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I recently thought about this exercise that states:

Does there exist a matrix $A \in M_2(\mathbb R)$ s.t.

$$(A^{-1} - 3 I_2)^t = 2 \left( \begin{array}{ccc} -1 & -2 \\ 1 & -5 \\ \end{array} \right)$$

But such a matrix $A$ is not uniquely defined?

I thought that the determinant of $A$ must have determinant = 3 to have a matrix that is not uniquely defined. But I am a bit lost on how to actually find it.

Anyone mind lending a hand?

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$$(A^{-1} - 3 I_2)^t = 2 \left( \begin{array}{ccc} -1 & -2 \\ 1 & -5 \\ \end{array} \right)$$

$$(A^{-1} - 3 I_2) = 2 \left( \begin{array}{ccc} -1 & 1 \\ -2 & -5 \\ \end{array} \right)=\begin{pmatrix} -2 & 2 \\ -4 & -10\end{pmatrix}$$

$$A^{-1} =\begin{pmatrix} 1 & 2 \\ -4 & -7\end{pmatrix}$$

$$A=\begin{pmatrix} 1 & 2 \\ -4 & -7\end{pmatrix}^{-1}$$

Are you able to compute $A$ explicitly?

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  • $\begingroup$ Yes thanks, so A is always obviously uniquely defined. Thanks again. $\endgroup$ – Monolite Jul 12 '16 at 22:57
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Why should the determinat of $A$ be $3$?

In this case you just take the transpose on both sides and bring the $3I_2$ term on the other side; this clearly defines uniquely $A^{-1}$

So the question is: do two (or more) different matrix exists such that their inverse is equal to $A^{-1}$, which you found? Basically, when is the inverse of a matrix unique?

This should be easy enough to answer; if the expression you find for $A^{-1}$ admits a unique inverse, then the matrix $A$ is uniquely defined

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