Matematica discreta Esempi

$2 x - 5 y + z = - 14$ , $x + 4 y - 6 z = - 7$ , $3 x - 4 y - 2 z = - 21$

Passaggio 1

Write the system as a matrix.

$[\begin{array}{c} 2 & - 5 & 1 & - 14 \\ 1 & 4 & - 6 & - 7 \\ 3 & - 4 & - 2 & - 21 \end{array}]$

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 $1$ .

$[\begin{array}{c} \frac{2}{2} & \frac{- 5}{2} & \frac{1}{2} & \frac{- 14}{2} \\ 1 & 4 & - 6 & - 7 \\ 3 & - 4 & - 2 & - 21 \end{array}]$

Passaggio 2.1.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{5}{2} & \frac{1}{2} & - 7 \\ 1 & 4 & - 6 & - 7 \\ 3 & - 4 & - 2 & - 21 \end{array}]$

Passaggio 2.2

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

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Passaggio 2.2.1

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

$[\begin{array}{c} 1 & - \frac{5}{2} & \frac{1}{2} & - 7 \\ 1 - 1 & 4 + \frac{5}{2} & - 6 - \frac{1}{2} & - 7 + 7 \\ 3 & - 4 & - 2 & - 21 \end{array}]$

Passaggio 2.2.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{5}{2} & \frac{1}{2} & - 7 \\ 0 & \frac{13}{2} & - \frac{13}{2} & 0 \\ 3 & - 4 & - 2 & - 21 \end{array}]$

Passaggio 2.3

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

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Passaggio 2.3.1

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

$[\begin{array}{c} 1 & - \frac{5}{2} & \frac{1}{2} & - 7 \\ 0 & \frac{13}{2} & - \frac{13}{2} & 0 \\ 3 - 3 \cdot 1 & - 4 - 3 (- \frac{5}{2}) & - 2 - 3 (\frac{1}{2}) & - 21 - 3 \cdot - 7 \end{array}]$

Passaggio 2.3.2

Semplifica $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{5}{2} & \frac{1}{2} & - 7 \\ 0 & \frac{13}{2} & - \frac{13}{2} & 0 \\ 0 & \frac{7}{2} & - \frac{7}{2} & 0 \end{array}]$

Passaggio 2.4

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

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Passaggio 2.4.1

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

$[\begin{array}{c} 1 & - \frac{5}{2} & \frac{1}{2} & - 7 \\ \frac{2}{13} \cdot 0 & \frac{2}{13} \cdot \frac{13}{2} & \frac{2}{13} (- \frac{13}{2}) & \frac{2}{13} \cdot 0 \\ 0 & \frac{7}{2} & - \frac{7}{2} & 0 \end{array}]$

Passaggio 2.4.2

Semplifica $R_{2}$ .

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

Passaggio 2.5

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

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Passaggio 2.5.1

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

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

Passaggio 2.5.2

Semplifica $R_{3}$ .

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

Passaggio 2.6

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

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Passaggio 2.6.1

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

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

Passaggio 2.6.2

Semplifica $R_{1}$ .

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

Passaggio 3

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

$x - 2 z = - 7$

$y - z = 0$

$0 = 0$

Passaggio 4

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

$(- 7 + 2 z, z, z)$