$\text{Adj}(A) = \begin{pmatrix} 1 & 4 & 4 \\ 2 & 1 & 7 \\ 1 & 1 & 3 \end{pmatrix}$, then $|A|=$
Step-by-Step Solution
Key Concept: For any invertible matrix $A$, the determinant of the adjugate matrix satisfies $|\text{Adj}(A)| = |A|^{n-1}$ in dimension $n$, leading to a polynomial equation for $|A|$.
We use the fundamental property that $\text{Adj}(A) \cdot A = |A| \cdot I$. Taking determinants of both sides: $|\text{Adj}(A)| \cdot |A| = |A|^3$. First, calculate $|\text{Adj}(A)| = 1(3-7) - 4(6-7) + 4(2-1) = -4 + 4 + 4 = 4$. Therefore $4 \cdot |A| = |A|^3$, which gives $|A|^3 - 4|A| = 0$, so $|A|(|A|^2 - 4) = 0$. This yields $|A| = 0, 2, \text{ or } -2$. Since $\text{Adj}(A)$ is non-zero, $|A| \neq 0$, leaving $|A| = \pm 2$.
Correct Answer: 3,4