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Your matrix acts on column vectors to the right and on row vectors to the left (there is some duality here but it is not important in first approach). So, it can be considered either as a linear map \mathbb{Q}^3\to \mathbb{Q}^10 or as a linear map \mathbb{10}^3\to \mathbb{Q}^3 , respectively.

It has rank 3, so in the first case, the kernel is trivial, in the second case the kernel has dimension 7.

Here is some code to understand what i just wrote:

sage: A = matrix(QQ, [[1, 0, 1], [1, 0, 0], [0, 1, 1], [0, 1, 0], [0, 0, 1], [-1, 0, 0], [0, 0, -1], [0, -1, 1], [0, -1, 0], [-1, 0, 1]])
sage: A
[ 1  0  1]
[ 1  0  0]
[ 0  1  1]
[ 0  1  0]
[ 0  0  1]
[-1  0  0]
[ 0  0 -1]
[ 0 -1  1]
[ 0 -1  0]
[-1  0  1]

sage: A.parent()
Full MatrixSpace of 10 by 3 dense matrices over Rational Field

sage: A.rank()
3

sage: A.right_kernel()
Vector space of degree 3 and dimension 0 over Rational Field
Basis matrix:
[]

sage: A.right_kernel().dimension()
0

sage: K = A.left_kernel()
sage: K
Vector space of degree 10 and dimension 7 over Rational Field
Basis matrix:
[ 1  0  0  0  0  0  0 -2  2  1]
[ 0  1  0  0  0  0  0 -1  1  1]
[ 0  0  1  0  0  0  0 -1  2  0]
[ 0  0  0  1  0  0  0  0  1  0]
[ 0  0  0  0  1  0  0 -1  1  0]
[ 0  0  0  0  0  1  0  1 -1 -1]
[ 0  0  0  0  0  0  1  1 -1  0]

sage: K.dimension()
7

sage: K.basis()
[
(1, 0, 0, 0, 0, 0, 0, -2, 2, 1),
(0, 1, 0, 0, 0, 0, 0, -1, 1, 1),
(0, 0, 1, 0, 0, 0, 0, -1, 2, 0),
(0, 0, 0, 1, 0, 0, 0, 0, 1, 0),
(0, 0, 0, 0, 1, 0, 0, -1, 1, 0),
(0, 0, 0, 0, 0, 1, 0, 1, -1, -1),
(0, 0, 0, 0, 0, 0, 1, 1, -1, 0)
]

sage: K.basis()[0]
(1, 0, 0, 0, 0, 0, 0, -2, 2, 1)

sage: K.basis()[0] * A
(0, 0, 0)