Matematica discreta Esempi

$2 x - 3 y = 14$ , $3 x + 2 y = - 5$

Passaggio 1

Write the system as a matrix.

$[\begin{array}{c} 2 & - 3 & 14 \\ 3 & 2 & - 5 \end{array}]$

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 $1$ .

$[\begin{array}{c} \frac{2}{2} & \frac{- 3}{2} & \frac{14}{2} \\ 3 & 2 & - 5 \end{array}]$

Passaggio 2.1.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{3}{2} & 7 \\ 3 & 2 & - 5 \end{array}]$

Passaggio 2.2

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

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Passaggio 2.2.1

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

$[\begin{array}{c} 1 & - \frac{3}{2} & 7 \\ 3 - 3 \cdot 1 & 2 - 3 (- \frac{3}{2}) & - 5 - 3 \cdot 7 \end{array}]$

Passaggio 2.2.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{3}{2} & 7 \\ 0 & \frac{13}{2} & - 26 \end{array}]$

Passaggio 2.3

Multiply each element of $R_{2}$ by $\frac{2}{13}$ to make the entry at $2, 2$ a $1$ .

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Passaggio 2.3.1

Multiply each element of $R_{2}$ by $\frac{2}{13}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{3}{2} & 7 \\ \frac{2}{13} \cdot 0 & \frac{2}{13} \cdot \frac{13}{2} & \frac{2}{13} \cdot - 26 \end{array}]$

Passaggio 2.3.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{3}{2} & 7 \\ 0 & 1 & - 4 \end{array}]$

Passaggio 2.4

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

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Passaggio 2.4.1

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

$[\begin{array}{c} 1 + \frac{3}{2} \cdot 0 & - \frac{3}{2} + \frac{3}{2} \cdot 1 & 7 + \frac{3}{2} \cdot - 4 \\ 0 & 1 & - 4 \end{array}]$

Passaggio 2.4.2

Semplifica $R_{1}$ .

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

Passaggio 3

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

$x = 1$

$y = - 4$

Passaggio 4

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

$(1, - 4)$