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Chapter 4: Problem 51

What is a "basic solution"? How might one find a basic solution of a given system of linear equations?

Short Answer

Expert verified
A basic solution is a solution to a system of linear equations where some variables are assigned the value zero, and others are determined by solving a smaller system. To find a basic solution, first convert the system into an augmented matrix and then into row-echelon form. Identify pivot variables and free variables, express pivot variables in terms of free variables, and create basic solutions by setting free variables to zero.

Step by step solution

01

Rewriting the system as an augmented matrix

Write the given system of linear equations as an augmented matrix. This will make it easier to see the relationships between the variables and the constants in the system. Augmented matrix is obtained by placing the constant terms of the equations to the right of the coefficients matrix, separated by a vertical line.
02

Convert the matrix to row-echelon form

Use elementary row operations (swapping rows, multiplying rows by nonzero constants, or adding/subtracting rows) to convert the augmented matrix to row-echelon form. A matrix is in row-echelon form if each row has at least one nonzero entry, the first nonzero entry in a row (called a pivot) is one, the pivots move strictly to the right as we move down the rows, and rows of zeroes (if any) are at the bottom.
03

Identify pivot variables and free variables

Once the matrix is in row-echelon form, identify the variables corresponding to the pivots (these are called pivot variables) and the variables that don't correspond to pivots (called free variables). The pivot variables are determined by the non-zero elements in each row of the row-echelon form, while the free variables can be assigned any value.
04

Express pivot variables in terms of free variables

For each pivot variable, express it in terms of the free variables by rewriting the corresponding equation in the row-echelon form, isolating the pivot variable on the left side.
05

Create the basic solutions by setting free variables to zero

To find the basic solutions of the system, individually set each of the free variables to zero while keeping the other free variables as parameters. For each setting, compute the values of the pivot variables expressed in terms of these values for the free variables. Each set of values for the pivot variables and the chosen zero-valued free variables forms a basic solution of the system. These are the steps to find the basic solutions of a given system of linear equations. Keep in mind that a basic solution is not necessarily unique in many cases, and not all systems of linear equations will have a basic solution.

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Most popular questions from this chapter

Solve the LP problems. If no optimal solution exists, indicate whether the feasible region is empty or the objective function is unbounded. \(\vee\) Minimize \(c=-x+2 y\) subject to \(\begin{aligned} y & \leq \frac{2 x}{3} \\\ x & \leq 3 y \\ y & \geq 4 \\ x & \geq 6 \\ x+y & \leq 16 . \end{aligned}\)

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If a linear programming problem has an unbounded, nonempty feasible region, then optimal solutions (A) must exist (B) may or may not exist (C) cannot exist

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