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For a diagonal matrix, each diagonal entry separately obeys $d^2+2d-15=(d-3)(d+5)=0$. Hence all four possibilities areFor the second form, $X^2=I$ givesBecause $I,X$ are linearly independent and $b\ne 0$, the $X$ coefficient forces $a=-1$; the $I$ coefficient then gives $b^2=16$. ThusBoth constants are nonzero, as required.

A plane through the three points also contains the origin exactly when their position vectors are linearly dependent. The scalar triple product isThus $\boxed{\gamma =9}$. The resulting plane is $\boxed{x-2y+z=0}$, which contains all three points and the origin. The points are not collinear: the differences $(3,3,3)$ and $(\gamma -1,6,4)$ cannot be proportional, since their last two components would require $6=4$. Therefore the plane used in the argument is well defined.

Rearrange as $y^{\prime }+(x-1)y=x-1$ and put $w=y-1$. The resulting linear differential equation is $w^{\prime }+(x-1)w=0$, soThe initial condition gives $C=1$. Its graph is a Gaussian function above the horizontal asymptote $y=1$, symmetric about $x=1$, with its maximum $y=2$ there. It rises for $x<1$ and falls for $x>1$, since $y^{\prime }=-(x-1)e^{-(x-1{)}^2/2}$.

Choose the cylinder centre's initial position as the coordinate origin, with downhill motion to the right and $0\le \alpha <\pi /2$. For rolling without slipping, the clockwise angular speed is $\omega =v/r$. A point initially vertically above the axis therefore hasA different origin adds constant coordinates without changing the following velocity test. Setting $\theta =vt/r$ gives $\dot{x}/v=\cos \alpha +(a/r)\cos \theta$ and $\dot{y}/v=-\sin \alpha -(a/r)\sin \theta$.

For strictly upward motion, $\dot{x}=0$ and $\dot{y}>0$. The first condition fixes $\cos \theta =-(r/a)\cos \alpha$; choose the negative sine branch to maximize $\dot{y}$. Positivity then requiresAt $a=r$ the relevant point is momentarily stationary, not moving upward. Thus a material point inside or on an ordinary cylinder cannot move exactly vertically upward in this motion. The mathematical extension $a>r$ describes the looping trochoid of a point beyond the rolling radius, illustrated in the figure.

Let $R=|\alpha |$, $S=|\beta |$ and $d=|\mathbf{b}-\mathbf{a}|$. For distinct centres, the two spherical surfaces intersect exactly whenStrict inequalities give a nondegenerate circle; equality gives a single tangency point. If $d=0$, equal radii give coincident spheres and unequal radii give no intersection, so the circle-centre formula below does not apply.

Subtracting the squared sphere equations gives the radical planeThe centre of the circle of intersection of two spheres is the projection of either sphere centre onto this plane. Write $\mathbf{c}=\mathbf{a}+\ell (\mathbf{b}-\mathbf{a})/d$; substitution gives $\ell =(d^2+{\alpha }^2-{\beta }^2)/(2d)$ and henceThe circle radius squared is $R^2-{\ell }^2$, providing another check of the intersection conditions.

Its centre equals $\mathbf{a}$ precisely when $\ell =0$, or $\boxed{{\beta }^2={\alpha }^2+d^2}$. For a positive first radius, the intersection is then a great circle of the first sphere. A triangle from its centre to an intersection point and then to the second centre is right-angled at $\mathbf{a}$, explaining the Pythagorean theorem relation geometrically.

Expanding the determinant along its first row givesThe required factorization isEach factor is a symmetric polynomial. The second is $\frac{1}{2}[(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]$, making its permutation symmetry especially transparent. This factorization holds over both real and complex coefficients.

Let $P$ have columns $a,b,c$ and let $G$ have columns $A,B,C$. The cross-product formulas defining the reciprocal basis giveprovided the original scalar triple product $T=\det P$ is nonzero. Taking determinants yields the reciprocal basis involution identityWhen $T=0$, the displayed reciprocal vectors are undefined, so there is no additional zero-denominator case to evaluate.