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Thursday, July 30, 2015

Matrices and determinants

Let $n$ be an integer greater or equal to $2$ and let $A, \; B$ be two $n \times n$ real matrices such that:

$$A^{-1} +B^{-1} = (A+B)^{-1}$$

Prove that $\det A = \det B$.

IMC 2015 1st problem

Solution



From the given equation we have that:

$$(A+B)(A^{-1} +B^{-1})=\mathbb{I}$$

hence $I+AB^{-1} + BA^{-1} = \mathbb{O}$. Multiplying from the right by $B$ it follows that:

$$B+A=-BA^{-1} B$$

Due to symmetry we also get that:

$$B+A=-AB^{-1}A$$.

Equating we have that $BA^{-1} B = AB^{-1} A$ and by taking dets. at the last equation we get the wanted result since we get $|A|^3 = |B|^3$ and for real matrices it follows that $|A|=|B|$ what we wanted.

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