# Linear algebra: determinant of the sum of a left circuit and a non-singular matrix over the field of order 2

For any matrix $$A = (a_ {ij}) in M_n ( mathbb {F} _2)$$denote $$bar {A} = ( bar {a_ {ij}}),$$ where $$bar {a} = 0$$ Yes $$a = 1$$ Y $$bar {a} = 1$$ Yes $$a = 0$$. Suppose $$n$$ It is a power of two. Leave $$A$$ (circulating left) and $$I & # 39;$$ be $$n times n$$ matrices about $$mathbb {F} _2$$ defined by,
$$X = begin {pmatrix} x_1 & x_2 & x_3 & ldots & x_ {n-1} & x_n \ x_2 & x_3 & x_4 & ldots & x_ {n} & x_1 \ x_3 & x_4 & x_5 & ldots & x_ {1} & x_2 \ & & & vdots \ x_ {n-1} & x_n & x_1 & ldots & x_ {n-3} & x_ {n-2} \ x_n & x_1 & x_2 & ldots & x_ {n-2} & x_ {n-1} end {pmatrix}, I & # 39; = begin {pmatrix} 0 & 0 & 0 & ldots & 0 & 0 \ 0 & 0 & 0 & ldots & 0 & 1 \ 0 & 0 & 0 & ldots & 1 & 1 \ & & & & vdots \ 0 & 0 & 1 & ldots & 1 & 1 \ 0 & 1 & 1 & ldots & 1 & 1 end {pmatrix}.$$
Then it seems that neither $$X + I & # 39;$$ or $$X + bar {I & # 39;$$ It's invertible. But I can not prove that. Is there a simple method to prove this?