio 3

Trova la forma ridotta a scala per righe di Echelon.

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Passaggio 3.1

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

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Passaggio 3.1.1

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

$[\begin{array}{c} \frac{5}{5} & \frac{- 1}{5} & \frac{10}{5} \\ 1 & - 2 & 8 \end{array}]$

Passaggio 3.1.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & 2 \\ 1 & - 2 & 8 \end{array}]$

Passaggio 3.2

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

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Passaggio 3.2.1

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

$[\begin{array}{c} 1 & - \frac{1}{5} & 2 \\ 1 - 1 & - 2 + \frac{1}{5} & 8 - 2 \end{array}]$

Passaggio 3.2.2

Semplifica $R_{2}$ .

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

Passaggio 3.3

Multiply each element of $R_{2}$ by $- \frac{5}{9}$ to make the entry at $2, 2$ a $1$ .

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Passaggio 3.3.1

Multiply each element of $R_{2}$ by $- \frac{5}{9}$ to make the entry at $2, 2$ a $1$ .

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

Passaggio 3.3.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{5} & 2 \\ 0 & 1 & - \frac{10}{3} \end{array}]$

Passaggio 3.4

Perform the row operation $R_{1} = R_{1} + \frac{1}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Passaggio 3.4.1

Perform the row operation $R_{1} = R_{1} + \frac{1}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{1}{5} \cdot 0 & - \frac{1}{5} + \frac{1}{5} \cdot 1 & 2 + \frac{1}{5} (- \frac{10}{3}) \\ 0 & 1 & - \frac{10}{3} \end{array}]$

Passaggio 3.4.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{4}{3} \\ 0 & 1 & - \frac{10}{3} \end{array}]$

Passaggio 4

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

$x = \frac{4}{3}$

$y = - \frac{10}{3}$

Passaggio 5

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

$(\frac{4}{3}, - \frac{10}{3})$