Matematica discreta Esempi

$3 x + 2 y = 1$ , $- 9 x - 6 y = - 3$

Passaggio 1

Write the system as a matrix.

$[\begin{array}{c} 3 & 2 & 1 \\ - 9 & - 6 & - 3 \end{array}]$

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 $\frac{1}{3}$ to make the entry at $1, 1$ a $1$ .

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Passaggio 2.1.1

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

$[\begin{array}{c} \frac{3}{3} & \frac{2}{3} & \frac{1}{3} \\ - 9 & - 6 & - 3 \end{array}]$

Passaggio 2.1.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{1}{3} \\ - 9 & - 6 & - 3 \end{array}]$

Passaggio 2.2

Perform the row operation $R_{2} = R_{2} + 9 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Passaggio 2.2.1

Perform the row operation $R_{2} = R_{2} + 9 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{1}{3} \\ - 9 + 9 \cdot 1 & - 6 + 9 (\frac{2}{3}) & - 3 + 9 (\frac{1}{3}) \end{array}]$

Passaggio 2.2.2

Semplifica $R_{2}$ .

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

Passaggio 3

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

$x + \frac{2}{3} y = \frac{1}{3}$

$0 = 0$

Passaggio 4

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

$(\frac{1}{3} - \frac{2 y}{3}, y)$