Past exam of the mathematical methods I course of the University of Oxford / 2019 / 12

The linear system has the matrix formThe first equation gives $3x_2+2x_3=-4$. Substitute it into the second to obtain $x_1=2+4x_2+2x_3=x_2-2$. Substitution into the third gives $x_2+x_3=-7$. ConsequentlyIt follows thatSubstitution verifies all three equations. Also $\det A=1\ne 0$, so this solution of a linear system is unique.

The characteristic polynomial of $M$ isThe corresponding eigenvectors may be chosen as $v_4=(1,2{)}^T$ and $v_5=(1,1{)}^T$. They are linearly independent, so write $\mathbf{x}=q_4(t)v_4+q_5(t)v_5$. Substitution in the system of ordinary differential equations givesIndependence of the eigenvectors yields two decoupled harmonic oscillator equations. Their four integration constants give the general real solutionBoth components can be differentiated directly to verify ${\mathbf{x}}^{\prime \prime }+M\mathbf{x}=0$. Orthogonality of the eigenvectors is unnecessary: diagonalizing this nonsymmetric matrix requires an invertible eigenbasis, not an orthonormal one.

In a real orthonormal basis, a matrix is a proper rotation matrix exactly when $R^{T}R=I$ and $\det R=1$. The first condition preserves dot products, and the second preserves orientation of a vector space.

Put $s=\sqrt{3}$, $p=(1+s)/3$, $q=(1-s)/3$ and $r=1/3$, so the rows of the given matrix are $(p,q,r),(r,p,q),(q,r,p)$. Their squared Euclidean norms and pairwise dot products areThus $R^{T}R=I$. For the sign of its determinant, use the circulant form:This proves that it is a rotation matrix. Its fixed axis is $(1,1,1{)}^T$, because each row sums to one. The rotation trace formula gives $\cos \theta =(\mathrm{tr}R-1)/2=\sqrt{3}/2$; the skew part selects a positive angle $\theta =\pi /6$ about the unit axis $(1,1,1{)}^T/\sqrt{3}$.

The original triangle has edge vectors $b-a=(-2,2,0{)}^T$ and $c-a=(-2,0,2{)}^T$. Its area is half the Euclidean norm of their cross product:The rotated vertices are twice the three columns of $R$:A rotation matrix preserves lengths and the angle between the two edge vectors, so it preserves their parallelogram area. More explicitly, $(Ru)\times (Rv)=R(u\times v)$ for a proper rotation matrix, and $R(1,1,1{)}^T=(1,1,1{)}^T$, so the rotated edge cross product is again $(4,4,4{)}^T$. Therefore