$A$ is a $3\times3$ matrix and $A^2+4 A−12 I =0 $. If det$(A+2I )>0$, what is $det(A+2I )$?

Question

I need to rearrange the following $A^2+4A-12I=0$, where $A$ is an unknown $3\times3$ matrix so that I can find det of $(A+2I)$

Answer 1

Notice $(A+2I)^{2}-16I=0$, so we get $\det^2(A+2I)=16^3$. The value is potisive and hence 64.

Answer 2

The equation $A^2+4A-12I=0$ is equivalent to $(A+2I)^2 = 16I$. Now use the product formula for determinants to solve for $\det(A+2I)$.

Answer 3

A has -6 and 2 and 0 as Eigen values. It can be a matrix with all elements zero except diagonal matrix with -6,2,0 in the diagonal. Now the determinant of A+2I should be greater than zero and it will be if 2 is at 1,1 and -6 at 3,3. This matrix satisfies all the conditions u asked. And determinant of A+2I is 4. Not the best method though.