Possible Determinant Values of Matrix from Adjoint Matrix

To find the possible value(s) of the determinant of the $3 \times 3$ matrix $P$, we use the property relating a matrix and its adjoint: $\det(\text{adj}(P)) = (\det(P))^{n-1}$

For an $n \times n$ matrix with $n = 3$, this property becomes: $\det(\text{adj}(P)) = (\det(P))^{3-1} = (\det(P))^2$

Given:

$$\text{adj}(P) = \begin{bmatrix} 1 & 4 & 4 \\ 2 & 1 & 7 \\ 1 & 1 & 3 \end{bmatrix}$$

Now, we evaluate the determinant of $\text{adj}(P)$:

$$\det(\text{adj}(P)) = \begin{vmatrix} 1 & 4 & 4 \\ 2 & 1 & 7 \\ 1 & 1 & 3 \end{vmatrix}$$

Expanding along the first row: $\det(\text{adj}(P)) = 1 \cdot (1 \cdot 3 - 7 \cdot 1) - 4 \cdot (2 \cdot 3 - 7 \cdot 1) + 4 \cdot (2 \cdot 1 - 1 \cdot 1)$ $\det(\text{adj}(P)) = 1(3 - 7) - 4(6 - 7) + 4(2 - 1)$ $\det(\text{adj}(P)) = 1(-4) - 4(-1) + 4(1)$ $\det(\text{adj}(P)) = -4 + 4 + 4 = 4$

Equating this to $(\det(P))^2$: $(\det(P))^2 = 4$

Taking the square root on both sides: $\det(P) = \pm 2$

Thus, the possible values of the determinant of $P$ are $-2$ and $2$.

Therefore, the correct options are A and D.