Below is a problem I did. However, I am getting the wrong answer.
Problem:
Find the inverse of the following matrix by finding its adjoint:
$$
\begin{bmatrix}
1 & 2 & 3 \\
2 & 5 & 3 \\
1 & 0 & 8 \\
\end{bmatrix} $$
Answer:
To find the inverse, we set up the following matrix and then transform it so the left hand side is the identity matrix.
\begin{align*}
\begin{bmatrix}
1 & 2 & 3 & 1 & 0 & 0 \\
2 & 5 & 3 & 0 & 1 & 0 \\
1 & 0 & 8 & 0 & 0 & 1 \\
\end{bmatrix} &\sim
\begin{bmatrix}
1 & 2 & 3 & 1 & 0 & 0 \\
0 & 1 & -3 & -2 & 1 & 0 \\
0 & -2 & 5 & -1 & 0 & 1 \\
\end{bmatrix} \\
&\sim
\begin{bmatrix}
1 & 2 & 3 & 1 & 0 & 0 \\
0 & 1 & -3 & -2 & 1 & 0 \\
0 & 0 & -1 & -5 & -2 & 1 \\
\end{bmatrix} \\
&\sim \begin{bmatrix}
1 & 2 & 3 & 1 & 0 & 0 \\
0 & 1 & -3 & -2 & 1 & 0 \\
0 & 0 & 1 & 5 & 2 & -1 \\
\end{bmatrix} \\
&\sim \begin{bmatrix}
1 & 2 & 3 & 1 & 0 & 0 \\
0 & 1 & 0 & 13 & -5 & -3 \\
0 & 0 & 1 & 5 & -3 & -1 \\
\end{bmatrix} \\
&\sim \begin{bmatrix}
1 & 0 & 3 & -25 & 10 & 6 \\
0 & 1 & 0 & 13 & -5 & -3 \\
0 & 0 & 1 & 5 & -3 & -1 \\
\end{bmatrix} \\
&\sim \begin{bmatrix}
1 & 0 & 0 & -40 & 19 & 9 \\
0 & 1 & 0 & 13 & -5 & -3 \\
0 & 0 & 1 & 5 & -3 & -1 \\
\end{bmatrix}
\end{align*}
Hence the answer is:
$$
\begin{bmatrix}
-40 & 19 & 9 \\
13 & -5 & -3 \\
5 & -3 & -1 \\
\end{bmatrix}
$$
I believe the correct answer is:
$$ \begin{bmatrix}
-40 & 16 & 9 \\
13 & -5 & -3 \\
5 & -2 & -1 \\
\end{bmatrix}
$$
Where did I go wrong?
Based upon comments from the group, I have updated my post. Here is my updated answer which is still wrong.
\begin{align*}
\begin{bmatrix}
1 & 2 & 3 & 1 & 0 & 0 \\
2 & 5 & 3 & 0 & 1 & 0 \\
1 & 0 & 8 & 0 & 0 & 1 \\
\end{bmatrix} &\sim
\begin{bmatrix}
1 & 2 & 3 & 1 & 0 & 0 \\
0 & 1 & -3 & -2 & 1 & 0 \\
0 & -2 & 5 & -1 & 0 & 1 \\
\end{bmatrix} \\
&\sim
\begin{bmatrix}
1 & 2 & 3 & 1 & 0 & 0 \\
0 & 1 & -3 & -2 & 1 & 0 \\
0 & 0 & -1 & -5 & -2 & 1 \\
\end{bmatrix} \\
&\sim \begin{bmatrix}
1 & 2 & 3 & 1 & 0 & 0 \\
0 & 1 & -3 & -2 & 1 & 0 \\
0 & 0 & 1 & 5 & 2 & -1 \\
\end{bmatrix} \\
&\sim \begin{bmatrix}
1 & 2 & 3 & 1 & 0 & 0 \\
0 & 1 & 0 & 13 & 7 & -3 \\
0 & 0 & 1 & 5 & 2 & -1 \\
\end{bmatrix} \\
&\sim \begin{bmatrix}
1 & 0 & 3 & -25 & -14 & 6 \\
0 & 1 & 0 & 13 & 7 & -3 \\
0 & 0 & 1 & 5 & 2 & -1 \\
\end{bmatrix} \\
&\sim \begin{bmatrix}
1 & 0 & 0 & -40 & -20 & 9 \\
0 & 1 & 0 & 13 & 7 & -3 \\
0 & 0 & 1 & 5 & 2 & -1 \\
\end{bmatrix} \\
\end{align*}
Hence the answer is:
$$
\begin{bmatrix}
-40 & 19 & 9 \\
13 & -5 & -3 \\
5 & -3 & -1 \\
\end{bmatrix}
$$
Best Answer
The method you have used to find the inverse is a fine method to find matrix inverses, however, you have a calculation error.
After your second set of row operations, you should have: \begin{bmatrix} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & -3 & -2 & 1 & 0 \\ 0 & 0 & -1 & -5 & 2 & 1 \\ \end{bmatrix}
Note the sign of the 2 in the last line vs. what you have above. This error, then accumulates down the following row operations.
However, you say you want to find the adjoint matrix.
In that case, if we call the column vectors of your original matrix $A,B,C$
The adjoint matrix is
\begin{bmatrix} B\times C\\C\times A\\A\times C \end{bmatrix}
Using the cross products of your column vectors.