Question:

Step 1: Calculate the determinant and inverse of matrix $A$. Given matrix $A$: $$A = \begin{pmatrix} 1 & 1 & -1 \\ 2 & 1 & 1 \\ 3 & 2 & 2 \end{pmatrix}$$ The determinant of $A$ is: $$|A| = 1(1 \cdot 2 - 1 \cdot 2) - 1(2 \cdot 2 - 1 \cdot 3) + (-1)(2 \cdot 2 - 1 \cdot 3)$$ $$|A| = 1(0) - 1(4-3) - 1(4-3)$$ $$|A| = 0 - 1(1) - 1(1) = -2$$ The cofactor matrix $C$ for $A$ is: $$C = \begin{pmatrix} +(1 \cdot 2 - 1 \cdot 2) & -(2 \cdot 2 - 1 \cdot 3) & +(2 \cdot 2 - 1 \cdot 3) \\ -(1 \cdot 2 - (-1) \cdot 2) & +(1 \cdot 2 - (-1) \cdot 3) & -(1 \cdot 2 - 1 \cdot 3) \\ +(1 \cdot 1 - (-1) \cdot 1) & -(1 \cdot 1 - (-1) \cdot 2) & +(1 \cdot 1 - 1 \cdot 2) \end{pmatrix}$$ $$C = \begin{pmatrix} 0 & -1 & 1 \\ -4 & 5 & 1 \\ 2 & -3 & -1 \end{pmatrix}$$ The adjugate matrix $Adj(A)$ is the transpose of the cofactor matrix: $$Adj(A) = C^T = \begin{pmatrix} 0 & -4 & 2 \\ -1 & 5 & -3 \\ 1 & 1 & -1 \end{pmatrix}$$ The inverse matrix $A^{-1}$ is $\frac{1}{|A|} Adj(A)$: $$A^{-1} = \frac{1}{-2} \begin{pmatrix} 0 & -4 & 2 \\ -1 & 5 & -3 \\ 1 & 1 & -1 \end{pmatrix}$$ Step 2: Solve for $X$ in the equation $AX=B$. Given matrix $B$: $$B = \begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix}$$ We find $X$ using $X = A^{-1}B$: $$X = \frac{1}{-2} \begin{pmatrix} 0 & -4 & 2 \\ -1 & 5 & -3 \\ 1 & 1 & -1 \end{pmatrix} \begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix}$$ $$X = \frac{1}{-2} \begin{pmatrix} (0)(-1) + (-4)(2) + (2)(3) \\ (-1)(-1) + (5)(2) + (-3)(3) \\ (1)(-1) + (1)(2) + (-1)(3) \end{pmatrix}$$ $$X = \frac{1}{-2} \begin{pmatrix} 0 - 8 + 6 \\ 1 + 10 - 9 \\ -1 + 2 - 3 \end{pmatrix} = \frac{1}{-2} \begin{pmatrix} -2 \\ 2 \\ -2 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$$ Step 3: Calculate the product $X B^T$. The transpose of $B$ is $B^T = \begin{pmatrix} -1 & 2 & 3 \end{pmatrix}$. $$X B^T = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} \begin{pmatrix} -1 & 2 & 3 \end{pmatrix}$$ $$X B^T = \begin{pmatrix} (1)(-1) & (1)(2) & (1)(3) \\ (-1)(-1) & (-1)(2) & (-1)(3) \\ (1)(-1) & (1)(2) & (1)(3) \end{pmatrix} = \begin{pmatrix} -1 & 2 & 3 \\ 1 & -2 & -3 \\ -1 & 2 & 3 \end{pmatrix}$$ Step 4: Calculate the trace of $X B^T$. The trace of a square matrix is the sum of the elements on its main diagonal. $$tr(X B^T) = (-1) + (-2) + (3)$$ $$tr(X B^T) = -1 - 2 + 3 = 0$$