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Prove that for every matrix $A$ in $K^{n\times n}$, where $K$ is a field, there exists a $B$ in $K^{n\times n}$ such that $AB = BA = (\det A) \times I$ ($I$ denotes the unit matrix).

Later edit: Sure, for $\det A \ne 0$, $B = A^*$ satisfies the equalities, where $A^*$ is the adjoint matrix of $A$. I would be interested in finding $B \ne O_n$ for $\det A = 0$

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    Just a minor remark: the quest for $B\neq0$ is hopeless when $n=0$. And then $A$ is not even singular in that case!2011-12-21

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When $A$ is nonsingular, the adjugate matrix $\operatorname{adj} A$ already works. However, if $A$ is singular, this approach fails because it is possible that $\operatorname{adj} A = 0$. Here's one alternate approach for this case.

Since the space of $n \times n$ matrices is finite dimensional, the infinite list of matrices $\{ I, A, A^2, \ldots \}$ is linearly dependent. Pick a linear dependence $ a_m A^{m} + a_{m-1} A^{m-1} + \cdots + a_{1} A^1 + a_0 I = 0 $ among the powers of $A$ such that $m$ is as small as possible, and define $B = a_m A^{m-1} + a_{m-1} A^{m-2} + \cdots + a_{1} I . \tag{$\dagger$}$

Now, first note that $-a_0 I = A B = B A$. Taking determinants on both sides, we have $(-1)^n a_0^n = 0$. So $a_0 = 0$, and hence $AB = AB = 0$. Finally, $(\dagger)$ says that $B$ is a linear combination of the first $m-1$ powers of $A$, and hence it is nonzero by our choice of $m$.

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    I understand now! Thank you so much!2011-12-21
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Here is another way of producing a non-zero $B$ when $A$ is singular.

Let $x$ be a non-zero column vector such that $Ax = 0$. Since the transpose $A^\mathrm T$ is also singular, we can find a $y \in K^n$ such that $(A^\mathrm T)y = 0$, i.e. $y^\mathrm TA = 0$. Then $B = xy^\mathrm T$ will work: if $x_i$ and $y_j$ are non-zero, then the $ij$-th component of $B$ is $x_iy_j \neq 0$. [Many thanks to Srivatsan for showing me that this is enough.]