Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 18

Since $A^2=-I$, the matrix inverse is $A^{-1}=-A$. The orthogonal matrix condition also gives $A^{-1}=A^{\mathsf{T}}$, so $A^{\mathsf{T}}=-A$. Thus $A$ is skew-symmetric, forcing $a=0$ and $c=-b$. Its columns must be unit vectors, giving $b^2=1$. Therefore the two possibilities areBoth satisfy the original conditions by direct multiplication. They are the two orthogonal skew-symmetric matrices in two dimensions, representing opposite quarter-turn rotations.

For either $A$ from part (i), chooseIndeed, using $A^2=-I$ and $A^{\mathsf{T}}=-A$,Its matrix trace is $-2/\sqrt{2}=-\sqrt{2}<0$. Explicitly, for the two $A$ matrices in the order displayed above, the corresponding orthogonal matrices areTheir negatives have positive matrix trace and would not meet the additional sign condition.

Each prediction is a linear combination of the constant parameter and the coefficient of $x_i$. Thus the design matrix isIts $i$th row multiplied by the parameter vector $(\alpha ,\beta {)}^{\mathsf{T}}$ gives the required prediction $\alpha +\beta x_i$.

Compute the two sides of the normal equations for linear least squares and divide by $n$:The first equation gives $\alpha =\overline{y}-\beta \overline{x}$. Substitute this into the second and solve:These simple linear regression coefficients are unique provided the observed distances are not all equal, sinceThe condition $n\ge 2$ alone does not guarantee this. If all $x_i=x_0$, the design matrix has matrix rank one and the valid alternative is the familyIn this rank-deficient ordinary least squares case, the fitted predictions are uniquely $\overline{y}$ but the two parameters cannot be separately determined.