Past exam of the mathematical methods I course of the University of Oxford / 2015 / 8

The Leibniz formula for determinants is unchanged by interchanging rows and columns, or equivalently replacing each permutation by its inverse. Thus

Use the multiplicativity of the determinant on $A{A}^{-1}=I$:The denominator is nonzero because $A$ is an invertible matrix.

Multiplying $A$ by $c$ multiplies each of its $n$ columns by $c$. The multilinearity of the determinant therefore givesThis includes $c=0$.

For an integer $c$, repeated multiplicativity of the determinant, with the inverse identity for negative powers, givesFor an arbitrary real $c$, an invertible real matrix does not by itself define a unique real matrix power. For example, the scalar matrix $(-1)$ has no real square root. If a matrix logarithm $L$ has been chosen with $e^L=A$ and $A^c$ is defined as $e^{cL}$, the matrix exponential determinant identity gives the precise formulaThis cannot always be replaced by an independently chosen scalar branch of $(\det A{)}^c$. For instance, $A=-I_2$ has the complex square root $i{I}_2$, whose determinant is $-1$, while the principal scalar square root of $\det A=1$ is $+1$. For a positive-definite symmetric matrix with its usual positive spectral power, all eigenvalues are positive and the familiar formula $(\det A{)}^c$ holds for every real $c$.