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Interchange the diagonal entries, negate the off-diagonal entries, and divide by the nonzero determinant. The matrix inverse isMultiplication by $A$ gives the identity matrix, since each diagonal product sum is $25-9=16$ and each off-diagonal product sum is zero.

At an intersection, the two position vectors agree, so $a-b=tn-sm$. The scalar triple product with $m,n$ is thereforesince a determinant with repeated columns vanishes. Geometrically the displacement of the two anchor points must lie in the same plane as the two line directions.

This is a necessary condition. If $m,n$ are not parallel, it is also sufficient, because their span is exactly the plane perpendicular to $m\times n$. For parallel directions the triple product vanishes automatically, even for distinct lines, so the condition alone cannot guarantee intersection.

Since $f>0$, a level $f=c$ can occur only for $c>0$. Taking its natural logarithm gives $xy=\ln c$. Thus $c>1$ produces rectangular hyperbolas in the first and third quadrants, $0<c<1$ produces them in the second and fourth quadrants, and $c=1$ consists of both coordinate axes.

The gradient $e^{xy}(y,x)$ vanishes only at $(0,0)$, where $f=1$. The Hessian matrix there is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, with negative determinant. Equivalently, $f(x,x)>1$ while $f(x,-x)<1$ for nonzero small $x$. Hence the only stationary point is a saddle point at $(0,0,1)$ on the graph; its location in the contour plane is the origin.

The characteristic polynomial, using the convention $\det (\lambda I-A)$, isIts zeros give the eigenvalues, including algebraic multiplicity:Their product is $45$ and their sum is $-1$, agreeing with the determinant and trace of the matrix.

Expand the determinant along the first row:Thus $\boxed{\det A=45}$. The signs in the middle cofactor account for its positive contribution.

Take determinants of $(CB{)}^{\mathsf{T}}=-B^{-1}C$. In order three, multiplication of a matrix by minus one changes its determinant's sign. HenceThe assumption $\det (B^{\mathsf{T}})\ne 0$ is the same as $\det B\ne 0$, so multiplication by $\det B$ givesSince the matrices are real, the first factor is strictly positive. Thus $\boxed{\det C=0}$, proving that $C$ is a singular matrix. This is singularity forced by an odd-dimensional transpose-inverse identity; no orthogonality of $B$ was used.

Collecting the coefficients gives the coefficient matrixA square homogeneous linear system has a nonzero solution precisely when its matrix is a singular matrix. Expanding the determinant down the first column,Hence the necessary and sufficient condition is $a=0$ or $b=c$ or $b=-c$. A zero determinant gives a nontrivial kernel of a linear map; otherwise multiplication by the matrix inverse would force the zero solution.

Apply the vector triple product identity and the two plane constraints:Put $d=b\times c\ne 0$. If $a$ is one intersection point, subtraction gives $(r-a)\times d=0$, so $r-a$ is parallel to $d$. Conversely, adding any multiple of $d$ preserves both plane constraints because $d$ is orthogonal to both normal vectors. The whole intersection is thereforeSuch an $a$ exists: choose $a=\alpha b+\beta c$ and solve the two-by-two linear system with determinant $|b{|}^2|c{|}^2-(b\cdot c{)}^2=|b\times c{|}^2>0$. This also proves sufficiency rather than assuming a nonempty intersection.

The trace is the sum of the eigenvalues, counting multiplicity, while the determinant is their product:A direct expansion along the middle row checks the latter: $(-2)(1-9)=16$.

The determinant is $2\cdot 4-3\cdot 1=5\ne 0$, so the matrix inverse exists. Using the two-by-two adjugate matrix formula givesMultiplication on either side gives the identity matrix, which also checks the signs of the off-diagonal entries.

Choose Cramer's rule, the permitted alternative to finding an inverse. The determinant isReplacing each column by the right-hand side givesConsequentlyThe nonzero determinant also proves uniqueness.

For a classical solution take $F,G$ twice differentiable, so that the second derivatives exist. Direct use of the chain rule giveswith the two profile arguments as specified. Thus $4y_{tt}=25y_{xx}$.

To prove generality, introduce characteristic coordinates $\xi =2x+5t$, $\eta =2x-5t$, whose Jacobian determinant is $-20\ne 0$. For $y=U(\xi ,\eta )$ the differential operator isThe equation becomes $U_{\xi \eta }=0$. Integrating once in each variable yieldsThis is the countertravelling decomposition of the one-dimensional wave equation, with wave speed $5/2$. Once differentiability alone is not sufficient for a classical second-order equation; the displayed argument uses the required second derivatives.

Direct multiplication gives $A^2=-I/4$. Therefore $A(-4A)=I=(-4A)A$, yieldingThe nonzero determinant ensures that this two-sided inverse is unique as a matrix inverse.

A real two-dimensional antisymmetric matrix has the form $A=\left(\begin{array}{cc}0& s\\ -s& 0\end{array}\right)$. Its determinant is $s^2=1/4$, and the specified positive upper-right entry selects $s=1/2$. Thus

Expanding the determinant along the first row givesThe matrix trace is the sum of the diagonal entries, $\mathrm{tr}A=1+4+1=6$. Since the determinant is the product of the eigenvalues, at least one eigenvalue is zero. Thus $A$ is singular; the determinant alone does not specify the two other eigenvalues.

The symmetry between the first and third coordinates makes $(1,0,-1{)}^{\mathsf{T}}$ an eigenvector, with eigenvalue $-6$. In the remaining invariant plane, write a vector as $(s,t,s{)}^{\mathsf{T}}$. Its image is $(8s-4t,-8s+4t,8s-4t{)}^{\mathsf{T}}$, so the reduced eigenvalue equation has determinantThe corresponding directions have $t=2s$ for $\lambda =0$ and $t=-s$ for $\lambda =12$. Normalizing givesTheir squared unnormalized norms are $2,6,3$. The pairwise unnormalized dot products are $1-1=0$, $1-1=0$, and $1-2+1=0$, so these unit eigenvectors are mutually orthogonal. The eigenvalues sum to $6$ and have product zero, agreeing with part (a).

The gradient equations factor asFor $y=0$ they give $x=\pm 5$. Otherwise $y=4x/3$, which gives $x=\pm 3$ and the matching $y=\pm 4$. Thus there are exactly four stationary points.

The Hessian matrix isAt $(5,0)$ it is $60I$, and at $(-5,0)$ it is $-60I$. At $(3,4)$ and $(-3,-4)$ its determinant is $-3600<0$. The second-derivative test therefore yieldsThe extrema are local: along $y=0$ the cubic is unbounded both above and below.

At a regular constrained stationary point, the Lagrange multiplier equations are $f_x=\lambda g_x$, $f_y=\lambda g_y$. Eliminating the multiplier givesThus the required Lagrange multiplier determinant condition in two variables isIf the constraint gradient vanishes, this determinant is zero automatically, although the regular multiplier theorem cannot be applied. The condition is necessary and is not sufficient to classify the constrained point. This addresses the original unnumbered task before part (a).

The same elimination at general $a$ gives the rowsThe coefficient determinant is $(a-3)(a-1)$. Only $a=1$ or $a=3$ can therefore fail to have a unique solution. At $a=3$, the second row reads $0=-2$, so there is no solution. At $a=1$, the last two rows are proportional and both specify $z=1$, leaving one free unknown; hence there are infinitely many solutions. Every other real $a$ gives a unique solution.

At $a=2$, eliminate the first column by $R_2\leftarrow R_2-2R_1$ and $R_3\leftarrow R_3-2R_1$ in the augmented matrix. The last two rows give $-z=-2$ and $-y+z=-1$. Hence $z=2$, $y=3$, and the first row then gives $x=-2$. ThusThe coefficient determinant at this parameter is nonzero, so this solution is unique.