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Linear System of Equations

Let us look at some examples of linear systems:

  1. Suppose a, b ∈ Consider the system ax = b.
    If a 6= 0 then the system has a unique solution

If a = 0 and:
a) b 6= 0 then the system has no solution.
b) b = 0 then the system has infinite number of solutions, namely all x ∈ R.

2. We now consider a system with 2 equations in 2 unknowns.
Consider the equation ax + by = c. If one of the coefficients, a or b is non-zero, then this linear equation represents a line in R2. Thus for the system
2017-07-18_22-05-51.jpg

the set of solutions is given by the points of intersection of the two lines. There are three cases to be considered. Each case is illustrated by an example.
a) Unique Solution

2017-07-18_22-08-30.jpg

  1. Observe that in this case, a1b2 − a2b1 6= 0.
    b) Infinite Number of Solutions
    x + 2y = 1 and 2x + 4y = 2. The set of solutions is
    (x, y)t = (1 − 2y, y)t = (1, 0)t + y(−2, 1)t
    with y arbitrary. In other words, both the equations represent the same line.
    Observe that in this case, a1b2 − a2b1 = 0, a1c2 − a2c1 = 0 and b1c2 − b2c1 = 0.
    c) No Solution
    x + 2y = 1 and 2x + 4y = 3. The equations represent a pair of parallel lines and hence there is no point of intersection.
    Observe that in this case, a1b2 − a2b1 = 0 but a1c2 − a2c1 6= 0.

3.

As a last example, consider 3 equations in 3 unknowns.

A linear equation ax + by + cz = d represent a plane in R3 provided (a, b, c) 6= (0, 0, 0). As in the case of 2 equations in 2 unknowns, we have to look at the points of intersection of the given three planes. Here again, we have three cases. The three cases are illustrated by examples.
a) Unique Solution
Consider the system x+y+z = 3, x+4y+2z = 7 and 4x+10y−z = 13. The unique solution to this system is (x, y, z)t = (1, 1, 1)t; i.e. the three planes intersect at a point.
b) Infinite Number of Solutions
Consider the system x + y + z = 3, x + 2y + 2z = 5 and 3x + 4y + 4z = 11. The set of solutions to this system is (x, y, z)t = (1, 2− z, z)t = (1, 2, 0)t + z(0,−1, 1)t, with z arbitrary:
the three planes intersect on a line.

  1. c) No Solution

The system x + y + z = 3, x + 2y + 2z = 5 and 3x + 4y + 4z = 13 has no solution. In this case, we get three parallel lines as intersections of the above planes taken two at a time.

Definition and a Solution Method

Linear System: A linear system of m equations in n unknowns x1, x2, . . . , xn is a set of equations of the form

a11x1 + a12x2 + · · · + a1nxn = b1

a21x1 + a22x2 + · · · + a2nxn = b2

 

am1x1 + am2x2 + · · · + amnxn = bm

where for 1 ≤ i ≤ n, and 1 ≤ j ≤ m; aij , bi ∈ R. Linear System (2.2.1) is called homogeneous if b1 = 0 = b2 = · · · = bm and non-homogeneous otherwise.

We rewrite the above equations in the form Ax = b, where

2017-07-18_22-14-18.jpg

The matrix A is called the coefficient matrix and the block matrix [A b] , is the augmented matrix of the linear system.

Remark: Observe that the ith row of the augmented matrix [A b] represents the ith equation and the jth column of the coefficient matrix A corresponds to coefficients of the jth variable xj . That is, for 1 ≤ i ≤ m and 1 ≤ j ≤ n, the entry aij of the coefficient matrix A corresponds to the ith equation and jth variable xj ..

For a system of linear equations Ax = b, the system Ax = 0 is called the associated homogeneous system.

Solution of a Linear System: A solution of the linear system Ax = b is a column vector y with entries y1, y2, . . . , yn such that the linear system.

That is, if yt = [y1, y2, . . . , yn] then Ay = b holds.

Note: The zero n-tuple x = 0 is always a solution of the system Ax = 0, and is called the trivial solution. A non-zero n-tuple x, if it satisfies Ax = 0, is called a non-trivial solution.

Example1 Let us solve the linear system x + 7y + 3z = 11, x + y + z = 3, and

4x + 10y − z = 13.

  1. The above linear system and the linear system
    2017-07-18_22-16-17.jpg    Interchange the first two equations.
  2. Eliminating x from 2nd and 3rd equation, we get the linear system
    2017-07-18_22-18-04.jpg   (obtained by subtracting the first  equation from the second equation.)
    2017-07-18_22-19-00.jpg  (obtained by subtracting 4 times the first equation from the third equation.)
  3. Eliminating y from the last two equations of system, we get the system
    obtained by subtracting the third equation from the second equation.
  4. Now, z = 1 implies
    2017-07-18_22-21-52.jpg
    and x = 3−(1+1) = 1. Or in terms of a vector, the set of solution is { (x, y, z)t : (x, y, z) = (1, 1, 1)}.

 

 

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