A problem is stated as such:
Solve the system of equations.
$\begin{bmatrix}x+2y=16\\3x-y=6\end{bmatrix}$
So, I started by partitioning the matrix up into its constant and coefficient parts.
$$\begin{bmatrix}1 & 2 \\ 3 & 1\end{bmatrix}$$ $$\begin{bmatrix}x\\y\end{bmatrix}$$ $$\begin{bmatrix}16\\6\end{bmatrix}$$
Then I calculated the determinant of the square matrix.
$$\begin{vmatrix}1&2\\3&1\end{vmatrix} = -5$$
Then I calculated the coefficient matrix of the square matrix.
$$(-1)^{1+1}\times1=1 \\ (-1)^{1+2}\times3=-3\\(-1)^{2+1}\times2=-2\\(-1)^{2+2}\times1=1\\ \begin{bmatrix}1&-3\\-2&1\end{bmatrix}$$
Transposing this coefficient matrix, I got the adjoint matrix.
$$\begin{bmatrix}1&-2\\-3&1\end{bmatrix}$$
Multiplying the adjoint matrix with the determinant, I got the inverse matrix of the original square matrix.
$$\begin{bmatrix}1&-2\\-3&1\end{bmatrix} \div -5 = \begin{bmatrix}-0.2&0.4\\0.6&-0.2\end{bmatrix}$$
Multiplying this last matrix, the inverse of the square matrix, with the matrix containing the constants $(16,6)$, I will get the $(x,y)$ matrix, only it will contain numerical values, thus revealing the values of the two unknowns.
$$\begin{bmatrix}-0.2&0.4\\0.6&-0.2\end{bmatrix} \times \begin{bmatrix}16\\6\end{bmatrix} = \begin{bmatrix}2.4\\8.4\end{bmatrix}$$
These values don't work within the original equations though. Where did I go wrong?
Your mistake is in the first equation. You have taken coefficient of $y$ in $3x-y=6$ as $1$ but should be $-1$. Thus the first matrix should be
$$\begin{bmatrix}1 & 2 \\ 3 & -1\end{bmatrix}$$
Now continue your answer and see you will get the right answer.