Friday, January 24 − Lecture 9 : Solutions of RREF systems of linear equations.
Concepts:
1. basic unknowns, free unknowns of a system of linear equations in RREF
2. rank of a coefficient matrix.
3. solutions to a system of linear equation in vector form
9.1 We now consider ways of representing a family of solutions for an RREF system of
linear equations. First note that a linear equation with n unknowns {x , x , …, 1 }2 n
a1 1+ a x2 2... + a x n n
can be written as a dot produce a ⋅x = b where a = (a , a1, …2 a ) ann x = (x , x ,1…, 2 ). n
If x1= u 1 x =2u , 2, x = u ns a nolution we can then say that u = (u , u , …, u )1is 2 n
solution to the linear equation a ⋅ x = b (now expressed as a vector equations) since
a ⋅ u = b.
9.2 Theorem − If a system of linear equations has more than one solution then it has
infinitely many solutions.
Proof : Consider the following general system of m linear equations with n unknowns:
a11 1+ a x12 2.. + a x =1n n 1
a21 1 a x 22 2. + a x = 2n n 2
...
am1 1 a x m2 2. + a x = bmn n m
th
The i equation can be written as (a , …, i1) in ⋅ (x1, …, x n = a i⋅x =b. i
Suppose u = (u , …1 u ) ann v = (v , …, 1 ) are nistinct solutions to this system. Let α ∈
ℝ. Then for i = 1 to m and any scalar α we have
Since u and v are distinct if α ≠ β then u + α(u – v) ≠ u + β(u – v). So the system has
infinitely many distinct solutions, as required. 9. 3 Example − In the following example we solve a system of 2 linear equations in 4
unknowns by applying basic principles illustrated in a previous example.
Solve the system
x + u + −4z = 36
y + z = 6
where x, u, y, z are variables and expressed the solution set using vectors.
Solution :
Note that this is a system in RREF. We can rewrite the system as
x = −u + 4z + 36
y = 0u − z + 6
So the solution can be presented as follows:
{(x, y, u, z) : (x, y, u, z) = (−u + 4z + 36, 0u − z + 6, 1u + 0z + 0, 0u + 1z + 0), u, z ∈ ℝ }
We see it has infinitely many solutions since the values of u and z are arbitrary.
It is interesting to note that the family of solutions for the above system can be
conveniently expressed as a vector equation.
where u and z are free to take on any values.
It will be important for the student to be able to express the solution set in the form of a
vector equation when the sy

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