$$A=\begin{bmatrix} 3 & -2 & -1 \\ 1 & 2 & 1 \\ -1 & 1 & 1 \end{bmatrix}, X= \begin{bmatrix} x \\ y \\ z \end{bmatrix}, B = \begin{bmatrix} 1 \\ 7 \\ 2\end{bmatrix}$$
\begin{align} AX=B A^{-1} & \text{(from right side)} \\ A^{-1} AX = A^{-1}B \\ EX = A^{-1}B \\ X = A^{-1}B \end{align}
So, now I have to find the inverse matrix.
$$A^{-1} = \frac{1}{\det(A)}* (\text{adj} A)^T$$
$$\det(A)=\begin{vmatrix} 3 & -2 & -1 \\ 1 & 2 & 1 \\ -1 & 1 & 1 \end{vmatrix}=4$$ $$\text{adj}(A)=\begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{bmatrix}$$
So, now that I found all the elements for the adjA, I get the following matrix: $$\text{adj}(A)=\begin{bmatrix} -1 & -2 & 3 \\ 1 & 2 & -1 \\ 0 & -4 & 8 \end{bmatrix}$$ $$(\text{adj}A)^T=\begin{bmatrix} -1 & 1 & 0 \\ -2 & 2 & -4 \\ 3 & -1 & 8 \end{bmatrix}$$
So, the inverse matrix is equal to: $$A^{-1}=\frac{1}{4} \begin{bmatrix} -1 & 1 & 0 \\ -2 & 2 & -4 \\ 3 & -1 & 8 \end{bmatrix}$$
When I do the math, I get the following matrix: $$A^{-1}= \begin{bmatrix} -\frac{1}{4} & \frac{1}{4} & 0 \\ -\frac{2}{4} & \frac{2}{4} & -\frac{4}{4} \\ \frac{3}{4} & -\frac{1}{4} & \frac{8}{4} \end{bmatrix}$$
Now the final equation $X = A^{-1} B$ is: $$A^{-1}= \begin{bmatrix} -\frac{1}{4} & \frac{1}{4} & 0 \\ -\frac{2}{4} & \frac{2}{4} & -\frac{4}{4} \\ \frac{3}{4} & -\frac{1}{4} & \frac{8}{4} \end{bmatrix} \begin{bmatrix} 1 \\ 7 \\ 2 \end{bmatrix} = \, ?$$
The solution is supposed to be $$X = \begin{bmatrix} 1 \\ 7 \\ 2 \end{bmatrix}$$
But I can't get that solution when I multiply my equations what did I do wrong?
The first entry in adj$A$ should be 1