Matematica discreta Esempi

$3 x + 7 y + 9 z = 0$ , $x + 2 y + 3 z = 2$ , $x + 4 y + z = 2$

Passaggio 1

Write the system as a matrix.

$[\begin{array}{c} 3 & 7 & 9 & 0 \\ 1 & 2 & 3 & 2 \\ 1 & 4 & 1 & 2 \end{array}]$

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 $1$ .

$[\begin{array}{c} \frac{3}{3} & \frac{7}{3} & \frac{9}{3} & \frac{0}{3} \\ 1 & 2 & 3 & 2 \\ 1 & 4 & 1 & 2 \end{array}]$

Passaggio 2.1.2

Semplifica $R_{1}$ .

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

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 $0$ .

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

Passaggio 2.2.2

Semplifica $R_{2}$ .

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

Passaggio 2.3

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

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Passaggio 2.3.1

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

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

Passaggio 2.3.2

Semplifica $R_{3}$ .

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

Passaggio 2.4

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

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Passaggio 2.4.1

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

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

Passaggio 2.4.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & \frac{7}{3} & 3 & 0 \\ 0 & 1 & 0 & - 6 \\ 0 & \frac{5}{3} & - 2 & 2 \end{array}]$

Passaggio 2.5

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

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Passaggio 2.5.1

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

$[\begin{array}{c} 1 & \frac{7}{3} & 3 & 0 \\ 0 & 1 & 0 & - 6 \\ 0 - \frac{5}{3} \cdot 0 & \frac{5}{3} - \frac{5}{3} \cdot 1 & - 2 - \frac{5}{3} \cdot 0 & 2 - \frac{5}{3} \cdot - 6 \end{array}]$

Passaggio 2.5.2

Semplifica $R_{3}$ .

$[\begin{array}{c} 1 & \frac{7}{3} & 3 & 0 \\ 0 & 1 & 0 & - 6 \\ 0 & 0 & - 2 & 12 \end{array}]$

Passaggio 2.6

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

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Passaggio 2.6.1

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

$[\begin{array}{c} 1 & \frac{7}{3} & 3 & 0 \\ 0 & 1 & 0 & - 6 \\ - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot 12 \end{array}]$

Passaggio 2.6.2

Semplifica $R_{3}$ .

$[\begin{array}{c} 1 & \frac{7}{3} & 3 & 0 \\ 0 & 1 & 0 & - 6 \\ 0 & 0 & 1 & - 6 \end{array}]$

Passaggio 2.7

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

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Passaggio 2.7.1

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

$[\begin{array}{c} 1 - 3 \cdot 0 & \frac{7}{3} - 3 \cdot 0 & 3 - 3 \cdot 1 & 0 - 3 \cdot - 6 \\ 0 & 1 & 0 & - 6 \\ 0 & 0 & 1 & - 6 \end{array}]$

Passaggio 2.7.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & \frac{7}{3} & 0 & 18 \\ 0 & 1 & 0 & - 6 \\ 0 & 0 & 1 & - 6 \end{array}]$

Passaggio 2.8

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

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Passaggio 2.8.1

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

$[\begin{array}{c} 1 - \frac{7}{3} \cdot 0 & \frac{7}{3} - \frac{7}{3} \cdot 1 & 0 - \frac{7}{3} \cdot 0 & 18 - \frac{7}{3} \cdot - 6 \\ 0 & 1 & 0 & - 6 \\ 0 & 0 & 1 & - 6 \end{array}]$

Passaggio 2.8.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 32 \\ 0 & 1 & 0 & - 6 \\ 0 & 0 & 1 & - 6 \end{array}]$

Passaggio 3

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

$x = 32$

$y = - 6$

$z = - 6$

Passaggio 4

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

$(32, - 6, - 6)$