Determinant and Eigens

The sum of diagonal entries is the same as the sum of the eigenvalues which is called trace of matrix. And the determinant of the matrix is equal to the product of the eigenvalues.

Since, the characteristic polynomial will be of degree 3 there will be 3 roots of the polynomial.

If $1+i$ is a solution then $1-i$ will also be its solution and let the third eigenvalue will be $\lambda_3$ . So we have $1+i+1-i+\lambda_3=0$ $\lambda_3=-2$ $|A|=(1+i)(1-i)(-2)$ $|A|=\boxed{-4}$