लीनियर एलजेब्रा उदाहरण

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

चरण 1

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

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

चरण 2

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

$x_{1} + 2 x_{2} + 2 x_{5} = 0$

$x_{3} - x_{5} = 0$

$x_{4} + 2 x_{5} = 0$

चरण 3

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

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

चरण 4

Write the solution as a linear combination of vectors.

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

चरण 5

Write as a solution set.

${x_{2} [\begin{array}{c} - 2 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}] + x_{5} [\begin{array}{c} - 2 \\ 0 \\ 1 \\ - 2 \\ 1 \end{array}] | x_{2}, x_{5} \in R}$