Matematica discreta Esempi

$2 x - 2 y = - 5$ , $2 y + z = 0$ , $2 x + z = - 7$

Passaggio 1

Write the system as a matrix.

$[\begin{array}{c} 2 & - 2 & 0 & - 5 \\ 0 & 2 & 1 & 0 \\ 2 & 0 & 1 & - 7 \end{array}]$

Passaggio 2

Trova la forma ridotta a scala per righe di Echelon.

Tocca per altri passaggi...

Passaggio 2.1

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

Tocca per altri passaggi...

Passaggio 2.1.1

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

$[\begin{array}{c} \frac{2}{2} & \frac{- 2}{2} & \frac{0}{2} & \frac{- 5}{2} \\ 0 & 2 & 1 & 0 \\ 2 & 0 & 1 & - 7 \end{array}]$

Passaggio 2.1.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & - 1 & 0 & - \frac{5}{2} \\ 0 & 2 & 1 & 0 \\ 2 & 0 & 1 & - 7 \end{array}]$

Passaggio 2.2

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

Tocca per altri passaggi...

Passaggio 2.2.1

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

$[\begin{array}{c} 1 & - 1 & 0 & - \frac{5}{2} \\ 0 & 2 & 1 & 0 \\ 2 - 2 \cdot 1 & 0 - 2 \cdot - 1 & 1 - 2 \cdot 0 & - 7 - 2 (- \frac{5}{2}) \end{array}]$

Passaggio 2.2.2

Semplifica $R_{3}$ .

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

Passaggio 2.3

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

Tocca per altri passaggi...

Passaggio 2.3.1

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

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

Passaggio 2.3.2

Semplifica $R_{2}$ .

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

Passaggio 2.4

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

Tocca per altri passaggi...

Passaggio 2.4.1

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

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

Passaggio 2.4.2

Semplifica $R_{3}$ .

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

Passaggio 2.5

Multiply each element of $R_{3}$ by $- \frac{1}{2}$ to make the entry at $3, 4$ a $1$ .

Tocca per altri passaggi...

Passaggio 2.5.1

Multiply each element of $R_{3}$ by $- \frac{1}{2}$ to make the entry at $3, 4$ a $1$ .

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

Passaggio 2.5.2

Semplifica $R_{3}$ .

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

Passaggio 2.6

Perform the row operation $R_{1} = R_{1} + \frac{5}{2} R_{3}$ to make the entry at $1, 4$ a $0$ .

Tocca per altri passaggi...

Passaggio 2.6.1

Perform the row operation $R_{1} = R_{1} + \frac{5}{2} R_{3}$ to make the entry at $1, 4$ a $0$ .

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

Passaggio 2.6.2

Semplifica $R_{1}$ .

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

Passaggio 2.7

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

Tocca per altri passaggi...

Passaggio 2.7.1

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

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

Passaggio 2.7.2

Semplifica $R_{1}$ .

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

Passaggio 3

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

$x + \frac{1}{2} z = 0$

$y + \frac{1}{2} z = 0$

$0 = 1$

Passaggio 4

The system is inconsistent so there is no solution.

$Nessuna soluzione$