Algèbre linéaire Exemples

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

Étape 1

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

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

Étape 2

Déterminez la forme d’échelon en ligne réduite.

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Étape 2.1

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

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Étape 2.1.1

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

$[\begin{array}{c} 1 + 2 \cdot 0 & - 2 + 2 \cdot 1 & - 4 + 2 \cdot 2 & - 5 + 2 \cdot 1 & 0 + 2 \cdot 0 & 0 + 2 \cdot 0 \\ 0 & 1 & 2 & 1 & 0 & 0 \end{array}]$

Étape 2.1.2

Simplifiez $R_{1}$ .

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

Étape 3

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

$x_{1} - 3 x_{4} = 0$

$x_{2} + 2 x_{3} + x_{4} = 0$

Étape 4

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

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{array}] = [\begin{array}{c} 3 x_{4} \\ - 2 x_{3} - x_{4} \\ x_{3} \\ x_{4} \\ x_{5} \end{array}]$

Étape 5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{array}] = x_{4} [\begin{array}{c} 3 \\ - 1 \\ 0 \\ 1 \\ 0 \end{array}] + x_{3} [\begin{array}{c} 0 \\ - 2 \\ 1 \\ 0 \\ 0 \end{array}] + x_{5} [\begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{array}]$

Étape 6

Write as a solution set.

${x_{4} [\begin{array}{c} 3 \\ - 1 \\ 0 \\ 1 \\ 0 \end{array}] + x_{3} [\begin{array}{c} 0 \\ - 2 \\ 1 \\ 0 \\ 0 \end{array}] + x_{5} [\begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{array}] | x_{3}, x_{4}, x_{5} \in R}$