Question:

Given the matrices: $$A = \begin{pmatrix} 1 & 1 & -1 \\ 2 & 1 & 1 \\ 3 & 2 & 2 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix}$$ The objective is to find the value of $\text{tr}(XB^T)$, where $X$ is the solution to the matrix equation $AX=B$. Step 1: Calculate the determinant of $A$. $$|A| = 1(1 \cdot 2 - 1 \cdot 2) - 1(2 \cdot 2 - 1 \cdot 3) + (-1)(2 \cdot 2 - 1 \cdot 3)$$ $$|A| = 1(0) - 1(4 - 3) - 1(4 - 3)$$ $$|A| = 0 - 1(1) - 1(1) = -2$$ Step 2: Calculate the adjoint of $A$. First, find the cofactor matrix $C$: $$C_{11} = (1 \cdot 2 - 1 \cdot 2) = 0$$ $$C_{12} = -(2 \cdot 2 - 1 \cdot 3) = -(4 - 3) = -1$$ $$C_{13} = (2 \cdot 2 - 1 \cdot 3) = (4 - 3) = 1$$ $$C_{21} = -(1 \cdot 2 - (-1) \cdot 2) = -(2 + 2) = -4$$ $$C_{22} = (1 \cdot 2 - (-1) \cdot 3) = (2 + 3) = 5$$ $$C_{23} = -(1 \cdot 2 - 1 \cdot 3) = -(2 - 3) = 1$$ $$C_{31} = (1 \cdot 1 - (-1) \cdot 1) = (1 + 1) = 2$$ $$C_{32} = -(1 \cdot 1 - (-1) \cdot 2) = -(1 + 2) = -3$$ $$C_{33} = (1 \cdot 1 - 1 \cdot 2) = (1 - 2) = -1$$ Thus, the cofactor matrix is: $$C = \begin{pmatrix} 0 & -1 & 1 \\ -4 & 5 & 1 \\ 2 & -3 & -1 \end{pmatrix}$$ The adjoint of $A$ is the transpose of the cofactor matrix: $$\text{Adj}(A) = C^T = \begin{pmatrix} 0 & -4 & 2 \\ -1 & 5 & -3 \\ 1 & 1 & -1 \end{pmatrix}$$ Step 3: Calculate the inverse of $A$. $$A^{-1} = \frac{1}{|A|} \text{Adj}(A) = \frac{1}{-2} \begin{pmatrix} 0 & -4 & 2 \\ -1 & 5 & -3 \\ 1 & 1 & -1 \end{pmatrix}$$ Step 4: Solve for $X$ in $AX=B$. $$X = A^{-1}B = \frac{1}{-2} \begin{pmatrix} 0 & -4 & 2 \\ -1 & 5 & -3 \\ 1 & 1 & -1 \end{pmatrix} \begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix}$$ $$X = \frac{1}{-2} \begin{pmatrix} (0)(-1) + (-4)(2) + (2)(3) \\ (-1)(-1) + (5)(2) + (-3)(3) \\ (1)(-1) + (1)(2) + (-1)(3) \end{pmatrix}$$ $$X = \frac{1}{-2} \begin{pmatrix} 0 - 8 + 6 \\ 1 + 10 - 9 \\ -1 + 2 - 3 \end{pmatrix} = \frac{1}{-2} \begin{pmatrix} -2 \\ 2 \\ -2 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$$ Step 5: Calculate $XB^T$. Given $X = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$ and $B^T = \begin{pmatrix} -1 & 2 & 3 \end{pmatrix}$. $$XB^T = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} \begin{pmatrix} -1 & 2 & 3 \end{pmatrix}$$ $$XB^T = \begin{pmatrix} (1)(-1) & (1)(2) & (1)(3) \\ (-1)(-1) & (-1)(2) & (-1)(3) \\ (1)(-1) & (1)(2) & (1)(3) \end{pmatrix} = \begin{pmatrix} -1 & 2 & 3 \\ 1 & -2 & -3 \\ -1 & 2 & 3 \end{pmatrix}$$ Step 6: Calculate the trace of $XB^T$. The trace of a square matrix is the sum of the elements on its main diagonal. $$\text{tr}(XB^T) = (-1) + (-2) + 3$$ $$\text{tr}(XB^T) = -3 + 3 = 0$$ The value of $\text{tr}(XB^T)$ is $0$.