# Can A matrix A have different cofactor matrices? If So then, can you have different inverses for one matrix?

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.

• 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. – user Oct 23 '18 at 13:59

Elementary row operations lead, in general, to a different matrix $$\bar A\neq A$$.
There is not reason that, if it exists, $$\bar A^{-1}=A^{-1}$$