Algèbre linéaire Exemples

$[\begin{array}{c} 0.4 & 0.1 & 0.5 \end{array}]$

Étape 1

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

$[\begin{array}{c} 0.4 & 0.1 & 0.5 & 0 \end{array}]$

Étape 2

Déterminez la forme d’échelon en ligne réduite.

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Étape 2.1

Multiply each element of $R_{1}$ by $\frac{1}{0.4}$ to make the entry at $1, 1$ a $1$ .

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Étape 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{0.4}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{0.4}{0.4} & \frac{0.1}{0.4} & \frac{0.5}{0.4} & \frac{0}{0.4} \end{array}]$

Étape 2.1.2

Simplifiez $R_{1}$ .

$[\begin{array}{c} 1 & 0.25 & 1.25 & 0 \end{array}]$

Étape 3

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

$x + 0.25 y + 1.25 z = 0$

Étape 4

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

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - 0.25 y - 1.25 z \\ y \\ z \end{array}]$

Étape 5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \\ z \end{array}] = y [\begin{array}{c} - 0.25 \\ 1 \\ 0 \end{array}] + z [\begin{array}{c} - 1.25 \\ 0 \\ 1 \end{array}]$

Étape 6

Write as a solution set.

${y [\begin{array}{c} - 0.25 \\ 1 \\ 0 \end{array}] + z [\begin{array}{c} - 1.25 \\ 0 \\ 1 \end{array}] | y, z \in R}$