Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 2 / 3

A homogeneous linear system has a nonzero solution exactly when its coefficient matrix is singular. Expansion of the determinant givesThus the allowed values are $\boxed{\alpha =1\text{ or }-8/5}$. Solving the first two equations and checking the third givesIn each case $z\ne 0$ gives a nonzero solution and spans the one-dimensional null space; $z=0$ gives only the zero vector.

Since $A^2=-{ϵ}^{2}I$, its even powers are $A^{2k}=(-1{)}^k{ϵ}^{2k}I$ and its odd powers are $A^{2k+1}=(-1{)}^k{ϵ}^{2k}A$. Splitting the matrix exponential into these two absolutely convergent subseries gives the matrix exponential for a scalar square:At $ϵ=0$, the coefficient $\sin ϵ/ϵ$ has limiting value one and $A=0$, so $S=I$. The result is a planar rotation through angle $-ϵ$.

Now $B^2={ϵ}^{2}I$, so splitting its matrix exponential into even and odd powers yields hyperbolic functions:The limiting coefficient at $ϵ=0$ is again one, and $T=I$. The eigenvectors $(1,1)$ and $(1,-1)$ have factors $e^{ϵ}$ and $e^{-ϵ}$, respectively, showing how this hyperbolic transformation differs from a rotation.