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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.

Using the ordered orthonormal basis chosen in (c), the column matrix isOther signs or another orthonormal choice within the repeated eigenspace are equally valid, provided the same choice is retained in the remaining parts.

Collect the coefficients of the three unknowns in rows. The linear system has matrix formThe next two parts solve this same system by the two requested methods.

The characteristic polynomial is $(1-\lambda )(4-\lambda )-4=\lambda (\lambda -5)$. Solving each eigenvalue equation gives the eigenvalues and eigenvectorsFor unit-length representatives divide each displayed direction by $\sqrt{5}$. The two eigendirections are orthogonal, consistently with the matrix being a symmetric matrix.

Use the order-reversal rule for the transpose of a product:Therefore the product of two orthogonal matrices is orthogonal. The argument does not require $A$ and $B$ to commute.

To invert the matrix, row-reduce the augmented matrix $(M\mid I)$. Subtracting row one from rows two and three gives left rows $(0,-4,1)$ and $(0,0,2)$, so the pivots are nonzero and back-substitution is possible. The resulting matrix inverse isMultiplying by $M$ verifies the identity matrix. This inverse is not $M^T$; indeedThus $M$ is not orthogonal.

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.

Collect the coefficients of the unknowns into the matrix and the constants into a column. The requested linear system isThe unknown coordinate $y$ here is a scalar component; the right-hand column is the vector denoted by the question's matrix-form notation.

The vectors $(1,1)$ and $(1,-1)$ are eigenvectors, with respective eigenvalues $\cosh \varphi +\sinh \varphi$ and $\cosh \varphi -\sinh \varphi$. The exponential definitions reduce these toAt $\varphi =0$ they coincide and the matrix is the identity.

A row's squared Euclidean norm is $[(1+\sqrt{3}{)}^2+1+(1-\sqrt{3}{)}^2]/6=3/2$, and the dot product of the two rows is $[2(1-3)+1]/6=-1/2$. Thus the row Gram matrix isFor the column Gram matrix, the first and third squared column norms are $4/3$, the middle one is $1/3$, and the respective cross-products are $1/3,-2/3,1/3$. HenceThe two products are square matrices of different sizes; their unequal dimensions are consistent with $A$ being rectangular.

For a matrix with equal diagonal entries, the symmetric and antisymmetric directions $(1,1)$ and $(1,-1)$ are eigenvectors. Direct multiplication of $A{A}^{\mathsf{T}}$ gives eigenvalues $3/2-1/2=1$ and $3/2+1/2=2$. Dividing each direction by its Euclidean norm givesThe two normalized eigenvectors are orthogonal; either may independently be replaced by its negative.

In $A^{\mathsf{T}}A$, every row sums to one, so $(1,1,1)$ is an eigenvector with eigenvalue $1$. The antisymmetric direction $(1,0,-1)$ gives eigenvalue $2$. A direction orthogonal to both is $(1,-2,1)$, and direct multiplication shows that it lies in the kernel of a linear map. Normalizing all three gives the complete systemThey form an orthonormal basis of ${\mathbb{R}}^3$. The extra zero eigenvalue reflects the one-dimensional kernel of a linear map for this rank-two rectangular matrix. The question's symbols ${\sigma }_m$ denote the eigenvalues of $A^{\mathsf{T}}A$; the nonzero singular values of $A$ are their square roots, namely $1$ and $\sqrt{2}$.

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).

Reading off coefficients gives the matrix equation

Multiplication of a matrix by a column vector forms the linear combination of its columns with coefficients from that vector. Hereso $A\mathbf{r}=x{\mathbf{c}}_1+y{\mathbf{c}}_2+z{\mathbf{c}}_3$. Equating this expression with $\mathbf{b}$ reproduces the original three linear equations.

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.

Successive substitution in the matrix recurrence relation gives $V_k=V_0{Q}^k$. In the matrix power of $Q=ED{E}^{-1}$, every adjacent $E^{-1}E$ cancels, so $Q^k=E{D}^k{E}^{-1}$. ConsequentlyHere $D^k=\mathrm{diag}({\lambda }_1^k,\dots ,{\lambda }_n^k)$ for $k\ge 1$, and $D^0=I$. The formula also gives $V_0$ correctly at $k=0$.

Put $d=c-1$. The matrix is real and symmetric, so orthogonality means $A^{\mathsf{T}}A=A^2=I$. Multiplication gives the scalar conditionsIf $d=0$, then $c=1$ and $a=\pm 1$; if $a=0$, then $d=\pm 1$ and $c=0$ or $2$. In both cases $b=\pm 1$. The eight parameter triples are thereforeThese conditions are sufficient as well as necessary, since they make every entry of $A^{\mathsf{T}}A$ equal to the corresponding entry of the identity matrix.

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

For any real square matrix $N$, the symmetric and antisymmetric parts of a matrix give $N=(N+N^T)/2+(N-N^T)/2$, whose summands are respectively a symmetric matrix and an antisymmetric matrix. Here it givesTransposing the first summand leaves it unchanged; transposing the second changes its sign. Entrywise addition recovers the original matrix. The decomposition is unique because a matrix that is both symmetric and antisymmetric must be zero.