0 个投票

I have a matrix:
syms m
A = [1 2 3 4;2 -1 1 1;-1 4 3 2;1 2 0 m]
with m is the parameter I use rref command:
rref(A)
ans =
[ 1, 0, 0, 0]
[ 0, 1, 0, 0]
[ 0, 0, 1, 0]
[ 0, 0, 0, 1]
the parameter m was lost. How I can convert matrix A to echelon form that the parameter m is still kept. thanks you very much.

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Matt J
Matt J 2014-1-5
编辑：Matt J 2014-1-5

0 个投票

I suspect, after experimenting with multiple m, that the echelon form of this particular matrix is independent of m, e.g.,
>> m=1; A = [1 2 3 4;2 -1 1 1;-1 4 3 2;1 2 0 m]; rref(A)
ans =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
>> m=10; A = [1 2 3 4;2 -1 1 1;-1 4 3 2;1 2 0 m]; rref(A)
ans =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1

更多回答（1 个）

Roger Stafford
Roger Stafford 2014-1-6

0 个投票

It is in the nature of the reduced row echelon form for your square matrix to be the identity matrix. That is because no matter what value m has, the matrix is always non-singular.
Suppose we alter one number in A:
A = [1 2 3 4;2 -1 1 1;1 4 3 2;1 2 0 m]
Then rref(A) still gives the appearance of being independent of the value of m. However there is one and only one value, m = -3.2, which makes the matrix singular and in that case the bottom row of rref(A) will become all zeros, showing that it can be affected by the value of m.
You can read about this form and its properties at:
http://en.wikipedia.org/wiki/Row_echelon_form

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