If A= \begin{bmatrix} 2 & 0 & 3 \\ 0 & 3 & 2 \\ -2 & 0 & -4 \\ \end{bmatrix} Then cofactor matix= \begin{bmatrix} -12 & -4 & 6 \\ 0 & -2 & 0 \\ -9 & -4 & 6 \\ \end{bmatrix}
But if I use elementary row operation (R3=R3+R1) on A,I can get
A= \begin{bmatrix}
2 & 0 & 3 \\
0 & 3 & 2 \\
0 & 0 & -1 \\
\end{bmatrix}
and cofactor matrix== \begin{bmatrix} -3 & 0 & 0 \\ 0 & -2 & 0 \\ -9 & -4 & 6 \\ \end{bmatrix}
Am I doing something wrong or can you have different cofactor matrices for the same matirx?
Also, since A inverse =1/det(A) * adj (A)
Since I will have different adj(A), then I will have different A inverse too. But I thought inverse was unique.
I'm so confused. I'm not sure where I'm wrong. Can someone help me out? Thanks.
Elementary row operations lead, in general, to a different matrix $\bar A\neq A$.
There is not reason that, if it exists, $\bar A^{-1}=A^{-1}$