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

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 $(A+2I)$

## closed as off-topic by Watson, naslundx, T. Bongers, heropup, HenrikSep 3 '16 at 19:21

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

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

• I stand corrected. Edited! – Theorem Sep 3 '16 at 13:44

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.

• If $A = \begin{bmatrix} -6 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix}$, then $A^2 +4A - 12I = \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 &-12 \end{bmatrix}.$ – thanasissdr Sep 3 '16 at 13:41
• I missed that. But it got eliminated as determinant of A+2I is not greater than 0. – jnyan Sep 3 '16 at 13:51