Matematika Berhingga Contoh

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

Langkah 1

Write the system as a matrix.

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

Langkah 2

Tentukan bentuk eselon baris yang dikurangi.

Ketuk untuk lebih banyak langkah...

Langkah 2.1

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

Ketuk untuk lebih banyak langkah...

Langkah 2.1.1

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

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

Langkah 2.1.2

Sederhanakan $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 9 \\ 0 & - 5 & 2 & - 11 \\ 1 & - 4 & 3 & - 2 \end{array}]$

Langkah 2.2

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

Ketuk untuk lebih banyak langkah...

Langkah 2.2.1

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

$[\begin{array}{c} 1 & 1 & 1 & 9 \\ 0 & - 5 & 2 & - 11 \\ 1 - 1 & - 4 - 1 & 3 - 1 & - 2 - 9 \end{array}]$

Langkah 2.2.2

Sederhanakan $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 9 \\ 0 & - 5 & 2 & - 11 \\ 0 & - 5 & 2 & - 11 \end{array}]$

Langkah 2.3

Multiply each element of $R_{2}$ by $- \frac{1}{5}$ to make the entry at $2, 2$ a $1$ .

Ketuk untuk lebih banyak langkah...

Langkah 2.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{5}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 1 & 1 & 9 \\ - \frac{1}{5} \cdot 0 & - \frac{1}{5} \cdot - 5 & - \frac{1}{5} \cdot 2 & - \frac{1}{5} \cdot - 11 \\ 0 & - 5 & 2 & - 11 \end{array}]$

Langkah 2.3.2

Sederhanakan $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 9 \\ 0 & 1 & - \frac{2}{5} & \frac{11}{5} \\ 0 & - 5 & 2 & - 11 \end{array}]$

Langkah 2.4

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

Ketuk untuk lebih banyak langkah...

Langkah 2.4.1

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

$[\begin{array}{c} 1 & 1 & 1 & 9 \\ 0 & 1 & - \frac{2}{5} & \frac{11}{5} \\ 0 + 5 \cdot 0 & - 5 + 5 \cdot 1 & 2 + 5 (- \frac{2}{5}) & - 11 + 5 (\frac{11}{5}) \end{array}]$

Langkah 2.4.2

Sederhanakan $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 9 \\ 0 & 1 & - \frac{2}{5} & \frac{11}{5} \\ 0 & 0 & 0 & 0 \end{array}]$

Langkah 2.5

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

Ketuk untuk lebih banyak langkah...

Langkah 2.5.1

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

$[\begin{array}{c} 1 - 0 & 1 - 1 & 1 + \frac{2}{5} & 9 - \frac{11}{5} \\ 0 & 1 & - \frac{2}{5} & \frac{11}{5} \\ 0 & 0 & 0 & 0 \end{array}]$

Langkah 2.5.2

Sederhanakan $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{7}{5} & \frac{34}{5} \\ 0 & 1 & - \frac{2}{5} & \frac{11}{5} \\ 0 & 0 & 0 & 0 \end{array}]$

Langkah 3

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

$x + \frac{7}{5} z = \frac{34}{5}$

$y - \frac{2}{5} z = \frac{11}{5}$

$0 = 0$

Langkah 4

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

$(\frac{34}{5} - \frac{7 z}{5}, \frac{11}{5} + \frac{2 z}{5}, z)$