The properties described in this question come from P. Grinfeld's book: "Introduction to Tensor Analysis and the Calculus of Moving Surfaces".
The Kronecker delta (or symbol) is defined by equation (4.73) as: $$ \delta^i_j=\left\{\begin{array}{r} 1,\forall i=j\\ 0,\forall i\neq j. \end{array}\right. $$
Later, in chapter 9, equation (9.27), the second order delta system $\delta^{ij}_{rs}$ is defined as a determinant: $$ \delta^{ij}_{rs} = \begin{vmatrix} \delta^i_r & \delta^i_s\\ \delta^j_r & \delta^j_s \end{vmatrix}. $$
Note that prior to this definition, an important remark is given: A particular entry of a delta system has value $1$ when the upper and lower are identical sets of distinct numbers related by an even permutation, $-1$ if the sets are related by an odd permutation, and $0$ otherwise.
Therefore, the object $\delta^{ij}_{rs}$ is skew-symmetric in its lower or upper indices, meaning that $\delta^{ij}_{sr}=-\delta^{ij}_{rs} = -\delta^{ji}_{sr} = \delta^{ji}_{rs}$.
My issue is with the contraction property with indices that appear twice, once as upper, and once as lower. For this, equation (9.22) states that: $$ \delta^{ij}_{rj}=2\delta^i_r. $$
However, taking the previous definition with the determinant operation,
$$ \delta^{ij}_{rj} = \begin{vmatrix} \delta^i_r & \delta^i_j\\ \delta^j_r & \delta^j_j \end{vmatrix} = \delta^i_r\delta^j_j - \delta^i_j\delta^j_r. $$
But according to the Kronecker delta definition above, $\delta^j_j=1$. Additionally, by contraction, the second term of the right hand side above becomes $\delta^i_j\delta^j_r = \delta^i_r$. Therefore, $$ \delta^{ij}_{rj} =\delta^i_r-\delta^i_r=0. $$
What am I doing wrong here? Is there some property that I am still missing in order to retrieve equation (9.22)?