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Collecting the coefficients in the given equation order yields the system of linear equations in matrix form

The matrix trace and determinant of the matrices in part (i) areSince $\mathrm{Tr}X=0$ and $\det X=2$, the extra conditions force $\alpha =0$ and ${\beta }^2=1$. HenceBoth choices are real, commute with $X$, and satisfy the required trace and determinant conditions.

Let $Y=\left(\begin{array}{cc}p& q\\ r& s\end{array}\right)$. The commutativity condition givesEquality is equivalent to $s=p$ and $q=-2r$. ThereforeEach displayed matrix commutes with $X$, since it is a polynomial in $X$. The centralizer of a nonscalar two by two matrix explains why this two-parameter form is typical.

The two orthogonal directions $(1,1{)}^T$ and $(1,-1{)}^T$ are preserved by this symmetric matrix. Direct multiplication, using the hyperbolic function identities $\cosh u\pm \sinh u=e^{\pm u}$, gives the eigenvalue–eigenvector pairsEach eigenvector can be multiplied by any nonzero scalar. For $u\ne 0$ these give the two one-dimensional eigenspaces. When $u=0$, the matrix is the identity: the only eigenvalue is $1$, with multiplicity two, and every nonzero vector is an eigenvector.

The vector $(1,-1,0{)}^T$ is an eigenvector with eigenvalue $4$. The complementary subspace of vectors with first two components equal is preserved by $A$: a vector $(s,s,t{)}^T$ maps to $(4s+2t,4s+2t,4s+2t{)}^T$. There, the eigenvalue equations reduce to the $2\times 2$ matrix $\left(\begin{array}{cc}4& 2\\ 4& 2\end{array}\right)$, whose characteristic polynomial is $\lambda (\lambda -6)$. For $\lambda =6$, $t=s$; for $\lambda =0$, $t=-2s$.

Normalising by the Euclidean norm, a complete orthonormal basis of eigenvectors isSign changes of any of these eigenvectors are equally valid. The eigenvalues also check part (a): their sum is $10$ and the sum of their squares is $52$.

Three vectors form a basis of three-dimensional space exactly when their scalar triple product is nonzero:It is the determinant of the matrix with these columns, so this condition is equivalent to linear independence and spanning the whole space.

Take the dot product of the expansion of $\mathbf{y}$ with $\mathbf{v}\times \mathbf{w}$. The terms containing $\mathbf{v}$ and $\mathbf{w}$ vanish, leaving $\mathbf{y}\cdot (\mathbf{v}\times \mathbf{w})=\alpha \Delta$. Cycling the same argument yieldsThe three cross-product numerators use the reciprocal basis.

In the nonsingular case the unique solution isGeometrically this is a single selected point of the straight line through $\mathbf{a}$ parallel to $\mathbf{c}$.

When $\mathbf{b}\cdot \mathbf{c}=1$ and $\mathbf{b}\cdot \mathbf{a}=0$, the solution set is that entire straight line, $\mathbf{a}+\lambda \mathbf{c}$. The remaining case has empty solution set. In matrix terms the equation is $(I-\mathbf{c}{\mathbf{b}}^T)\mathbf{x}=\mathbf{a}$, a rank-one matrix update of the identity matrix.

The displayed matrix is not orthogonal: its first column has squared Euclidean norm three, whereas every column of an orthogonal matrix must have squared length one.

For the general assertion, put $C=AB$. In the Einstein summation convention, orthogonality gives $A_{ki}{A}_{kj}={\delta }_{ij}$ and $B_{ki}{B}_{kj}={\delta }_{ij}$. The entries of $C$ satisfyThis is exactly $C^{T}C=I$, proving that the matrix product $AB$ is orthogonal. The Kronecker delta contracts the two middle indices without changing the multiplication order of the matrices.

For $a\ne 1,5/2$, the matrix is invertible, so there is exactly one solution. To exhibit it, subtract the first linear equation from the third to obtain $(a-1)x=2$, then subtract twice the second from the first to obtain $-x+(5-2a)z=-3$. Back-substitution givesAt $a=1$, the first and third linear equations have identical left-hand sides but right-hand sides one and three, so there is no solution.

At $a=5/2$, subtracting the first from the third forces $x=4/3$, while the first minus twice the second forces $x=3$. These contradict each other, so there is again no solution. Thus there is one solution for $a\ne 1,5/2$, no solution for either exceptional value, and no parameter value with infinitely many solutions.

The matrix form is

The coefficient matrix has determinant $-1$, so it is an invertible matrix. Its homogeneous linear system therefore has only the zero solution:One can also subtract the first equation from the second to get $z=0$, then subtract the first from the third to get $y+2z=0$, giving $y=0$ and finally $x=0$. Thus there are no nonzero solutions.

Substituting $\beta =1$ and $\gamma =1-\alpha$ into the matrix givesIts row sums are $4,4,6$, checking that $A_{\alpha }\mathbf{c}=\mathbf{b}$ for every real $\alpha$.

Differentiate the first ordinary differential equation and substitute the second:The original first equation also gives $bv=u^{\prime }-au$, soTherefore the requested coefficients areThey are the trace and determinant of the coefficient matrix. This derivation never divides by $b$ and is valid even if $b=0$. In that degenerate case the scalar second-order equation is a necessary consequence, while the original constraint $u^{\prime }=au$ must also be retained when reconstructing the coupled system.

Suppose $B\mathbf{x}=0$. The given quadratic form bound then givesso $\mathbf{x}=0$. Thus the kernel of $B$ is trivial. A square matrix with trivial kernel is an invertible matrix, proving the assertion. No symmetry assumption on $B$ is needed.

The first coordinate direction is already an eigenvector of $A$ with eigenvalue $4$. On the last two coordinates, the matrix is $\left(\begin{array}{cc}2& 4\\ 4& 2\end{array}\right)$, with characteristic polynomialThus the three eigenvalues are $4,6,-2$. For $\lambda =6$, the last two components are equal; for $\lambda =-2$, they are opposite. Normalising by the Euclidean norm gives the complete listEach sign-reversed vector is an equally valid choice. The three displayed eigenvectors are mutually orthogonal, as expected for a real symmetric matrix with distinct eigenvalues.

The coefficient matrix, unknown vector and right-hand side are

The reversed product is the dot product, a scalar represented by a $1\times 1$ matrix:

The outer product of the column vector with its matrix transpose is the $2\times 2$ matrix

The matrix issoHere $\Vert \mathbf{p}{\Vert }^2=2$, so $P=I-2\hat{\mathbf{p}}{\hat{\mathbf{p}}}^T$ with $\hat{\mathbf{p}}=\mathbf{p}/\sqrt{2}$. This is a Householder transformation: reflection in the plane $x_1+x_2=0$. The component parallel to the normal vector $\mathbf{p}$ changes sign, while the perpendicular component is fixed.

The second matrix acts byApplying $P$ from part (e) then givesThe Householder transformation $Q$ reflects in the plane $x_1-x_2=0$. Its normal vector $\mathbf{q}$ is perpendicular to $\mathbf{p}$, so the two reflecting planes meet at a right angle along the $x_3$ axis. Their composition is a rotation through $\pi$ about the $x_3$ axis, as is also immediate from the displayed rotation matrix.