Given $ \det A=2$ and $3I+2A=A^2$; how to find $\det(3I+A)$?

Question

Let $A$ be a $3 \times 3$ matrix with $\det (A)= 2 $ and $3I + 2A= A^2$. Find $\det(3I+A)$.

I know the final answer is 16 according to the book, but I can not find a way to solve the question. Here is how I tried to simplify the question:

$$3I + A= 3I + 2A - A = A^2 - A = A(A-I)$$

so we can say: $$\det(3I + A)= \det(A).\det(A-I)$$

But I can not find the value for $\det(A-I)$

Answer 1

Notice that

$$(3I+A)^2 = 9I + 6A + A^2 = 3(3I+2A) + A^2 = 4A^2$$

Edit: This implies

$$\text{det}(3I+A)^2 = \text{det}(4A^2) = 4^3\text{det}(A)^2 = 256$$

so the determinant must be one of $\pm 16$.

Answer 2

\begin{align} \det(A-I)\det(A+I)&=\det\left((A-I)(A+I)\right)\\ &=\det (A^2-I)\\ &=\det (2A+2I)=2^3\det(A+I) \end{align} where in former equality we used from assumption. So if $\det(A+I)\neq 0$ then $$\det(A-I)=2^3$$ and so $\det(A) \det(A-I)=2\times 8=16$.

Answer 3

Let it be clear : No matrix $A$ exists with these constraints.

Indeed, if one assumes that $V \neq 0$ is such that $AV=\lambda V$, then applying relationship $A^2-2A-3I=0$ to this vector $V$, one gets

$$\lambda^2 V-2\lambda V-3V=0 \iff \lambda^2-2\lambda-3=0$$

Therefore, the only possible eigenvalues are : $\lambda = -1$ or $3$.

Thus, the different possibilities for the spectrum of $A$ are :

$(-1, -1, -1), \ (-1, -1, 3), \ (-1, 3, 3), \ (3, 3, 3), \ \ \tag{1}$

But the determinant is the product of eigenvalues, and none of the four products issued from (1) is equal to $2$.