Aljabar Linear Contoh

$[\begin{array}{c} 20 & 16 \\ 5 & 4 \\ - 15 & - 12 \\ - 20 & - 16 \end{array}]$

Langkah 1

Write as an augmented matrix for $A x = 0$ .

$[\begin{array}{c} 20 & 16 & 0 \\ 5 & 4 & 0 \\ - 15 & - 12 & 0 \\ - 20 & - 16 & 0 \end{array}]$

Langkah 2

Tentukan bentuk eselon baris yang dikurangi.

Ketuk untuk lebih banyak langkah...

Langkah 2.1

Multiply each element of $R_{1}$ by $\frac{1}{20}$ to make the entry at $1, 1$ a $1$ .

Ketuk untuk lebih banyak langkah...

Langkah 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{20}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{20}{20} & \frac{16}{20} & \frac{0}{20} \\ 5 & 4 & 0 \\ - 15 & - 12 & 0 \\ - 20 & - 16 & 0 \end{array}]$

Langkah 2.1.2

Sederhanakan $R_{1}$ .

$[\begin{array}{c} 1 & \frac{4}{5} & 0 \\ 5 & 4 & 0 \\ - 15 & - 12 & 0 \\ - 20 & - 16 & 0 \end{array}]$

Langkah 2.2

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

Ketuk untuk lebih banyak langkah...

Langkah 2.2.1

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

$[\begin{array}{c} 1 & \frac{4}{5} & 0 \\ 5 - 5 \cdot 1 & 4 - 5 (\frac{4}{5}) & 0 - 5 \cdot 0 \\ - 15 & - 12 & 0 \\ - 20 & - 16 & 0 \end{array}]$

Langkah 2.2.2

Sederhanakan $R_{2}$ .

$[\begin{array}{c} 1 & \frac{4}{5} & 0 \\ 0 & 0 & 0 \\ - 15 & - 12 & 0 \\ - 20 & - 16 & 0 \end{array}]$

Langkah 2.3

Perform the row operation $R_{3} = R_{3} + 15 R_{1}$ to make the entry at $3, 1$ a $0$ .

Ketuk untuk lebih banyak langkah...

Langkah 2.3.1

Perform the row operation $R_{3} = R_{3} + 15 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{4}{5} & 0 \\ 0 & 0 & 0 \\ - 15 + 15 \cdot 1 & - 12 + 15 (\frac{4}{5}) & 0 + 15 \cdot 0 \\ - 20 & - 16 & 0 \end{array}]$

Langkah 2.3.2

Sederhanakan $R_{3}$ .

$[\begin{array}{c} 1 & \frac{4}{5} & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ - 20 & - 16 & 0 \end{array}]$

Langkah 2.4

Perform the row operation $R_{4} = R_{4} + 20 R_{1}$ to make the entry at $4, 1$ a $0$ .

Ketuk untuk lebih banyak langkah...

Langkah 2.4.1

Perform the row operation $R_{4} = R_{4} + 20 R_{1}$ to make the entry at $4, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{4}{5} & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ - 20 + 20 \cdot 1 & - 16 + 20 (\frac{4}{5}) & 0 + 20 \cdot 0 \end{array}]$

Langkah 2.4.2

Sederhanakan $R_{4}$ .

$[\begin{array}{c} 1 & \frac{4}{5} & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]$

Langkah 3

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

$x + \frac{4}{5} y = 0$

$0 = 0$

Langkah 4

Write a solution vector by solving in terms of the free variables in each row.

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - \frac{4 y}{5} \\ y \end{array}]$

Langkah 5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \end{array}] = y [\begin{array}{c} - \frac{4}{5} \\ 1 \end{array}]$

Langkah 6

Write as a solution set.

${y [\begin{array}{c} - \frac{4}{5} \\ 1 \end{array}] | y \in R}$