Matematica discreta Esempi

$x + 2 y + z = 82$ , $7 y + 3 z = 63$

Passaggio 1

Write the system as a matrix.

$[\begin{array}{c} 1 & 2 & 1 & 82 \\ 0 & 7 & 3 & 63 \end{array}]$

Passaggio 2

Trova la forma ridotta a scala per righe di Echelon.

Tocca per altri passaggi...

Passaggio 2.1

Multiply each element of $R_{2}$ by $\frac{1}{7}$ to make the entry at $2, 2$ a $1$ .

Tocca per altri passaggi...

Passaggio 2.1.1

Multiply each element of $R_{2}$ by $\frac{1}{7}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 2 & 1 & 82 \\ \frac{0}{7} & \frac{7}{7} & \frac{3}{7} & \frac{63}{7} \end{array}]$

Passaggio 2.1.2

Semplifica $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 1 & 82 \\ 0 & 1 & \frac{3}{7} & 9 \end{array}]$

Passaggio 2.2

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

Tocca per altri passaggi...

Passaggio 2.2.1

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

$[\begin{array}{c} 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & 1 - 2 (\frac{3}{7}) & 82 - 2 \cdot 9 \\ 0 & 1 & \frac{3}{7} & 9 \end{array}]$

Passaggio 2.2.2

Semplifica $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{1}{7} & 64 \\ 0 & 1 & \frac{3}{7} & 9 \end{array}]$

Passaggio 3

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

$x + \frac{1}{7} z = 64$

$y + \frac{3}{7} z = 9$

Passaggio 4

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

$(64 - \frac{z}{7}, 9 - \frac{3 z}{7}, z)$