Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 12 / b

An antisymmetric matrix satisfies $A^T=-A$. Each diagonal entry is therefore its own negative, so both diagonal entries vanish. Consequently

A real two-dimensional antisymmetric matrix has the form $A=\left(\begin{array}{cc}0& s\\ -s& 0\end{array}\right)$. Its determinant is $s^2=1/4$, and the specified positive upper-right entry selects $s=1/2$. Thus

Direct multiplication gives $A^2=-I/4$. Therefore $A(-4A)=I=(-4A)A$, yieldingThe nonzero determinant ensures that this two-sided inverse is unique as a matrix inverse.

Multiply $A^2=-I/4$ by $A$ once more:No additional diagonalization is needed.

Repeatedly using the quadratic identity givesIn entries these are $(-1{)}^n{4}^{-n}\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$ and $(-1{)}^n{2}^{-(2n+1)}\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, respectively. This is the basic reduction for functions of a two-dimensional antisymmetric matrix.

The even-power formula reduces every summand to a scalar multiple of $I$:Both series have absolute convergence. Apply the logarithm expansion ${\sum }_{n\ge 1}{z}^n/n=-\log (1-z)$ for $|z|<1$ to obtainThus the off-diagonal entries vanish and both diagonal entries have the required value.