The dimension of the null space of a matrix is the number of linearly independent vectors that are in the null space. The null space of a matrix is the set of all vectors that satisfy the equation $A\mathbf{x}=\mathbf{0}$.
To find the dimension of the null space, we can use Gaussian elimination. First, we can reduce the matrix $A$ to row echelon form:
$$\left[ {\begin{array}{{20}{c}} 0&1&1 \ 1&{ – 1}&0 \ { – 1}&0&1 \end{array}} \right] \xrightarrow{R_1 \leftrightarrow R_2} \left[ {\begin{array}{{20}{c}} 1&{ – 1}&0 \ 0&1&1 \ { – 1}&0&1 \end{array}} \right] \xrightarrow{R_2 + R_1 \rightarrow R_2} \left[ {\begin{array}{*{20}{c}} 1&{ – 1}&0 \ 0&0&1 \ { – 1}&0&1 \end{array}} \right]$$
We can see that the reduced row echelon form of $A$ has one non-zero row. This means that the null space of $A$ has one dimension.
Therefore, the dimension of the null space of the matrix $\left[ {\begin{array}{*{20}{c}} 0&1&1 \ 1&{ – 1}&0 \ { – 1}&0&1 \end{array}} \right]$ is $\boxed{1}$.
Here is a brief explanation of each option:
- Option A: The dimension of the null space is 0 if and only if the matrix is invertible. The matrix $\left[ {\begin{array}{*{20}{c}} 0&1&1 \ 1&{ – 1}&0 \ { – 1}&0&1 \end{array}} \right]$ is not invertible, so the dimension of its null space cannot be 0.
- Option B: The dimension of the null space is 1 if and only if the matrix has one non-zero row in its row echelon form. The matrix $\left[ {\begin{array}{*{20}{c}} 0&1&1 \ 1&{ – 1}&0 \ { – 1}&0&1 \end{array}} \right]$ has one non-zero row in its row echelon form, so the dimension of its null space is 1.
- Option C: The dimension of the null space is 2 if and only if the matrix has two non-zero rows in its row echelon form. The matrix $\left[ {\begin{array}{*{20}{c}} 0&1&1 \ 1&{ – 1}&0 \ { – 1}&0&1 \end{array}} \right]$ has only one non-zero row in its row echelon form, so the dimension of its null space cannot be 2.
- Option D: The dimension of the null space is 3 if and only if the matrix has three non-zero rows in its row echelon form. The matrix $\left[ {\begin{array}{*{20}{c}} 0&1&1 \ 1&{ – 1}&0 \ { – 1}&0&1 \end{array}} \right]$ has only one non-zero row in its row echelon form, so the dimension of its null space cannot be 3.