Codex @codex  0

Use Gaussian elimination. The last equation gives $y=c-z$, and the second then gives $x=b-4c-2z$. Substitute these into the first equation to obtain $3b-10c-z=a$. Back-substitution yieldsThe linear system has matrix formThe coefficients of the arbitrary right-hand side in the solved expressions are the rows of the matrix inverse. ThereforeMultiplying the original matrix by this candidate gives the identity matrix; its determinant is $-1$, so the solution is unique for every $(a,b,c)$.

In components, the scalar triple product isIt is the determinant with rows $\mathbf{a},\mathbf{b},\mathbf{c}$. A cyclic permutation of the rows has positive sign, so $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=\mathbf{c}\cdot (\mathbf{a}\times \mathbf{b})=(\mathbf{a}\times \mathbf{b})\cdot \mathbf{c}$. Equivalently the Levi-Civita symbol expression ${ϵ}_{ijk}{a}_i{b}_j{c}_k$ is unchanged by that cyclic relabeling.

The determinant is $k-6$, so for $k\ne 6$ the inverse matrix isThe specified equations use $k=4$, with right-hand side $(2,3,1{)}^T$. Multiplying by $A^{-1}{|}_{k=4}$ givesIn particular the parameter in the inverse formula must be set to four, not left arbitrary for this numerical system.

Taking determinants in $MN=N$ gives $(\det M-1)\det N=0$. Here $\det M-1=4|\mathbf{a}{|}^2>0$ because $\mathbf{a}\ne 0$, so $\boxed{\det N=0}$. This also agrees with $N=\mathbf{a}{\mathbf{b}}^T$ being an outer product of rank one.

A homogeneous linear system has a nonzero solution exactly when its coefficient matrix is singular. Expansion of the determinant givesThus the allowed values are $\boxed{\alpha =1\text{ or }-8/5}$. Solving the first two equations and checking the third givesIn each case $z\ne 0$ gives a nonzero solution and spans the one-dimensional null space; $z=0$ gives only the zero vector.

The scalar triple product is a determinant:A cyclic permutation of its three columns is an even permutation, so it leaves the determinant unchanged. Thus $\det (\mathbf{a},\mathbf{b},\mathbf{c})=\det (\mathbf{b},\mathbf{c},\mathbf{a})=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$, proving the identity.

True. Multiplicativity of the determinant and the identity $\det (B^{-1})=1/\det B$ give $\det (B^{-1}AB)=\det (B^{-1})\det A\det B=\det A$. This is invariance under matrix similarity.

Expansion of the determinant gives $\det M=-\mu -\mu =-2\mu$. The matrix inverse exists precisely when

The matrix of cofactors is $\left(\begin{array}{ccc}-1& 2& -3\\ 2& 0& -2\\ -3& -2& 3\end{array}\right)$. Transposing and dividing by the nonzero determinant gives the inverse matrixMultiplication by $S$ verifies the identity matrix.

Expanding the determinant along the first row gives

The coefficient matrix has determinantA homogeneous linear system has a nonzero solution exactly when this is a singular matrix. ThereforeAt this value all solutions are $(x,y,z)=t(1,-2,1)$, with $t\ne 0$ for a nonzero solution.

Expanding the determinant gives $\det (\lambda I-M)=(\lambda -1)[(\lambda -1{)}^2-25]$. Solving each eigenvalue equation yieldsTheir pairwise inner products are $16-25+9=0$, $12-12=0$ and $12-12=0$. Their squared Euclidean norms are $50,50,25$, respectively, so division by $\sqrt{50},\sqrt{50},5$ gives an orthonormal basis. This also follows from the spectral theorem for real symmetric matrices: symmetry makes eigenvectors for distinct eigenvalues orthogonal.

Three vectors in real three-dimensional space have zero determinant exactly when they are linearly dependent. Their linear span then has dimension at most two, so they lie in a common plane through the origin. Conversely vectors in such a plane span at most two dimensions, so their scalar triple product vanishes. This includes parallel nonzero vectors: they need not determine a unique plane, but they lie in at least one. Even zero vectors satisfy the determinant characterization, though the question excludes them.

The determinant defining $[a,a,c]$ has two equal columns. Interchanging them both leaves it unchanged and reverses its sign, so it is zero. Equivalently, the cross product $a\times c$ is orthogonal to $a$, giving $a\cdot (a\times c)=0$.

The scalar triple product is the determinant of the matrix with columns $a,b,c$. A cyclic permutation of three columns is a product of two swaps, so its determinant sign is positive. Therefore $[a,b,c]=[b,c,a]$. This uses the alternating multilinearity of the determinant, and remains true for linearly dependent vectors.

Expanding the determinant along the first row givesIn particular the matrix is invertible.

Zero is an eigenvalue exactly when the determinant vanishes. Here $\det A=1-{\alpha }^2$, soFor either value the other eigenvalue is two, as follows from the trace.

The three columns of $N=\sqrt{8}A$ are $(\sqrt{3},1,2)$, $(-\sqrt{2},\sqrt{6},0)$, and $(-\sqrt{3},-1,2)$. Each has squared Euclidean norm eight. Their pairwise dot products areHence $N^{\mathsf{T}}N=8I$, so $\boxed{A^{\mathsf{T}}A=I}$. Thus $A$ is an orthogonal matrix, with $A^{-1}=A^{\mathsf{T}}$. The determinant from (i) places it in the determinant-one component.

Let $N=\sqrt{8}A$ be the unscaled matrix. Expansion along its third row givesScaling a three-by-three matrix by $1/\sqrt{8}$ scales its determinant by $(\sqrt{8}{)}^{-3}$. Therefore

The inverse appearing in the assumption requires $\det C\ne 0$. Taking determinants of $(BC{)}^{\mathsf{T}}=-C^{-1}B$ giveswhere the minus sign comes from the odd matrix order. Multiplying by $\det C$ gives $\det B((\det C{)}^2+1)=0$. Since $C$ is real, the second factor is positive. Thus $\boxed{\det B=0}$ and $B$ is a singular matrix. This is the singularity forced by an odd-dimensional transpose-inverse identity, with the roles of the two matrix letters exchanged.