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I am slightly stuck on this seemlingly simple problem that I encoutered as part of a problem to show that the orthogonality condition of $M_{2\times2}$ matrices given by $\sum_i a_{ij}a_{jk} = \delta_{jk}$, where the $a_{ij}$ are the matrix entries, implies that $\det(M) = \pm 1$.

If $b(1-a^2) = a(1-b^2)$, show that $a=b$ or $a=1, b=-1$ or $a=-1, b =1$ is the only solutions where $a,b \in \mathbb R$ or give a counter-example if the statement is false.

I am thinking of testing the intervals $a,b \in (-\infty,-1); a,b \in (1,\infty)$ and $a,b \in [-1,1]$ seperately and showing in each case that if $a > b$ or $b > a$ then equality cannot hold. However, this requires testing lots of cases and I thought there might be a simpler way.

A hint will help a lot.

Thank You.

Floris
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2 Answers2

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Hint: By rearranging and factoring, notice that: \begin{align*} (ba^2 - ab^2) - (b - a) &= 0 \\ ab(a - b) + (a - b) &= 0 \\ (ab + 1)(a - b) &= 0 \\ \end{align*}

Adriano
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Technically speaking your equation has also the solution $a=-1/b$ with $a,b \neq 0$...

Umberto
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