Matematica discreta Esempi

$2 x + 3 y = 11$ , $- 4 x - 6 y = - 22$

Passaggio 1

Write the system as a matrix.

$[\begin{array}{c} 2 & 3 & 11 \\ - 4 & - 6 & - 22 \end{array}]$

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 $\frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

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Passaggio 2.1.1

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

$[\begin{array}{c} \frac{2}{2} & \frac{3}{2} & \frac{11}{2} \\ - 4 & - 6 & - 22 \end{array}]$

Passaggio 2.1.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & \frac{11}{2} \\ - 4 & - 6 & - 22 \end{array}]$

Passaggio 2.2

Perform the row operation $R_{2} = R_{2} + 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Passaggio 2.2.1

Perform the row operation $R_{2} = R_{2} + 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{2} & \frac{11}{2} \\ - 4 + 4 \cdot 1 & - 6 + 4 (\frac{3}{2}) & - 22 + 4 (\frac{11}{2}) \end{array}]$

Passaggio 2.2.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & \frac{11}{2} \\ 0 & 0 & 0 \end{array}]$

Passaggio 3

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

$x + \frac{3}{2} y = \frac{11}{2}$

$0 = 0$

Passaggio 4

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

$(\frac{11}{2} - \frac{3 y}{2}, y)$