x and y are terrible variable names. length1 and length2 would be better.
matrix = [[-1]*(x) for val in range (0,y)]
You are not using val in this list comprehension. Convention is to use _ for throw-away, variables needed for syntax reasons, but not actually used.
0 is the default minimum in range() statements. If you loop from 0 to some limit, you don't need to mention the 0; you can just use range(y).
You are never reading the -1 value anywhere. The value is always overwritten by another value before it is read. To make this clearer, store None instead of -1 in the matrix you are creating.
matrix = [ [None] * x for _ in range(y) ]
Using i-1 < 0 is an awkward way of writing i == 0. Similarly, i-1 >= 0 can be written simply as i > 0, or perhaps even i, since non-zero values are "Truthy".
The following is awkward and hard to understand. 6 statements, 4 assignments, two conditionals. What does it do? What does it mean?
val1 = 0
val2 = 0
if i-1 >= 0:
val1 = matrix[i-1][j]
if j-1 >= 0:
val2 = matrix[i][j-1]
Python has a x if cond else y trinary operator, which may help simplify and clarify the code.
val1 = matrix[i-1][j] if i > 0 else 0
val2 = matrix[i][j-1] if j > 0 else 0
That a lot more concise. Two statements which look the similar; the differences should be clear, and it should be easier to understand what those differences mean.
for i in range(y):
for j in range(x):
if word1[j] == word2[i]:
matrix[i][j] = 1
if i > 0 and j > 0:
maxtrix[i][j] += matrix[i-1][j-1]
else:
val1 = matrix[i-1][j] if i > 0 else 0
val2 = matrix[i][j-1] if j > 0 else 0
matrix[i][j] = max(val1, val2)
The statement return matrix[y-1][x-1] returns the last column of the last row. You don't actually need to know the dimensions of the matrix for this. Simply return matrix[-1][-1].
After you generate row 1, you no longer need row 0 of the matrix. After you generate row 2, you no longer need row 1 of the matrix. After you generate row 3, you no longer need row 2 of the matrix. This means you could solve the problem in \$O(m)\$ memory, instead of \$O(m*v)\$ memory, by simply maintaining a prev_row and next_row, instead of an entire matrix.
