Question

Consider the set of equations

5x₁ − 4x₂ + 3x₃ + x₄ = 3

2x₁ + x₂ + 5x₃ − 3x₄ = 0

x₁ + 6x₂ − 4x₃ + 2x₄ = 15

where x₁ = 1, x₂ = 2, x₃ = 1, x₄ = 3 is a feasible solution. Is this solution feasible solution

If not, reduce this feasible solution to two different basic feasible solution.

Answer 1

$5x_1 − 4x_2 + 3x_3 + x_4 = 3...(1) \\2x_1 + x_2 + 5x_3 − 3x_4 = 0 ...(2) \\x_1 + 6x_2 − 4x_3 + 2x_4 = 15 ...(3)$

Put $x_1 = 1, x_2 = 2, x_3 = 1, x_4 = 3$ in (1), we get

$5\times 1-4\times2+3\times1+3=3 \\\Rightarrow 3=3$

Put $x_1 = 1, x_2 = 2, x_3 = 1, x_4 = 3$ in (2), we get

$2\times1+2+5\times1-3\times3=0 \\\Rightarrow 0=0$

Put $x_1 = 1, x_2 = 2, x_3 = 1, x_4 = 3$ in (3), we get

$1+6\times2-4\times1+2\times3=15 \\\Rightarrow 15=15$

$\therefore$ $x_1 = 1, x_2 = 2, x_3 = 1, x_4 = 3$ is a feasible solution.