Math 2250-1
Tues Sept 18 - Filled in version of notes
3.3 The structure of the solution sets to linear algebraic systems of equations, based on reduced row
echelon form computations.
> restart :
Exercise 1 What four conditions are necessary for a matrix to be in reduced row echelon form?
1) each non-zero row starts with a "1" (called the leading 1 for that row.)
2) as you go down non-zero rows, the leading 1's move strictly to the right
3) if you look at the column of any (row's) leading 1, all the other entries in that column are zero.
4) the rows that are all 0's are at the bottom of the matrix....i.e. after the non-zero rows.
Exercise 2 Are the following matrices in reduced row echelon form or not? Explain.
a) no .... leading 1 in second row is to the left of the leading 1 in the first row
0 0 1 2
0 1 0 K1
0 0 0
0
0 1 0
2
b) yes
0 0 1 K8
0 0 0 0
c) no ... there is a leading 1 for the 3rd row, but the column entries for that leading 1 are not zero....the
correct rref form for this matrix would just have zeroes in those other column 4 entries, since you can add
multiples of row 3 to rows 1 and 2 to bring this about.
1 0 1 2
0 1 3 K1
0 0 0
1
d) yes
1 0 2 0
0 1 3 0
0 0 0 1
e) yes
1 0 0
2
0 1 0 K3
0 0 1
7
Exercise 2 Coefficient matrix taken from problem #19, section 3.3, page 174. (You are assigned similar
problems).
> with LinearAlgebra :
> A d Matrix 3, 5, 2, 7,K10,K19, 13,
1, 3,K4, K8, 6,
1, 0, 2, 1, 3 ;
2 7 K10 K19 13
1 3
K4
K8
6
1 0
2
1
3
A :=
(1)
> ReducedRowEchelonForm A ;
1 0
2
1 3
0 1 K2 K3 1
0 0
0
(2)
0 0
Let's consider three different linear systems for which A is the coefficient matrix. In the first one, the right
hand sides are all zero (what we call the "homogeneous" problem), and I have carefully picked the other
two right hand sides. The three right hand sides are separated by the dividing line below:
2 7 K10 K19 13 0 7 7
1 3
K4
K8
6 0 0 3 .
1 0
2
1 3 0 0 0
We'll try solving three linear systems at once!
> b1 d Vector 0, 0, 0 :
b2 d Vector 7, 0, 0 :
b3 d Vector 7, 3, 0 :
C d A b1 b2 b3 ; # very augmented matrix
ReducedRowEchelonForm C ;
2 7 K10 K19 13 0 7 7
C :=
1 3
K4
K8
6 0 0 3
1 0
2
1
3 0 0 0
1 0
2
1 3 0 0 0
0 1 K2 K3 1 0 0 1
0 0
0
(3)
0 0 0 1 0
2a) Find the solution sets for each of the three systems:
In class we did this carefully on the blackboard. As it turns out, Maple also backsolves from the reduced
row echelon forms of the reduced augmented matrices. It labels the free parameters according the
column from which they originated:
> LinearSolve A, b1 ;
(4)
K2 _t3 K _t4 K 3 _t5
2 _t3 C 3 _t4 K _t5
_t3
(4)
_t4
_t5
> LinearSolve A, b2 ; # this is the inconsistent system having no solutions.
Error, (in LinearAlgebra:-LinearSolve) inconsistent system
> LinearSolve A, b3 ; #Maple is using t1's instead of t's because it used t's in the
#previous computation
K2 _t13 K _t14 K 3 _t15
2 _t13 C 3 _t14 K _t15 C 1
_t13
(5)
_t14
_t15
Important conceptual questions:
2b) Which of these three solutions could you have written down just from the reduced row echelon form
of A, i.e. without using the augmented matrix? Why?
The first one, i.e. the homogenous problem. because if you start with an augmented column of zeroes, it
stays zero as you do elementary row operations. So you can just append a column of zeroes to the
reduced row echelon form of A, and backsolve to get the explicit solutions.
2c) Linear systems in which right hand side vectors equal zero are called homogeneous linear systems.
Otherwise they are called inhomogeneous or nonhomogeneous. Notice that the general solution to the
consistent inhomogeneous system is the sum of a particular solution to it, together with the general
solution to the homogeneous system!!! Was this an accident? It's related to an important general concept
which will keep coming up in the rest of the course.
Exercise 3) The reduced row echelon form of a (non-augmented) matrix A can tell us a lot about the
solution set to the linear system Ax=b . (Do you remember this "matrix times vector" notation for writing
a linear system?) Consider the matrix A below, and answer all questions:
> A d Matrix 2, 5, 2, 7,K10,K19, 13, 1, 3,K4,K8, 6 ;
ReducedRowEchelonForm A ;
2 7 K10 K19 13
A :=
1 3 K4 K8 6
1 0
2
1 3
(6)
0 1 K2 K3 1
> Vector
b1 , b2
;
b1
(7)
b2
3a) Is the homogeneous problem Ax=0 always solvable?
yes. in fact there infinitely many solutions, with 3 free parameters. We could backsolve
as usual, since the original and reduced augmented matrix would just have zeroes in the right-hand side
column vector.
3b) Is the inhomogeneous problem Ax=b solvable no matter the choice of b?
YES: The augmented matrix
2 7 K10 K19 13 b1
1 3
K4
K8
6 b2
will reduce to a matrix of the form
1 0
2
1 3 c1
0 1 K2 K3 1 c2
and we can just backsolve as before...e.g..
x5 = t, x4 = r, x3 = p, x2 = c2 C 2 r C 3 p K t, x1 = c1 K 2 p K r K 3 t . We can always backsolve to find
solutions as there isn't a row in the reduced augmented matrix that has all zero coefficients (to the left of
the dividing bar), but a non-zero entry to the right of it.
3c) How many solutions are there? How many free parameters are there in the solution? How does this
number relate to the reduced row echelon form of A?
infinitely many, 3 free parameters....which is exactly the number of columns which do not have a leading 1
(coming from some row).
Exercise 4) Now consider the matrix B and similar questions:
> B d Matrix 3, 2, 1, 2,K1, 3, 4, 2 ;
ReducedRowEchelonForm B ;
1 2
B :=
K1 3
4 2
1 0
0 1
(8)
0 0
4a) How many solutions to the homogeneous problem Bx = 0?
one solution. x2 = 0, x1 = 0 .
4b) Is the inhomogeneous problem Bx = b solvable for every right side vector b?
NO. if we started with a right side b, and it turned into a c vector in the reduced row echelon form, we
could only solve the problem if c3 = 0. To be more precise, if we started with the augmented matrix
1 2 b1
K1 3 b2
4 2 b
3
it would reduce to
1 0 c1
0 1 c2 .
0 0 c
3
If c3 s 0 (it would = 1 if were in reduced row echelon form in this case), then there is no solution. And
there are b-vectors for which c3 s0, since we could start such a reduced form and reverse the elementary
row ops that created it, in order to create a b-vector for an inconsistent system with the beginning A
matrix.
4c) When the inhomogeneous problem is solvable, how many solutions does it have?
ONE. when you back solve, (after c3 = 0, you'd get x2 = c2 , x1 = c1 .
Exercise 5) Square matrices (i.e number of rows equals number of columns) with 1's down the diagonal
which runs from the upper left to lower right corner are special. They are called identity matrices.
> C d Matrix 4, 4, 1, 0,K1, 1, 22,K1, 3, 5, 7, 4, 6, 2, 3, 5, 7, 13 ;
ReducedRowEchelonForm C ;
1 0 K1 1
C :=
22 K1
3
5
2
7
4
6
3
5
7 13
1 0 0 0
0 1 0 0
(9)
0 0 1 0
0 0 0 1
>
5a) How many solutions to the homogeneous problem Cx = 0 ?
one. The zero vector always the homogenous problem, and it's the only one when you backsolve this
system.
5b) Is the inhomogeneous problem Cx = b solvable for every choice of b?
YES
5c) How many solutions?
ONE. In fact, the solution is exactly the "c" vector in the 5th column of the would arise from the reduced
row echelon form computation
b
c
1 0 K1 1 1
1 0 0 0 1
22 K1 3 5 b2
0 1 0 0 c2
/
.
7 4 6 2 b3
0 0 1 0 c3
3
5
7 13 b
4
0 0 0 1 c
4
Exercise 6: What are your general conclusions?
6a) What conditions on the reduced row echelon form of the matrix A guarantee that the homogeneous
equation Ax = 0 has infinitely many solutions?
6b) What conditions on the dimensions of A (i.e. number of rows and number of columns) always force
infinitely many solutions to the homogeneous problem?
6c) What conditions on the reduced row echelon form of A guarantee that solutions x to Ax = b are always
unique (if they exist)?
6d) If A is a square matrix (m=n), what can you say about the solution set to Ax = b when
∗ The reduced row echelon form of A is the identity matrix?
∗ The reduced row echelon form of A is not the identity matrix?