Use Gaussian elimination with partial pivoting, choosing a largest absolute entry in the uneliminated part of each pivot column. The first pivot is two in row one, with multipliers $1/2,1/2,1/5$ for rows two, three and four. The second-column candidates are then $-1,0,-2/5$, so row two stays as pivot and the remaining nonzero multiplier is $2/5$ for row four. The augmented matrix at this stage isThe third pivot is three, so swap rows three and four. The final multiplier is $-1/4$, leaving last pivot $-2$. Carrying that swap through the previously stored multipliers givesThe transformed right side is $(3,1,0,0{)}^T$. Back substitution gives $z=0$, $y=0$, $x=-1$ and $w=1$, hence $\boxed{(w,x,y,z)=(1,-1,0,0)}$.

For the eigenvalue claim, put $S=P^{T}L$. It is invertible because a permutation matrix and a unit lower triangular matrix are invertible. The factorization gives $A=SU$, soSimilar matrices have the same characteristic polynomial and therefore the same eigenvalues with algebraic multiplicities. This proves the claim without changing the prescribed elimination method.

The normal equations for linear least squares are $X^{T}X\hat{\theta }=X^{T}Y$. Under full column rank, $\hat{\theta }=(X^{T}X{)}^{-1}{X}^{T}Y$. Since $\mathrm{Cov}(Y)={\sigma }^{2}I$,Here $\bar{x}=0$ and $\bar{z}=2$, so the design matrix isIts columns are pairwise orthogonal and have squared lengths $6,28,4$. Consequently $X^{T}X=\mathrm{diag}(6,28,4)$, givingBoth estimators are linear functions of the Jointly Gaussian observation vector, so they are Jointly Gaussian. Zero covariance then implies independence. This final implication uses the normal error model, whereas the variance calculation itself needs only its second moments.

The Markov property says that conditional on the present state, the next-state law does not depend on the earlier history. For this time-homogeneous chain, $\mathbb{P}(X_{n+1}=j\mid X_0=i_0,\dots ,X_n=i)=p_{ij}$ whenever the conditioning event has positive probability.

States communicate when each is accessible from the other by a finite positive-probability path; the equivalence classes are the communicating classes. A class is closed if no one-step transition leaves it, and open otherwise. Probability nonnegativity adds the implicit restriction $p\ge 0$ to the printed upper bound.

At $p=0$, every state is absorbing, so the classes are the singletons, all closed. For $0<p\le 1/2$, every neighbouring pair communicates, so the whole state space is one closed class, making the chain irreducible. There are no open classes in either case.

For $m=3$ and $p>0$, detailed balance along the path gives $2p{\pi }_0=p{\pi }_1$, ${\pi }_1={\pi }_2$, and $p{\pi }_2=2p{\pi }_3$. After normalization,For $0<p<1/2$, the self-loops have positive probability; the finite irreducible chain is aperiodic, so ${\mathbb{P}}_i(X_n=j)\to {\pi }_j$ for all fixed starting states and targets. For $p=1/2$, every step changes parity. The chain has period two, and for each fixed $i,j$ one parity of times has probability zero while ${\pi }_j>0$. Thus the stated convergence from a fixed starting state holds exactly when $0<p<1/2$. This is the stationary law of a reflecting nearest-neighbor walk on a finite path.

The inverse transformation is $x=2u+v$, $y=v$, with absolute Jacobian determinant two. The original positivity constraints become $v>0$ and $2u+v>0$. ThusIntegrating out $v$ givesThis is the centred Laplace distribution with scale one.

Gauss law states that for outward-oriented boundary $S=\partial V$,For a continuously differentiable field and continuous density, the divergence theorem turns the left side into ${\int }_V\nabla \cdot \mathbf{E}$. Thus the differential law implies the integral law. Conversely, if the integral law holds for every suitable volume, the integral of $\nabla \cdot \mathbf{E}-\rho /{ϵ}_0$ vanishes on every small ball. A nonzero continuous value would keep one strict sign in some small ball, a contradiction. Hence $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$. Weaker regularity gives the corresponding almost-everywhere or distributional formulation.

For the radial field, the flux on a radius-$r$ sphere is $4\pi r^{2}E(r)$, while its enclosed charge is $4\pi {\int }_0^r{s}^2\rho (s)ds$. ThereforeThis uses the stated density as the full charge distribution, with no additional point charge at the origin. Since it vanishes beyond $a$, the total charge isand for $r>a$ the field is $\boxed{\mathbf{E}=Q{\mathbf{e}}_r/(4\pi {ϵ}_0{r}^2)}$.

Put $A(r)=r^2/(r-2m{)}^2$ and $B(r)=r^3/(r-2m)$, so $L=A{\dot{r}}^2+B{\dot{\phi }}^2$. Direct differentiation gives $A^{\prime }=-4mr/(r-2m{)}^3$ and $B^{\prime }=2r^2(r-3m)/(r-2m{)}^2$. The Euler-Lagrange equations are $2A\ddot{r}+A^{\prime }{\dot{r}}^2-B^{\prime }{\dot{\phi }}^2=0$ and $(d/dt)(2B\dot{\phi })=0$, equivalentlyThe cyclic angular coordinate and time independence give the two conserved quantitiesThe expression for $E$ equals $L$ here because $L$ is homogeneous quadratic in the velocities.

For a constant radius and nonzero angular motion, the radial equation requires $(r_{\ast }-3m){\dot{\phi }}^2=0$, so $\boxed{r_{\ast }=3m}$. This lies outside $2m$, and the angular equation then gives $\phi =\Omega t+{\phi }_0$, with any nonzero constant $\Omega$. This is the photon sphere radius in the stated coordinates. If both velocities are zero, the displayed variational equations also admit a stationary configuration at any radius; the moving circular photon solution is the nonzero-speed case just found.