Algèbre linéaire Exemples

[\begin{array}{c} 1 & - 1 & - 2 \\ - 1 & - 1 & - 3 + k \end{array}]

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

Étape 1

Nullity is the dimension of the null space, which is the same as the number of free variables in the system after row reducing. The free variables are the columns without pivot positions.

Étape 2

Déterminez la forme d’échelon en ligne réduite.

Appuyez ici pour voir plus d’étapes...

Étape 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

$0$ .

Appuyez ici pour voir plus d’étapes...

Étape 2.1.1

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 & - 1 & - 2 \\ - 1 + 1 \cdot 1 & - 1 - 1 & - 3 + k - 2 \end{array}]

$[\begin{array}{c} 1 & - 1 & - 2 \\ - 1 + 1 \cdot 1 & - 1 - 1 & - 3 + k - 2 \end{array}]$

Étape 2.1.2

Simplifiez

R_{2}

$R_{2}$ .

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

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

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

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

Étape 2.2

Multiply each element of

R_{2}

$R_{2}$ by

- \frac{1}{2}

$- \frac{1}{2}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

Appuyez ici pour voir plus d’étapes...

Étape 2.2.1

Multiply each element of

R_{2}

$R_{2}$ by

- \frac{1}{2}

$- \frac{1}{2}$ to make the entry at

2, 2

$2, 2$ a

1

$1$ .

[\begin{array}{c} 1 & - 1 & - 2 \\ - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot - 2 & - \frac{1}{2} (k - 5) \end{array}]

$[\begin{array}{c} 1 & - 1 & - 2 \\ - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot - 2 & - \frac{1}{2} (k - 5) \end{array}]$

Étape 2.2.2

Simplifiez

R_{2}

$R_{2}$ .

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

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

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

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

Étape 2.3

Perform the row operation

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

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

1, 2

$1, 2$ a

0

$0$ .

Appuyez ici pour voir plus d’étapes...

Étape 2.3.1

Perform the row operation

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

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

1, 2

$1, 2$ a

0

$0$ .

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

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

Étape 2.3.2

Simplifiez

R_{1}

$R_{1}$ .

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

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

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

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

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

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

Étape 3

The pivot positions are the locations with the leading

1

$1$ in each row. The pivot columns are the columns that have a pivot position.

Pivot Positions:

a_{11}

$a_{11}$ and

a_{22}

$a_{22}$

Pivot Columns:

1

$1$ and

2

$2$

Étape 4

The nullity is the number of columns without a pivot position in the row reduced matrix.

1

$1$