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?