Algèbre linéaire Exemples

[\begin{array}{c} 5 & - 10 & 25 & - 20 \\ 1 & - 2 & 5 & - 4 \\ 3 & - 6 & 15 & 12 \end{array}]

$[\begin{array}{c} 5 & - 10 & 25 & - 20 \\ 1 & - 2 & 5 & - 4 \\ 3 & - 6 & 15 & 12 \end{array}]$

Étape 1

Write as an augmented matrix for

A x = 0

$A x = 0$ .

[\begin{array}{c} 5 & - 10 & 25 & - 20 & 0 \\ 1 & - 2 & 5 & - 4 & 0 \\ 3 & - 6 & 15 & 12 & 0 \end{array}]

$[\begin{array}{c} 5 & - 10 & 25 & - 20 & 0 \\ 1 & - 2 & 5 & - 4 & 0 \\ 3 & - 6 & 15 & 12 & 0 \end{array}]$

Étape 2

Déterminez la forme d’échelon en ligne réduite.

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Étape 2.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{5}

$\frac{1}{5}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

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Étape 2.1.1

Multiply each element of

R_{1}

$R_{1}$ by

\frac{1}{5}

$\frac{1}{5}$ to make the entry at

1, 1

$1, 1$ a

1

$1$ .

[\begin{array}{c} \frac{5}{5} & \frac{- 10}{5} & \frac{25}{5} & \frac{- 20}{5} & \frac{0}{5} \\ 1 & - 2 & 5 & - 4 & 0 \\ 3 & - 6 & 15 & 12 & 0 \end{array}]

$[\begin{array}{c} \frac{5}{5} & \frac{- 10}{5} & \frac{25}{5} & \frac{- 20}{5} & \frac{0}{5} \\ 1 & - 2 & 5 & - 4 & 0 \\ 3 & - 6 & 15 & 12 & 0 \end{array}]$

Étape 2.1.2

Simplifiez

R_{1}

$R_{1}$ .

[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 1 & - 2 & 5 & - 4 & 0 \\ 3 & - 6 & 15 & 12 & 0 \end{array}]

$[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 1 & - 2 & 5 & - 4 & 0 \\ 3 & - 6 & 15 & 12 & 0 \end{array}]$

[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 1 & - 2 & 5 & - 4 & 0 \\ 3 & - 6 & 15 & 12 & 0 \end{array}]

$[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 1 & - 2 & 5 & - 4 & 0 \\ 3 & - 6 & 15 & 12 & 0 \end{array}]$

Étape 2.2

Perform the row operation

R_{2} = R_{2} - R_{1}

$R_{2} = R_{2} - R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

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Étape 2.2.1

Perform the row operation

R_{2} = R_{2} - R_{1}

$R_{2} = R_{2} - R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 1 - 1 & - 2 + 2 & 5 - 5 & - 4 + 4 & 0 - 0 \\ 3 & - 6 & 15 & 12 & 0 \end{array}]

$[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 1 - 1 & - 2 + 2 & 5 - 5 & - 4 + 4 & 0 - 0 \\ 3 & - 6 & 15 & 12 & 0 \end{array}]$

Étape 2.2.2

Simplifiez

R_{2}

$R_{2}$ .

[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 3 & - 6 & 15 & 12 & 0 \end{array}]

$[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 3 & - 6 & 15 & 12 & 0 \end{array}]$

[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 3 & - 6 & 15 & 12 & 0 \end{array}]

$[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 3 & - 6 & 15 & 12 & 0 \end{array}]$

Étape 2.3

Perform the row operation

R_{3} = R_{3} - 3 R_{1}

$R_{3} = R_{3} - 3 R_{1}$ to make the entry at

3, 1

$3, 1$ a

0

$0$ .

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Étape 2.3.1

Perform the row operation

R_{3} = R_{3} - 3 R_{1}

$R_{3} = R_{3} - 3 R_{1}$ to make the entry at

3, 1

$3, 1$ a

0

$0$ .

[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 3 - 3 \cdot 1 & - 6 - 3 \cdot - 2 & 15 - 3 \cdot 5 & 12 - 3 \cdot - 4 & 0 - 3 \cdot 0 \end{array}]

$[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 3 - 3 \cdot 1 & - 6 - 3 \cdot - 2 & 15 - 3 \cdot 5 & 12 - 3 \cdot - 4 & 0 - 3 \cdot 0 \end{array}]$

Étape 2.3.2

Simplifiez

R_{3}

$R_{3}$ .

[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 24 & 0 \end{array}]

$[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 24 & 0 \end{array}]$

[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 24 & 0 \end{array}]

$[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 24 & 0 \end{array}]$

Étape 2.4

Swap

R_{3}

$R_{3}$ with

R_{2}

$R_{2}$ to put a nonzero entry at

2, 4

$2, 4$ .

[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 0 & 0 & 0 & 24 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}]

$[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ 0 & 0 & 0 & 24 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}]$

Étape 2.5

Multiply each element of

R_{2}

$R_{2}$ by

\frac{1}{24}

$\frac{1}{24}$ to make the entry at

2, 4

$2, 4$ a

1

$1$ .

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Étape 2.5.1

Multiply each element of

R_{2}

$R_{2}$ by

\frac{1}{24}

$\frac{1}{24}$ to make the entry at

2, 4

$2, 4$ a

1

$1$ .

[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ \frac{0}{24} & \frac{0}{24} & \frac{0}{24} & \frac{24}{24} & \frac{0}{24} \\ 0 & 0 & 0 & 0 & 0 \end{array}]

$[\begin{array}{c} 1 & - 2 & 5 & - 4 & 0 \\ \frac{0}{24} & \frac{0}{24} & \frac{0}{24} & \frac{24}{24} & \frac{0}{24} \\ 0 & 0 & 0 & 0 & 0 \end{array}]$

Étape 2.5.2

Simplifiez

R_{2}

$R_{2}$ .

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

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

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

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

Étape 2.6

Perform the row operation

R_{1} = R_{1} + 4 R_{2}

$R_{1} = R_{1} + 4 R_{2}$ to make the entry at

1, 4

$1, 4$ a

0

$0$ .

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Étape 2.6.1

Perform the row operation

R_{1} = R_{1} + 4 R_{2}

$R_{1} = R_{1} + 4 R_{2}$ to make the entry at

1, 4

$1, 4$ a

0

$0$ .

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

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

Étape 2.6.2

Simplifiez

R_{1}

$R_{1}$ .

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

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

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

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

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

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

Étape 3

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

x_{1} - 2 x_{2} + 5 x_{3} = 0

$x_{1} - 2 x_{2} + 5 x_{3} = 0$

x_{4} = 0

$x_{4} = 0$

0 = 0

$0 = 0$

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} 2 x_{2} - 5 x_{3} \\ x_{2} \\ x_{3} \\ 0 \end{array}]

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

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

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

Étape 6

Write as a solution set.

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

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