लीनियर एलजेब्रा उदाहरण

$[\begin{array}{c} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 1 \end{array}]$

चरण 1

Write as an augmented matrix for $A x = 0$ .

$[\begin{array}{c} 1 & 2 & 3 & 4 & 0 \\ 0 & 1 & 2 & 3 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{array}]$

चरण 2

घटी हुई पंक्ति के सोपानक रूप का पता करें.

और स्टेप्स के लिए टैप करें…

चरण 2.1

Perform the row operation $R_{2} = R_{2} - 3 R_{3}$ to make the entry at $2, 4$ a $0$ .

और स्टेप्स के लिए टैप करें…

चरण 2.1.1

Perform the row operation $R_{2} = R_{2} - 3 R_{3}$ to make the entry at $2, 4$ a $0$ .

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

चरण 2.1.2

$R_{2}$ को सरल करें.

$[\begin{array}{c} 1 & 2 & 3 & 4 & 0 \\ 0 & 1 & 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{array}]$

चरण 2.2

Perform the row operation $R_{1} = R_{1} - 4 R_{3}$ to make the entry at $1, 4$ a $0$ .

और स्टेप्स के लिए टैप करें…

चरण 2.2.1

Perform the row operation $R_{1} = R_{1} - 4 R_{3}$ to make the entry at $1, 4$ a $0$ .

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

चरण 2.2.2

$R_{1}$ को सरल करें.

$[\begin{array}{c} 1 & 2 & 3 & 0 & 0 \\ 0 & 1 & 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{array}]$

चरण 2.3

Perform the row operation $R_{1} = R_{1} - 2 R_{2}$ to make the entry at $1, 2$ a $0$ .

और स्टेप्स के लिए टैप करें…

चरण 2.3.1

Perform the row operation $R_{1} = R_{1} - 2 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & 3 - 2 \cdot 2 & 0 - 2 \cdot 0 & 0 - 2 \cdot 0 \\ 0 & 1 & 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{array}]$

चरण 2.3.2

$R_{1}$ को सरल करें.

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

चरण 3

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

$x_{1} - x_{3} = 0$

$x_{2} + 2 x_{3} = 0$

$x_{4} = 0$

चरण 4

Write a solution vector by solving in terms of the free variables in each row.

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

चरण 5

Write the solution as a linear combination of vectors.

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

चरण 6

Write as a solution set.

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