Solution
= Solution
The middle row is $-1/2$ times the sum of the first and third rows, so the rows are linearly dependent and the <determinant> is zero. The <matrix trace> is the sum of diagonal entries, $1+4+1=6$. A <transpose> preserves the <determinant>, while the <determinant> of a product is the product of its <determinants>. Consequently
$$
\boxed{\det A=0,\quad\operatorname{Tr}A=6,\quad\det(A^{\mathsf T})=0,\quad\det(A^2)=0.}
$$