Algèbre linéaire Exemples

[\begin{matrix} 0 & 10 & 20 & 30 25 & 45 & 65 & 85 \end{matrix}]

$[\begin{array}{c} 0 & 10 & 20 & 30 \\ 25 & 45 & 65 & 85 \end{array}]$

Étape 1

Write as an augmented matrix for

A x = 0

$A x = 0$ .

[\begin{matrix} 0 & 10 & 20 & 30 & 0 25 & 45 & 65 & 85 & 0 \end{matrix}]

$[\begin{array}{c} 0 & 10 & 20 & 30 & 0 \\ 25 & 45 & 65 & 85 & 0 \end{array}]$

Étape 2

Déterminez la forme d’échelon en ligne réduite.

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Étape 2.1

Swap

R_{2}

$R_{2}$ with

R_{1}

$R_{1}$ to put a nonzero entry at

1, 1

$1, 1$ .

[\begin{matrix} 25 & 45 & 65 & 85 & 0 0 & 10 & 20 & 30 & 0 \end{matrix}]

$[\begin{array}{c} 25 & 45 & 65 & 85 & 0 \\ 0 & 10 & 20 & 30 & 0 \end{array}]$

Étape 2.2

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{25}

$\frac{1}{25}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

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Étape 2.2.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{25}

$\frac{1}{25}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

[\begin{matrix} \frac{25}{25} & \frac{45}{25} & \frac{65}{25} & \frac{85}{25} & \frac{0}{25} 0 & 10 & 20 & 30 & 0 \end{matrix}]

$[\begin{array}{c} \frac{25}{25} & \frac{45}{25} & \frac{65}{25} & \frac{85}{25} & \frac{0}{25} \\ 0 & 10 & 20 & 30 & 0 \end{array}]$

Étape 2.2.2

Simplifiez

R_{1}

$R_{1}$ .

[\begin{matrix} 1 & \frac{9}{5} & \frac{13}{5} & \frac{17}{5} & 0 0 & 10 & 20 & 30 & 0 \end{matrix}]

$[\begin{array}{c} 1 & \frac{9}{5} & \frac{13}{5} & \frac{17}{5} & 0 \\ 0 & 10 & 20 & 30 & 0 \end{array}]$

Étape 2.3

Multiply each element of

$R_{2}$ by

$\frac{1}{10}$ to make the entry at

$2, 2$ a

$1$ .

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Étape 2.3.1

Multiply each element of

$R_{2}$ by

$\frac{1}{10}$ to make the entry at

$2, 2$ a

$1$ .

$[\begin{array}{c} 1 & \frac{9}{5} & \frac{13}{5} & \frac{17}{5} & 0 \\ \frac{0}{10} & \frac{10}{10} & \frac{20}{10} & \frac{30}{10} & \frac{0}{10} \end{array}]$

Étape 2.3.2

Simplifiez

$R_{2}$ .

$[\begin{array}{c} 1 & \frac{9}{5} & \frac{13}{5} & \frac{17}{5} & 0 \\ 0 & 1 & 2 & 3 & 0 \end{array}]$

Étape 2.4

Perform the row operation

$R_{1} = R_{1} - \frac{9}{5} R_{2}$ to make the entry at

$1, 2$ a

$0$ .

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Étape 2.4.1

Perform the row operation

$R_{1} = R_{1} - \frac{9}{5} R_{2}$ to make the entry at

$1, 2$ a

$0$ .

$[\begin{array}{c} 1 - \frac{9}{5} \cdot 0 & \frac{9}{5} - \frac{9}{5} \cdot 1 & \frac{13}{5} - \frac{9}{5} \cdot 2 & \frac{17}{5} - \frac{9}{5} \cdot 3 & 0 - \frac{9}{5} \cdot 0 \\ 0 & 1 & 2 & 3 & 0 \end{array}]$

Étape 2.4.2

Simplifiez

$R_{1}$ .

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

Étape 3

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

$x_{1} - x_{3} - 2 x_{4} = 0$

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

Étape 4

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

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

Étape 5

Write the solution as a linear combination of vectors.

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

Étape 6

Write as a solution set.

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