Matematica discreta Esempi

$3 x - 2 y = 7$ , $9 x - 6 y = 14$

Passaggio 1

Write the system as a matrix.

$[\begin{array}{c} 3 & - 2 & 7 \\ 9 & - 6 & 14 \end{array}]$

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 $1$ .

$[\begin{array}{c} \frac{3}{3} & \frac{- 2}{3} & \frac{7}{3} \\ 9 & - 6 & 14 \end{array}]$

Passaggio 2.1.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{7}{3} \\ 9 & - 6 & 14 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{7}{3} \\ 9 - 9 \cdot 1 & - 6 - 9 (- \frac{2}{3}) & 14 - 9 (\frac{7}{3}) \end{array}]$

Passaggio 2.2.2

Semplifica $R_{2}$ .

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

Passaggio 2.3

Multiply each element of $R_{2}$ by $- \frac{1}{7}$ to make the entry at $2, 3$ a $1$ .

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Passaggio 2.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{7}$ to make the entry at $2, 3$ a $1$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & \frac{7}{3} \\ - \frac{1}{7} \cdot 0 & - \frac{1}{7} \cdot 0 & - \frac{1}{7} \cdot - 7 \end{array}]$

Passaggio 2.3.2

Semplifica $R_{2}$ .

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

Passaggio 2.4

Perform the row operation $R_{1} = R_{1} - \frac{7}{3} R_{2}$ to make the entry at $1, 3$ a $0$ .

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Passaggio 2.4.1

Perform the row operation $R_{1} = R_{1} - \frac{7}{3} R_{2}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - \frac{7}{3} \cdot 0 & - \frac{2}{3} - \frac{7}{3} \cdot 0 & \frac{7}{3} - \frac{7}{3} \cdot 1 \\ 0 & 0 & 1 \end{array}]$

Passaggio 2.4.2

Semplifica $R_{1}$ .

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

Passaggio 3

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

$x - \frac{2}{3} y = 0$

$0 = 1$

Passaggio 4

The system is inconsistent so there is no solution.

$Nessuna soluzione$