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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.

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    $\begingroup$ You are making confusion with the row operation concept. The row operations preserve the solution for $Ax=b$, the row space, the determinant (when we combine row without multiply by scalars $\neq 1$) but they leads to completely different matrices. $\endgroup$ – user Oct 23 '18 at 13:59
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Elementary row operations lead, in general, to a different matrix $\bar A\neq A$.

There is not reason that, if it exists, $\bar A^{-1}=A^{-1}$

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