लीनियर एलजेब्रा उदाहरण

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

चरण 1

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

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

चरण 2

घटी हुई पंक्ति के सोपानक रूप का पता करें.

और स्टेप्स के लिए टैप करें…

चरण 2.1

Multiply each element of $R_{3}$ by $- 1$ to make the entry at $3, 5$ a $1$ .

और स्टेप्स के लिए टैप करें…

चरण 2.1.1

Multiply each element of $R_{3}$ by $- 1$ to make the entry at $3, 5$ a $1$ .

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

चरण 2.1.2

$R_{3}$ को सरल करें.

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

चरण 3

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

$x_{1} + 4 x_{2} + 3 x_{4} = 0$

$x_{3} - 3 x_{4} = 0$

$x_{5} = 0$

चरण 4

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

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{array}] = [\begin{array}{c} - 4 x_{2} - 3 x_{4} \\ x_{2} \\ 3 x_{4} \\ x_{4} \\ 0 \end{array}]$

चरण 5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{array}] = x_{2} [\begin{array}{c} - 4 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}] + x_{4} [\begin{array}{c} - 3 \\ 0 \\ 3 \\ 1 \\ 0 \end{array}]$

चरण 6

Write as a solution set.

${x_{2} [\begin{array}{c} - 4 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}] + x_{4} [\begin{array}{c} - 3 \\ 0 \\ 3 \\ 1 \\ 0 \end{array}] | x_{2}, x_{4} \in R}$