Álgebra linear Exemplos

$[\begin{array}{c} 5 & - 8 \\ - 10 & 16 \end{array}]$

Etapa 1

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

$[\begin{array}{c} 5 & - 8 & 0 \\ - 10 & 16 & 0 \end{array}]$

Etapa 2

Encontre a forma escalonada reduzida por linhas.

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Etapa 2.1

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

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Etapa 2.1.1

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

$[\begin{array}{c} \frac{5}{5} & \frac{- 8}{5} & \frac{0}{5} \\ - 10 & 16 & 0 \end{array}]$

Etapa 2.1.2

Simplifique $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{8}{5} & 0 \\ - 10 & 16 & 0 \end{array}]$

Etapa 2.2

Perform the row operation $R_{2} = R_{2} + 10 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Etapa 2.2.1

Perform the row operation $R_{2} = R_{2} + 10 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{8}{5} & 0 \\ - 10 + 10 \cdot 1 & 16 + 10 (- \frac{8}{5}) & 0 + 10 \cdot 0 \end{array}]$

Etapa 2.2.2

Simplifique $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{8}{5} & 0 \\ 0 & 0 & 0 \end{array}]$

Etapa 3

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

$x - \frac{8}{5} y = 0$

$0 = 0$

Etapa 4

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

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{8 y}{5} \\ y \end{array}]$

Etapa 5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \end{array}] = y [\begin{array}{c} \frac{8}{5} \\ 1 \end{array}]$

Etapa 6

Write as a solution set.

${y [\begin{array}{c} \frac{8}{5} \\ 1 \end{array}] | y \in R}$