Reduced Row Echelon of a Matrix containing unknown constants in Matlab - matlab

Question

the problem:
>> syms('a', 'b', 'c');
>>A =
[ -2, 3, 1, a]
[ 1, 1, -1, b]
[ 0, 5, -1, c]
>>rref(A)
ans =
[ 1, 0, -4/5, 0]
[ 0, 1, -1/5, 0]
[ 0, 0, 0, 1]
The problem I'm having is that I need the answer to be in terms of a b and c, so I need something like:
ans =
[1, 0, 0, a+2b-c]
[0, 1, 0, 3a-c]
[0, 0, 1, a+b+c]
Is there any way to get Matlab to accomplish this?

Answer 1

You just need to convert the problem to a relevant form in order to use rref properly, that is, instead of :
[1, 0, 0, a+2b-c]
[0, 1, 0, 3a-c]
[0, 0, 1, a+b+c]
where the x,y,z form the unity matrix block,
write:
syms x y z
D =
[ 1, 2, -1, x]
[ 3, 0, -1, y]
[ 1, 1, 1, z]
rref(D)
ans =
[ 1, 0, 0, (3*y)/10 - x/10 + z/5]
[ 0, 1, 0, (2*x)/5 - y/5 + z/5]
[ 0, 0, 1, (3*z)/5 - y/10 - (3*x)/10]

Answer 2

Matlab actually gave the correct answer for your example, which may seem surprising as it does not depend on a, b or c; if you use another example, e.g.
B = [1, 2, a; 2, 1, b];
rref(B)
then the answer will depend on a and b as expected.
Here I just explain why the reduced row echelon form of your A is indeed
[ 1, 0, -4/5, 0]
[ 0, 1, -1/5, 0]
[ 0, 0, 0, 1]
by showing some of the important steps of the row reduction. Matlab starts with A, and reduces it to
[ 1, 0, -4/5, (3*b)/5 - a/5]
[ 0, 1, -1/5, a/5 + (2*b)/5]
[ 0, 0, 0, a + 2*b - c]
it then divides the last row by (a + 2*b - c) to get
[ 1, 0, -4/5, (3*b)/5 - a/5]
[ 0, 1, -1/5, a/5 + (2*b)/5]
[ 0, 0, 0, 1]
and then it proceeds to subtract that row from the rows above it to give
[ 1, 0, -4/5, 0]
[ 0, 1, -1/5, 0]
[ 0, 0, 0, 1]
Matlab assumed that a + 2*b - c is not equal to zero (try replacing c with a + 2*b, then rref(A) will give you a different answer). I know of no way to force Matlab not to divide by a + 2*b - c, i.e. to make Matlab assume that a + 2*b + c may be equal to zero.

Answer 3

You want the elimination matrix that gives you rref(A).
But What is rref(A)?
UpperTri=rref(A)=[LastStep]...[Step3][Step2][Step1]*A.
That is a series of steps that reduces A to an upper triangle or the best one possible.
Matlab had,
[ 1, 0, -4/5, (3*b)/5 - a/5]
[ 0, 1, -1/5, a/5 + (2*b)/5]
[ 0, 0, 0, 1]
Then Matlab just rolled it up and you no longer have the steps. So it appears as:
[ 1, 0, -4/5, 0]
[ 0, 1, -1/5, 0]
[ 0, 0, 0, 1]
If you want the series of steps you want the elimination matrix with the series of steps so recorded it using an Identity matrix. Instead of:
A =
[ -2, 3, 1, a]
[ 1, 1, -1, b]
[ 0, 5, -1, c]
Use
>> A = [ -2, 3, 1, 1 0 0; 1, 1, -1, 0 1 0; 0, 5, -1, 0 0 1]
A =
-2 3 1 1 0 0
1 1 -1 0 1 0
0 5 -1 0 0 1
>> E=rref(A)
E =
1.0000 0 -0.8000 0 1.0000 -0.2000
0 1.0000 -0.2000 0 0 0.2000
0 0 0 1.0000 2.0000 -1.0000
This is like [A][I], Now [rref Steps matrix]*[A][I]=[rref(A)][rref Steps matrix]
Now E=[rref(A)][rref Steps matrix] =
E =
1.0000 0 -0.8000 0 1.0000 -0.2000
0 1.0000 -0.2000 0 0 0.2000
0 0 0 1.0000 2.0000 -1.0000
By inspection E1=[rref Steps matrix]=
E1 =
0 1.0000 -0.2000
0 0 0.2000
1.0000 2.0000 -1.0000
so now:
E1 =
0 1.0000 -0.2000
0 0 0.2000
1.0000 2.0000 -1.0000
>> B=[a;b;c;]
B =
a
b
c
>> B1=E1*B
B1 =
b - c/5
c/5
a + 2*b - c
To check:
A1 = [ -2, 3, 1, a; 1, 1, -1, b; 0, 5, -1, c]
A1 =
[ -2, 3, 1, a]
[ 1, 1, -1, b]
[ 0, 5, -1, c]
>> A2=E1*A1
A2 =
[ 1, 0, -4/5, b - c/5]
[ 0, 1, -1/5, c/5]
[ 0, 0, 0, a + 2*b - c]
Note a + 2b - c =0, therefore c=a+2b, therefore sub c, (c/5)= ((a+2b)/5), and (b-c/5)=(5b-(a+2b))/5
therefore
[(-a+3*b)/5; (a+2*b)/5; a + 2*b - c ]