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Let A be the matrix 1 0 1, 1 1 0, 0 1 0 defined over GF(2).
We're going to find the inverse matrix of A.
Remember that GF(2) is the field consisting of only two elements: 0 and 1.
Aaddition and multiplication are the same as for integers
except that 1 + 1 is 0.
So the process of finding the inverse matrix of A is the same as before.
We augment this matrix A with
the 3-by-3 identity matrix, and then we perform role reduction until we
obtain a matrix in reduced row-echelon form.
So the first thing we are going to do is we are
going to add the first row to the second row.
So the first and third row stay the same.
Now 1 + 1 is 0
so I get a 0 here. And 0 + 1 is 1
and 1 + 0 is 1.
And here I have 1 + 0
which is 1. 0 + 1 is 1 and 0 + 0 is 0.
Now I'm going to add the second row to the third row.
Next, we're going to try to eliminate these 1's.
And I'm going to do that by adding row 3 to row 1
as well as row 2. So I'm going to perform two operations at once.
And we have a matrix in reduced row-echelon form.
So these three columns would give us A inverse.
And you can indeed check that
this is the inverse matrix of A by multiplying the two matrices.
You should get the identity matrix.