Finding the eigenvalues Given a $3 \times 3$ matrix with real entries such that $\det (A) = 6$ and $\mathrm{trace}( A )= 0$. 
If $\det (A+I)=0$ where $I$ is a $3 \times 3$ identity matrix , then eigenvalues of $A$ are $-1$, $-2$, $3$. 
I am confused. Can we conclude from the given data that $-1$ is one of the eigenvalues. Rest follows as trace = sum of eigenvalues and determinant = product of eigenvalues.
Please advice.  
 A: $\lambda$ is an eigenvalue of $A$ if and only if $\lambda + \mu$ is an eigenvalue of $A+\mu I$. To see this, note that $A\mathbf{v}=\lambda\mathbf{v}$ if and only if $A\mathbf{v}+\mu\mathbf{v}=(\lambda+\mu)\mathbf{v}$, if and only if $(A+\mu I)\mathbf{v}=(\lambda+\mu)\mathbf{v}$.
The fact that the determinant of $A+I$ is $0$ tells you that $0$ is an eigenvalue of $A+I$, and therefore, that $-1$ is an eigenvalue of $A$ (so that $-1+1=0$ will be an eigenvalue of $A+I$).
Once you know that $-1$ is an eigenvalue of $A$, you know that the other two eigenvalues add up to $1$ (so the trace will be $0$), and have a product of $-6$ (so the determinant will be $6$). So you are looking for $r_1$ and $r_2$ with $r_1+r_2=1$, $r_1r_2=-6$. This gives you the quadratic
$$(\lambda-r_1)(\lambda-r_2) = \lambda^2 -\lambda - 6$$
whose roots are $3$ and $-2$, giving you the other two eigenvalues. 
A: Since $\det(A+I) = 0$, the homogeneous system of equations $(A+I)x = 0$ has a non-trivial solution, i.e. $Ax = -Ix = -1x$ for some non-zero vector $x$.
A: This is built into the definition of eigenvalue. We have $\lambda$ is an eigenvalue of $A$  iff there is a non-zero $v$ such that $Av=\lambda v$, meaning that $(A-\lambda I)v=0$. That is the case precisely if $A-\lambda I$ is not invertible, which holds precisely if the determinant of $A-\lambda I$ is $0$. Put $\lambda=-1$. 
