1. **State the problem:** We are given matrices $$A = \begin{bmatrix}1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4\end{bmatrix}, \quad B = \begin{bmatrix}-3 \\ -6 \\ -9\end{bmatrix}$$ and the equation $$AX = B$$ where $$X$$ is the unknown vector to find. 2. **Formula and approach:** To solve for $$X$$, we use the formula $$X = A^{-1}B$$ if $$A$$ is invertible. 3. **Check if $$A$$ is invertible by calculating its determinant:** $$\det(A) = 1 \times \begin{vmatrix}1 & 2 \\ 2 & 4\end{vmatrix} - 2 \times \begin{vmatrix}-1 & 2 \\ 1 & 4\end{vmatrix} + 3 \times \begin{vmatrix}-1 & 1 \\ 1 & 2\end{vmatrix}$$ Calculate minors: $$\begin{vmatrix}1 & 2 \\ 2 & 4\end{vmatrix} = (1)(4) - (2)(2) = 4 - 4 = 0$$ $$\begin{vmatrix}-1 & 2 \\ 1 & 4\end{vmatrix} = (-1)(4) - (2)(1) = -4 - 2 = -6$$ $$\begin{vmatrix}-1 & 1 \\ 1 & 2\end{vmatrix} = (-1)(2) - (1)(1) = -2 - 1 = -3$$ So, $$\det(A) = 1 \times 0 - 2 \times (-6) + 3 \times (-3) = 0 + 12 - 9 = 3$$ Since $$\det(A) = 3 \neq 0$$, $$A$$ is invertible. 4. **Find the inverse $$A^{-1}$$:** Calculate the matrix of cofactors, then transpose to get adjugate, then divide by determinant. Cofactors matrix: $$C = \begin{bmatrix} +\begin{vmatrix}1 & 2 \\ 2 & 4\end{vmatrix} & -\begin{vmatrix}-1 & 2 \\ 1 & 4\end{vmatrix} & +\begin{vmatrix}-1 & 1 \\ 1 & 2\end{vmatrix} \\ -\begin{vmatrix}2 & 3 \\ 2 & 4\end{vmatrix} & +\begin{vmatrix}1 & 3 \\ 1 & 4\end{vmatrix} & -\begin{vmatrix}1 & 2 \\ 1 & 2\end{vmatrix} \\ +\begin{vmatrix}2 & 3 \\ 1 & 2\end{vmatrix} & -\begin{vmatrix}1 & 3 \\ -1 & 2\end{vmatrix} & +\begin{vmatrix}1 & 2 \\ -1 & 1\end{vmatrix} \end{bmatrix}$$ Calculate each minor: $$\begin{vmatrix}1 & 2 \\ 2 & 4\end{vmatrix} = 0$$ $$\begin{vmatrix}-1 & 2 \\ 1 & 4\end{vmatrix} = -6$$ $$\begin{vmatrix}-1 & 1 \\ 1 & 2\end{vmatrix} = -3$$ $$\begin{vmatrix}2 & 3 \\ 2 & 4\end{vmatrix} = (2)(4) - (3)(2) = 8 - 6 = 2$$ $$\begin{vmatrix}1 & 3 \\ 1 & 4\end{vmatrix} = (1)(4) - (3)(1) = 4 - 3 = 1$$ $$\begin{vmatrix}1 & 2 \\ 1 & 2\end{vmatrix} = (1)(2) - (2)(1) = 2 - 2 = 0$$ $$\begin{vmatrix}2 & 3 \\ 1 & 2\end{vmatrix} = (2)(2) - (3)(1) = 4 - 3 = 1$$ $$\begin{vmatrix}1 & 3 \\ -1 & 2\end{vmatrix} = (1)(2) - (3)(-1) = 2 + 3 = 5$$ $$\begin{vmatrix}1 & 2 \\ -1 & 1\end{vmatrix} = (1)(1) - (2)(-1) = 1 + 2 = 3$$ So, $$C = \begin{bmatrix}0 & 6 & -3 \\ -2 & 1 & 0 \\ 1 & -5 & 3\end{bmatrix}$$ 5. **Adjugate matrix $$\text{adj}(A)$$ is the transpose of $$C$$:** $$\text{adj}(A) = C^T = \begin{bmatrix}0 & -2 & 1 \\ 6 & 1 & -5 \\ -3 & 0 & 3\end{bmatrix}$$ 6. **Calculate inverse:** $$A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = \frac{1}{3} \begin{bmatrix}0 & -2 & 1 \\ 6 & 1 & -5 \\ -3 & 0 & 3\end{bmatrix}$$ 7. **Find $$X = A^{-1}B$$:** $$X = \frac{1}{3} \begin{bmatrix}0 & -2 & 1 \\ 6 & 1 & -5 \\ -3 & 0 & 3\end{bmatrix} \begin{bmatrix}-3 \\ -6 \\ -9\end{bmatrix}$$ Calculate the product inside first: First row: $$0 \times (-3) + (-2) \times (-6) + 1 \times (-9) = 0 + 12 - 9 = 3$$ Second row: $$6 \times (-3) + 1 \times (-6) + (-5) \times (-9) = -18 - 6 + 45 = 21$$ Third row: $$-3 \times (-3) + 0 \times (-6) + 3 \times (-9) = 9 + 0 - 27 = -18$$ So, $$X = \frac{1}{3} \begin{bmatrix}3 \\ 21 \\ -18\end{bmatrix} = \begin{bmatrix}1 \\ 7 \\ -6\end{bmatrix}$$ **Final answer:** $$X = \begin{bmatrix}1 \\ 7 \\ -6\end{bmatrix}$$