Find the determinant of matrix $A$

Question

Let A be a 3 × 3 real matrix with zero diagonal entries. If $1 + i$ is an eigenvalue of A, the determinant of $A$ equals-

I know the trace of the matrix is sum of eigen value but couldn't solve it.

Answer 1

If the entries are real, then $i + 1$ can only arise as the root of a polynomial with real coefficients. Such roots appear as conjugate pairs. So you know what 2 of the roots are. The last one is such as to make their sum equal to the trace.

This should be sufficient for you to be able to solve it.

Answer 2

Since $i+1$ is an eigenvalue, as already noticed by @PrimeMover and @Bungo, $1-i$ is also an eigenvalue. Then, we know that the trace, which is $0$, is equal to the sum of eigenvalues. Thus: $$ 2+ \lambda_3 = 0 $$ Therefore, the other eigenvalue is $-2$. Now recall that: $$ \det A= \prod_{i=1}^{3} \lambda_i = -2(1+i)(1-i)= -4 $$