Past exam of the mathematical methods I course of the University of Oxford / 2017 resit / 10

Matrix multiplication is associative, soThe uniqueness of the matrix inverse then givesThe order is reversed; $A^{-1}{B}^{-1}$ does not generally work because matrix multiplication need not be commutative.

A symmetric matrix has $A^T=A$, without conjugation. One invertible example with nonreal off-diagonal entries isIt is symmetric but not Hermitian, illustrating the importance of the distinction over $\mathbb{C}$.

A Hermitian matrix has $A^{\ast }=A$. ChooseIts off-diagonal entries are complex conjugates and its diagonal entries are real, so the defining condition holds. It is also invertible.

A unitary matrix has $A^{\ast }A=I$. For example,This example is invertible, has nonreal off-diagonal entries, and preserves the standard Hermitian inner product. Although it is also symmetric, that additional property is not required for unitarity.

The quadratic form has symmetric matrixNormalized eigenvectors for $3$ and $-2$ are $(2,1)/\sqrt{5}$ and $(-1,2)/\sqrt{5}$. They form the columns of the rotation matrixThus the new $x^{\prime }$ axis is rotated counterclockwise by $\theta =\arctan (1/2)$, withThe opposite signs of the eigenvalues classify each positive level as a hyperbola. For the original level six its standard equation is $x^{\prime 2}/2-y^{\prime 2}/3=1$; for the final level one it is $3x^{\prime 2}-2y^{\prime 2}=1$. The printed change of level therefore changes the scale, but not the principal axes or asymptotes.

Setting the homogeneous quadratic part to zero gives $y^{\prime }=\pm \sqrt{3/2}{x}^{\prime }$. In the original coordinates put $y=mx$ in $2x^2+4xy-y^2=0$, obtainingThese are the slopes in the original $(x,y)$ axes; $\pm \sqrt{3/2}$ are the slopes in the rotated axes, so they must not be interchanged.