Matematika Berhingga Contoh

$y - 3 x = 10$ , $2 y + 2 x = - 4$

Langkah 1

Move variables to the left and constant terms to the right.

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Langkah 1.1

Susun kembali $y$ dan $- 3 x$ .

$- 3 x + y = 10$

$2 y + 2 x = - 4$

Langkah 1.2

Susun kembali $2 y$ dan $2 x$ .

$- 3 x + y = 10$

$2 x + 2 y = - 4$

$- 3 x + y = 10$

$2 x + 2 y = - 4$

Langkah 2

Write the system as a matrix.

$[\begin{array}{c} - 3 & 1 & 10 \\ 2 & 2 & - 4 \end{array}]$

Langkah 3

Tentukan bentuk eselon baris yang dikurangi.

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Langkah 3.1

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

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Langkah 3.1.1

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

$[\begin{array}{c} - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot 1 & - \frac{1}{3} \cdot 10 \\ 2 & 2 & - 4 \end{array}]$

Langkah 3.1.2

Sederhanakan $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{10}{3} \\ 2 & 2 & - 4 \end{array}]$

Langkah 3.2

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

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Langkah 3.2.1

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

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{10}{3} \\ 2 - 2 \cdot 1 & 2 - 2 (- \frac{1}{3}) & - 4 - 2 (- \frac{10}{3}) \end{array}]$

Langkah 3.2.2

Sederhanakan $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{10}{3} \\ 0 & \frac{8}{3} & \frac{8}{3} \end{array}]$

Langkah 3.3

Multiply each element of $R_{2}$ by $\frac{3}{8}$ to make the entry at $2, 2$ a $1$ .

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Langkah 3.3.1

Multiply each element of $R_{2}$ by $\frac{3}{8}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{10}{3} \\ \frac{3}{8} \cdot 0 & \frac{3}{8} \cdot \frac{8}{3} & \frac{3}{8} \cdot \frac{8}{3} \end{array}]$

Langkah 3.3.2

Sederhanakan $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{3} & - \frac{10}{3} \\ 0 & 1 & 1 \end{array}]$

Langkah 3.4

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

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Langkah 3.4.1

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

$[\begin{array}{c} 1 + \frac{1}{3} \cdot 0 & - \frac{1}{3} + \frac{1}{3} \cdot 1 & - \frac{10}{3} + \frac{1}{3} \cdot 1 \\ 0 & 1 & 1 \end{array}]$

Langkah 3.4.2

Sederhanakan $R_{1}$ .

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

Langkah 4

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

$x = - 3$

$y = 1$

Langkah 5

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

$(- 3, 1)$