Algèbre linéaire Exemples

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

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

Étape 1

Write as an augmented matrix for

A x = 0

$A x = 0$ .

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

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

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{- 8}{5} & \frac{0}{5} \\ - 10 & 16 & 0 \end{array}]

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

Étape 2.1.2

Simplifiez

R_{1}

$R_{1}$ .

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

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

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

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

Étape 2.2

Perform the row operation

R_{2} = R_{2} + 10 R_{1}

$R_{2} = R_{2} + 10 R_{1}$ 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} + 10 R_{1}

$R_{2} = R_{2} + 10 R_{1}$ to make the entry at

2, 1

$2, 1$ a

0

$0$ .

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

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

Étape 2.2.2

Simplifiez

R_{2}

$R_{2}$ .

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

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

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

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

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

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

Étape 3

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

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

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

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

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

Étape 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}]

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

Étape 6

Write as a solution set.

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

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