How to solve system of equations using inverse matrix? System of equations is the following:
$$x + 4y + 2z = 10$$
$$4x - 3y+0z = 6$$
$$2x + 2y + 2z = 14$$
Here is my solution:
$$det(A) = 1 *(-3 * 2 - 0 * 2) -4 * (4 * 2 - 0 * 2) + 2 * (4 * 2 - (-3) * 2)$$
$$= -6 -32 + 28$$
$$= -10$$
$$
    +\begin{pmatrix}
    -3 & 0 \\
    2 & 2 \\
    \end{pmatrix}
$$
$$
    -\begin{pmatrix}
    4 & 0 \\
    2 & 2 \\
    \end{pmatrix}
$$
$$
    +\begin{pmatrix}
    4 & -3 \\
    2 & 2 \\
    \end{pmatrix}
$$
$$
    -\begin{pmatrix}
    4 & 2 \\
    2 & 2 \\
    \end{pmatrix}
$$
$$
    +\begin{pmatrix}
    1 & 2 \\
    2 & 2 \\
    \end{pmatrix}
$$
$$
    -\begin{pmatrix}
    1 & 4 \\
    2 & 2 \\
    \end{pmatrix}
$$
$$
    +\begin{pmatrix}
    4 & 2 \\
    -3 & 0 \\
    \end{pmatrix}
$$
$$
    -\begin{pmatrix}
    1 & 2 \\
    4 & 0 \\
    \end{pmatrix}
$$
$$
    +\begin{pmatrix}
    1 & 4 \\
    4 & -3 \\
    \end{pmatrix}
$$
Above equals to:
$$
    \begin{pmatrix}
    -6 & -8 & 14 \\
    -4 & -2 & 6 \\
    -6 & -8 & -15\\
    \end{pmatrix}
$$
Then I multiply it by 1/-10 and the result is:
$$
    \begin{pmatrix}
    0,6 & 0,4 & 0,6 \\
    0,8 & 0,2 & -0,8 \\
    -1,4 & -0,6 & 1,5\\
    \end{pmatrix}
$$
Then I multiply it by:
$$
    \begin{pmatrix}
    10 \\
    4 \\
    16\\
    \end{pmatrix}
$$
Result is:
$$
    \begin{pmatrix}
    6 & 2,4 & 8,4 \\
    8 & 1,2 & -11,2 \\
    -14 & -3,6 & 21\\
    \end{pmatrix}
$$
Which results in:
$$
    \begin{pmatrix}
    16,8 \\
    -2\\
    3,4\\
    \end{pmatrix}
$$
So according to this logic:
$$x = 16,8$$
$$y = -2$$
$$z = 3,4$$
However when I test this solution it is incorrect, can anyone tell me what I'm doing wrong? Thanks.

There are mistakes in the last 3 determinants.
$+\begin{vmatrix}4 & 2 \\-3 & 0\end{vmatrix} = 0 + 6 = 6$
$-\begin{vmatrix}1 & 2 \\4 & 0\end{vmatrix} = -(0-8)=+8$
$+\begin{vmatrix}1 & 4 \\4 & -3\end{vmatrix} = -3 - 16 = -19$
Try using these values.
