# Find the determinant of matrix $A$

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.

• What do you mean by "zero diagonal entries"? All zeros on the main diagonal? – imranfat May 23 '20 at 9:39
• Hint: If $1+i$ is an eigenvalue, then so is $1-i$. – Bungo May 23 '20 at 9:39

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.
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: $$\det A= \prod_{i=1}^{3} \lambda_i = -2(1+i)(1-i)= -4$$