Matematica discreta Esempi

$4 x - 2 y - 5 z = - 51$ , $2 x - 2 y - 2 z = - 42$ , $4 x - 2 y + 4 z = - 24$

Passaggio 1

Write the system as a matrix.

$[\begin{array}{c} 4 & - 2 & - 5 & - 51 \\ 2 & - 2 & - 2 & - 42 \\ 4 & - 2 & 4 & - 24 \end{array}]$

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 $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

Tocca per altri passaggi...

Passaggio 2.1.1

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

$[\begin{array}{c} \frac{4}{4} & \frac{- 2}{4} & \frac{- 5}{4} & \frac{- 51}{4} \\ 2 & - 2 & - 2 & - 42 \\ 4 & - 2 & 4 & - 24 \end{array}]$

Passaggio 2.1.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{5}{4} & - \frac{51}{4} \\ 2 & - 2 & - 2 & - 42 \\ 4 & - 2 & 4 & - 24 \end{array}]$

Passaggio 2.2

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

Tocca per altri passaggi...

Passaggio 2.2.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{5}{4} & - \frac{51}{4} \\ 2 - 2 \cdot 1 & - 2 - 2 (- \frac{1}{2}) & - 2 - 2 (- \frac{5}{4}) & - 42 - 2 (- \frac{51}{4}) \\ 4 & - 2 & 4 & - 24 \end{array}]$

Passaggio 2.2.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{5}{4} & - \frac{51}{4} \\ 0 & - 1 & \frac{1}{2} & - \frac{33}{2} \\ 4 & - 2 & 4 & - 24 \end{array}]$

Passaggio 2.3

Perform the row operation $R_{3} = R_{3} - 4 R_{1}$ to make the entry at $3, 1$ a $0$ .

Tocca per altri passaggi...

Passaggio 2.3.1

Perform the row operation $R_{3} = R_{3} - 4 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{5}{4} & - \frac{51}{4} \\ 0 & - 1 & \frac{1}{2} & - \frac{33}{2} \\ 4 - 4 \cdot 1 & - 2 - 4 (- \frac{1}{2}) & 4 - 4 (- \frac{5}{4}) & - 24 - 4 (- \frac{51}{4}) \end{array}]$

Passaggio 2.3.2

Semplifica $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{5}{4} & - \frac{51}{4} \\ 0 & - 1 & \frac{1}{2} & - \frac{33}{2} \\ 0 & 0 & 9 & 27 \end{array}]$

Passaggio 2.4

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

Tocca per altri passaggi...

Passaggio 2.4.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{5}{4} & - \frac{51}{4} \\ - 0 & - - 1 & - \frac{1}{2} & - - \frac{33}{2} \\ 0 & 0 & 9 & 27 \end{array}]$

Passaggio 2.4.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{5}{4} & - \frac{51}{4} \\ 0 & 1 & - \frac{1}{2} & \frac{33}{2} \\ 0 & 0 & 9 & 27 \end{array}]$

Passaggio 2.5

Multiply each element of $R_{3}$ by $\frac{1}{9}$ to make the entry at $3, 3$ a $1$ .

Tocca per altri passaggi...

Passaggio 2.5.1

Multiply each element of $R_{3}$ by $\frac{1}{9}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{5}{4} & - \frac{51}{4} \\ 0 & 1 & - \frac{1}{2} & \frac{33}{2} \\ \frac{0}{9} & \frac{0}{9} & \frac{9}{9} & \frac{27}{9} \end{array}]$

Passaggio 2.5.2

Semplifica $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{5}{4} & - \frac{51}{4} \\ 0 & 1 & - \frac{1}{2} & \frac{33}{2} \\ 0 & 0 & 1 & 3 \end{array}]$

Passaggio 2.6

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

Tocca per altri passaggi...

Passaggio 2.6.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{5}{4} & - \frac{51}{4} \\ 0 + \frac{1}{2} \cdot 0 & 1 + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & \frac{33}{2} + \frac{1}{2} \cdot 3 \\ 0 & 0 & 1 & 3 \end{array}]$

Passaggio 2.6.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{5}{4} & - \frac{51}{4} \\ 0 & 1 & 0 & 18 \\ 0 & 0 & 1 & 3 \end{array}]$

Passaggio 2.7

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

Tocca per altri passaggi...

Passaggio 2.7.1

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

$[\begin{array}{c} 1 + \frac{5}{4} \cdot 0 & - \frac{1}{2} + \frac{5}{4} \cdot 0 & - \frac{5}{4} + \frac{5}{4} \cdot 1 & - \frac{51}{4} + \frac{5}{4} \cdot 3 \\ 0 & 1 & 0 & 18 \\ 0 & 0 & 1 & 3 \end{array}]$

Passaggio 2.7.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & 0 & - 9 \\ 0 & 1 & 0 & 18 \\ 0 & 0 & 1 & 3 \end{array}]$

Passaggio 2.8

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

Tocca per altri passaggi...

Passaggio 2.8.1

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

$[\begin{array}{c} 1 + \frac{1}{2} \cdot 0 & - \frac{1}{2} + \frac{1}{2} \cdot 1 & 0 + \frac{1}{2} \cdot 0 & - 9 + \frac{1}{2} \cdot 18 \\ 0 & 1 & 0 & 18 \\ 0 & 0 & 1 & 3 \end{array}]$

Passaggio 2.8.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 18 \\ 0 & 0 & 1 & 3 \end{array}]$

Passaggio 3

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

$x = 0$

$y = 18$

$z = 3$

Passaggio 4

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

$(0, 18, 3)$