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Expanding the determinant along the first row givesThe eigenvalues and normalized spatial eigenvectors, in increasing frequency order, areFor each mode both signs of ${\omega }_j$ give the same cosine displacement. An arbitrary overall mode amplitude remains free. These vectors are the $N=3$ instances of the sine eigenvectors of the fixed-end spring chain.

The inverse fails to exist exactly when at least one eigenvalue is zero. In that case a nonzero eigenvector lies in the kernel of a linear map, so $A$ is not one-to-one. Equivalently, its determinant ${\prod }_n{\lambda }_n$ is zero. Conversely, if every eigenvalue is nonzero, the preceding construction gives an inverse.

Suppress the held-fixed subscripts only within this matrix display. Collecting the preceding operator coefficients givesIf $J=\partial (x,y,z)/\partial (W,\theta ,\varphi )$ uses the usual convention of coordinate functions as rows, the displayed coefficient matrix is $J^{\mathsf{T}}$. Its determinant is $W^2{\sec }^2\theta \tan \theta >0$ on the given domain, so the inverse operator transformation exists everywhere there.

The middle row is $-1/2$ times the sum of the first and third rows, so the rows are linearly dependent and the determinant is zero. The matrix trace is the sum of diagonal entries, $1+4+1=6$. A transpose preserves the determinant, while the determinant of a product is the product of its determinants. Consequently

The matrix is symmetric under interchange of the first and third coordinates. For an antisymmetric vector $(1,0,-1)$, direct multiplication gives $A(1,0,-1{)}^{\mathsf{T}}=-6(1,0,-1{)}^{\mathsf{T}}$. For a symmetric vector $(p,q,p)$, the linear transformation reduces toThe reduced characteristic polynomial is $\lambda (\lambda -12)$. At $\lambda =0$, its equations give $q=2p$; at $\lambda =12$, they give $q=-p$. Normalizing the three eigenvectors givesTheir norms are one. Their unnormalized pairwise dot products are $1-1=0$, $1-1=0$ and $1-2+1=0$, proving mutual orthogonality. The eigenvalues sum to six and have product zero, agreeing with the matrix trace and determinant from part (a).

Subtract twice the second row from the third row when computing the determinant. The last row becomes $(0,0,\mu -6)$, soIts zeros give exactly the parameter values with a nontrivial null space:

For $\mathbf{b}=0$, a nonzero coefficient vector $(x,y,z)$ with $x{\mathbf{c}}_1+y{\mathbf{c}}_2+z{\mathbf{c}}_3=0$ is precisely a linear dependence of the three columns. Three vectors in three dimensions are linearly dependent exactly when the determinant of their column matrix, or their scalar triple product, is zero. ThusFor this square homogeneous linear system, the condition is also sufficient.

The position vectors end at the midpoints of $BC$, $CA$ and $AB$, respectively. The drawing places $E$, $F$ and $G$ on those edges and shows the arrows from the origin. The volume of their parallelepiped is the absolute scalar triple product:Equivalently, this is the absolute determinant of the matrix whose columns are the three vectors. Since it is nonzero, the vectors are linearly independent. Three linearly independent vectors in the three-dimensional vector space ${\mathbb{R}}^3$ form a basis.

Let $D_{\beta }=\mathrm{diag}({\beta }_1,\dots ,{\beta }_n)$ and $D_{\gamma }=\mathrm{diag}({\gamma }_1,\dots ,{\gamma }_n)$. The specified entries mean $C=D_{\beta }A{D}_{\gamma }$. Multiplication of determinants therefore givesThis is also the direct row- and column-scaling rule, and it includes zero scaling factors.

Multiplying an $n\times n$ matrix by $\alpha$ scales every row, giving a factor ${\alpha }^n$ in its determinant. Also $|A^{-1}|=1/|A|$, since $A{A}^{-1}=I$. HenceOnly $A$ needs to be invertible; the result remains valid when $\alpha =0$.

The determinant is multiplicative under matrix multiplication:This also holds when $B$ is singular, in which case both sides vanish.

The eigenvector equations $Q{\mathbf{e}}_i={\lambda }_i{\mathbf{e}}_i$ combine column by column into $QE=ED$. Since $|E|\ne 0$, multiplication by the inverse matrix gives the diagonalizationThe nonzero determinant of $E$ ensures that its columns form an eigenbasis.

This is a generalized eigenvalue problem. A nonzero vector lies in the kernel of $A-\lambda B$ precisely when its determinant vanishes:HenceThe corresponding eigenvectors can be taken as $(1,-1{)}^T$ and $(1,1{)}^T$, respectively, which also checks that both values give nontrivial solutions.

A three-dimensional rotation matrix is an orthogonal matrix with determinant $+1$. Here $\det A=b(a^2-d^2)$. For the branch $d=0$, the determinant condition gives $b=1$; for the branch $a=0$, it gives $b=-1$. HenceThe first is the identity rotation; the remaining three are half-turns. Their fixed axes are respectively the $y$ axis, the direction $(1,0,-1)$ and the direction $(1,0,1)$. The identity is included under the usual rotation matrix convention.

A square homogeneous linear system has a nonzero solution exactly when its determinant vanishes. Expanding along the second row givesConsequentlyFor an explicit check, the null space contains $(3,-1,1{)}^{\mathsf{T}}$ at $\lambda =3$ and $(3,1,-1{)}^{\mathsf{T}}$ at $\lambda =-3$.

Expanding the determinant along the first row givesUnder the matrix trace condition $a^2=3$, this simplifies to $\det M=b(3-b)$. A square matrix is invertible exactly when its determinant is nonzero. Thus

Among the two orthogonal matrices from part (d)(i), $d=-1$ gives $L=-I$, a rotation matrix through angle $\pi$. For $d=1$, the determinant is $-1$, so the matrix reverses orientation and is a reflection rather than a rotation:It is the reflection matrix across the $y$ axis.

For $\lambda \ne \pm 2$, the reduced equation gives $y=(3-\lambda )/(4-{\lambda }^2)$. Back substitution into $x=1-\lambda y$ and $z=1-4y$ gives all solutions:The nonzero determinant proves that there are no other solutions at these parameter values.

The determinant is $\det L=4-{\lambda }^2$, so there is exactly one solution whenever $\lambda \ne \pm 2$. A zero determinant does not alone distinguish inconsistency from infinitely many solutions, so check those values separately.

Eliminate $x$ and $z$ using $x=1-\lambda y$ and $z=1-4y$. The remaining equation isFor $\lambda =2$ this reads $0=1$, and for $\lambda =-2$ it reads $0=5$. Both are inconsistent. Consequentlyand each permitted value has a unique solution.

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.