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(B) Let $C$ be the matrix of cofactors of $A$. Given $C = \begin{bmatrix} 1 & -2 & 1 \ 4 & -5 & -2 \ -2 & 4 & 1 \end{bmatrix}$.We know that the adjo

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$3$

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(B) Let $C$ be the matrix of cofactors of $A$. Given $C = \begin{bmatrix} 1 & -2 & 1 \ 4 & -5 & -2 \ -2 & 4 & 1 \end{bmatrix}$.We know that the adjoint of $A$, denoted by $\operatorname{adj} A$, is the transpose of the cofactor matrix $C$.So, $\operatorname{adj} A = C^T = \begin{bmatrix} 1 & 4 & -2 \ -2 & -5 & 4 \ 1 & -2 & 1 \end{bmatrix}$.We use the property $|\operatorname{adj} A| = |A|^{n-1}$, where $n$ is the order of the matrix.Here $n = 3$, so $|\operatorname{adj} A| = |A|^{3-1} = |A|^2$.Now, calculate the determinant of $\operatorname{adj} A$:$|\operatorname{adj} A| = 1((-5)(1) - (4)(-2)) - 4((-2)(1) - (1)(4)) + (-2)((-2)(-2) - (1)(-5))$$|\operatorname{adj} A| = 1(-5 + 8) - 4(-2 - 4) - 2(4 + 5)$$|\operatorname{adj} A| = 1(3) - 4(-6) - 2(9)$$|\operatorname{adj} A| = 3 + 24 - 18 = 9$.Since $|A|^2 = 9$, we have $|A| = \pm 3$.Thus, a possible value of the determinant of $A$ is $3$.

If $A$ is a $3 \times 3$ matrix and the matrix obtained by replacing the elements of $A$ with their corresponding cofactors is $\begin{bmatrix} 1 & -2 & 1 \\ 4 & -5 & -2 \\ -2 & 4 & 1 \end{bmatrix}$, then a possible value of the determinant of $A$ is

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