लीनियर एलजेब्रा उदाहरण

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

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

चरण 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.

चरण 2

घटी हुई पंक्ति के सोपानक रूप का पता करें.

और स्टेप्स के लिए टैप करें…

चरण 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

$0$ .

और स्टेप्स के लिए टैप करें…

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

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

चरण 2.1.2

R_{2}

$R_{2}$ को सरल करें.

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

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

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

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

चरण 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

$1$ .

और स्टेप्स के लिए टैप करें…

चरण 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{matrix} 1 & - 1 & - 2 - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot - 2 & - \frac{1}{2} (k - 5) \end{matrix}]

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

चरण 2.2.2

R_{2}

$R_{2}$ को सरल करें.

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

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

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

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

चरण 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

$0$ .

और स्टेप्स के लिए टैप करें…

चरण 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{matrix} 1 + 0 & - 1 + 1 \cdot 1 & - 2 - \frac{k}{2} + \frac{5}{2} 0 & 1 & - \frac{k}{2} + \frac{5}{2} \end{matrix}]

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

चरण 2.3.2

R_{1}

$R_{1}$ को सरल करें.

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

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

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

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

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

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

चरण 3

The pivot positions are the locations with the leading

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

Pivot Positions:

$a_{11}$ and

$a_{22}$

Pivot Columns:

$1$ and

$2$

चरण 4

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

$1$