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.
$$A^{-1} +B^{-1} = (A+B)^{-1}$$
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.
No comments:
Post a Comment