लीनियर एलजेब्रा उदाहरण

$[\begin{array}{c} 1 & 0 & - 5 & 0 & - 14 \\ 0 & 1 & 2 & 0 & - 7 \\ 0 & 0 & 0 & 1 & 7 \\ 0 & 0 & 0 & 0 & 8 \end{array}]$

चरण 1

घटी हुई पंक्ति के सोपानक रूप का पता करें.

और स्टेप्स के लिए टैप करें…

चरण 1.1

Multiply each element of $R_{4}$ by $\frac{1}{8}$ to make the entry at $4, 5$ a $1$ .

और स्टेप्स के लिए टैप करें…

चरण 1.1.1

Multiply each element of $R_{4}$ by $\frac{1}{8}$ to make the entry at $4, 5$ a $1$ .

$[\begin{array}{c} 1 & 0 & - 5 & 0 & - 14 \\ 0 & 1 & 2 & 0 & - 7 \\ 0 & 0 & 0 & 1 & 7 \\ \frac{0}{8} & \frac{0}{8} & \frac{0}{8} & \frac{0}{8} & \frac{8}{8} \end{array}]$

चरण 1.1.2

$R_{4}$ को सरल करें.

$[\begin{array}{c} 1 & 0 & - 5 & 0 & - 14 \\ 0 & 1 & 2 & 0 & - 7 \\ 0 & 0 & 0 & 1 & 7 \\ 0 & 0 & 0 & 0 & 1 \end{array}]$

चरण 1.2

Perform the row operation $R_{3} = R_{3} - 7 R_{4}$ to make the entry at $3, 5$ a $0$ .

और स्टेप्स के लिए टैप करें…

चरण 1.2.1

Perform the row operation $R_{3} = R_{3} - 7 R_{4}$ to make the entry at $3, 5$ a $0$ .

$[\begin{array}{c} 1 & 0 & - 5 & 0 & - 14 \\ 0 & 1 & 2 & 0 & - 7 \\ 0 - 7 \cdot 0 & 0 - 7 \cdot 0 & 0 - 7 \cdot 0 & 1 - 7 \cdot 0 & 7 - 7 \cdot 1 \\ 0 & 0 & 0 & 0 & 1 \end{array}]$

चरण 1.2.2

$R_{3}$ को सरल करें.

$[\begin{array}{c} 1 & 0 & - 5 & 0 & - 14 \\ 0 & 1 & 2 & 0 & - 7 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}]$

चरण 1.3

Perform the row operation $R_{2} = R_{2} + 7 R_{4}$ to make the entry at $2, 5$ a $0$ .

और स्टेप्स के लिए टैप करें…

चरण 1.3.1

Perform the row operation $R_{2} = R_{2} + 7 R_{4}$ to make the entry at $2, 5$ a $0$ .

$[\begin{array}{c} 1 & 0 & - 5 & 0 & - 14 \\ 0 + 7 \cdot 0 & 1 + 7 \cdot 0 & 2 + 7 \cdot 0 & 0 + 7 \cdot 0 & - 7 + 7 \cdot 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}]$

चरण 1.3.2

$R_{2}$ को सरल करें.

$[\begin{array}{c} 1 & 0 & - 5 & 0 & - 14 \\ 0 & 1 & 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}]$

चरण 1.4

Perform the row operation $R_{1} = R_{1} + 14 R_{4}$ to make the entry at $1, 5$ a $0$ .

और स्टेप्स के लिए टैप करें…

चरण 1.4.1

Perform the row operation $R_{1} = R_{1} + 14 R_{4}$ to make the entry at $1, 5$ a $0$ .

$[\begin{array}{c} 1 + 14 \cdot 0 & 0 + 14 \cdot 0 & - 5 + 14 \cdot 0 & 0 + 14 \cdot 0 & - 14 + 14 \cdot 1 \\ 0 & 1 & 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}]$

चरण 1.4.2

$R_{1}$ को सरल करें.

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

चरण 2

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: $a_{11}, a_{22}, a_{34},$ and $a_{45}$

Pivot Columns: $1, 2, 4,$ and $5$