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True for these square matrices. The determinant multiplication identity gives $\det (XY)=\det X\det Y$. Since $XY$ is invertible, its determinant is nonzero; both factors are therefore nonzero, so both matrices have matrix inverses.

Equivalently, $Y$ must be injective and $X$ must be surjective, and either property implies invertibility for an endomorphism of a finite-dimensional vector space. The equal square dimensions matter: a rectangular matrix and a right inverse can have identity product without being mutually invertible square matrices.

The characteristic polynomial is $\det (\lambda I-A)={\lambda }^2-\lambda -1$. Its distinct eigenvalues are $\alpha =(1+\sqrt{5})/2$ and $\beta =(1-\sqrt{5})/2$. For either eigenvalue $\lambda$, the vector $(\lambda ,1{)}^{\top }$ is an eigenvector, since $A(\lambda ,1{)}^{\top }=(\lambda +1,\lambda {)}^{\top }=\lambda (\lambda ,1{)}^{\top }$.

ChooseThe determinant of $S$ is $\alpha -\beta =\sqrt{5}\ne 0$, soIts columns being eigenvectors gives $AS=SD$, and left multiplication by $S^{-1}$ proves $S^{-1}AS=D$.

Replacing row one by $R_1-R_2-R_3-R_4$ does not change the determinant: adding a multiple of another row preserves a determinant. It produces first row $(-2,0,0,0)$, whose cofactor minor is the three-dimensional identity matrix. Thus the determinant is $-2$.

The columns are homogeneous coordinate vectors $(1,v_i{)}^{\top }$ of the four vertices. Subtracting the first column from each other column preserves the determinant and reduces its magnitude to the absolute determinant of the three edge vectors based at $v_1$. That absolute determinant is the parallelepiped volume. A tetrahedron occupies one sixth of that parallelepiped, so $V=|-2|/3!=1/3$ in cubic coordinate units.

A unit vector along the line is $d=(3,4{)}^{\top }/5$. Reflection preserves the orthogonal projection $d{d}^{\top }u$ onto that line and negates its perpendicular component. Therefore $Au=d{d}^{\top }u-(u-d{d}^{\top }u)$ andAs checks, $A(3,4{)}^{\top }=(3,4{)}^{\top }$, $A(-4,3{)}^{\top }=-(-4,3{)}^{\top }$, and $A^2=I$. Its determinant is minus one, as expected for a planar reflection matrix.

Let $f_j$ be the determinant of the leading $j\times j$ principal submatrix. Expansion along the last row gives an $a_j{f}_{j-1}$ term and a $c_{j-1}$ term with negative cofactor sign. In the latter minor, the final column has only $b_{j-1}$ nonzero, and deleting that row and column leaves the leading $(j-2)\times (j-2)$ principal submatrix. Thus the tridiagonal determinant recurrence isHere $f_0=1$ is the empty determinant convention, needed to include $f_2=a_1{a}_2-b_1{c}_1$. The derivation uses cofactor expansion, so the recurrence relation remains valid when coefficients or intermediate determinants vanish; it never divides by a pivot element.

Rearrange the identity intoTaking determinants gives $\det A\det (B-A-I)=1$. Hence $\det A\ne 0$, so $A$ is an invertible matrix, with $A^{-1}=B-A-I$. This uses that the matrices are finite and square: a right inverse then implies a two-sided inverse. Applying the identity to a vector in the kernel without first establishing commutation would not justify cancelling an $A$ on the wrong side. The argument is a case of invertibility forced by a polynomial product identity.

The stationary point equations give $y=x$ and $x^3=x$, hence $(0,0),(1,1),(-1,-1)$. At $(0,0)$ the Hessian matrix has negative determinant $-16$, so $(0,0)$ is a saddle point. At each of $(1,1)$ and $(-1,-1)$, the Hessian matrix has positive leading minors $12$ and $32$, proving a strict local minimum. More strongly,Thus $(1,1)$ and $(-1,-1)$ are exactly the global minima, with value $3$. There are no local maxima, since any differentiable interior local maximum would be a stationary point, and all three have been classified.

The determinant is invariant under transpose, while multiplying all $n$ rows by $-1$ multiplies it by $(-1{)}^n$. Thus, for odd integer $n$,Over the real numbers, this forces $\det A=0$, so $A$ is a singular matrix. This is the odd-dimensional skew-symmetric singularity theorem.

The interior grid is $x=1,2,3$, with fixed-end displacements $u_0=u_4=0$. HenceUse the monic characteristic polynomial convention $p(\lambda )=\det (\lambda I-A)$, where $I$ is the identity matrix. Expansion of the determinant along the first row givesThe eigenvalues are the roots of a polynomial $p(\lambda )=0$. With the alternative determinant convention $\det (A-\lambda I)$, the characteristic polynomial is $-p(\lambda )$ because the matrix has odd size; its roots of a polynomial are unchanged.

Since $L$ is an invertible matrix, $Av=0$ is equivalent to $Uv=0$. The last nontrivial row gives $v_3=v_4$; the second gives $v_2=v_4$; and the first gives $2v_1=v_2+v_4$, so $v_1=v_4$. A basis of the null space is thereforeIts single basis vector has Euclidean norm $2$, so the unit null space basis vector is $(1,1,1,1{)}^T/2$.

The pivot elements of $U$ lie in columns $1,2,3$. Take these columns from the original matrix $A$ to obtain a basis of its column space:The fourth column is minus the sum of these three columns, and the leading three-by-three matrix minor has nonzero determinant $-4$, confirming linear independence.

Elementary row operations preserve the row space. Thus the nonzero rows of $U$ give a basis of the row space of $A$:These basis vectors need not have unit Euclidean norm or be orthogonal. Their Euclidean norms are $\sqrt{6},\sqrt{2},2\sqrt{2}$, respectively; the three column basis vectors have Euclidean norms $2\sqrt{2},\sqrt{6},2\sqrt{2}$.

The hemisphere graph $u_0=\pm \sqrt{1-|\mathbf{u}{|}^2}$ of SU(2) as the three-sphere has induced metric $g_{ij}={\delta }_{ij}+u_i{u}_j/u_0^2$, whose determinant is $1/u_0^2$. Its volume density is therefore $d^{3}u/|u_0|$. Fixed left or right multiplication preserves the Euclidean norm on the four Pauli coefficients, so the round volume is Haar measure. Both hemispheres have volume ${\pi }^2$, giving total volume $2{\pi }^2$ and normalized density $d^{3}u/(2{\pi }^2|u_0|)$ on each hemisphere. The apparent equatorial divergence is an integrable coordinate singularity.

A perfect matching is a matching in a graph that covers every vertex exactly once. Its edges have no shared endpoints, and every vertex is an endpoint of one selected edge. It is useful when a collection must be divided into disjoint pairs with no element left over.

In a bipartite graph with equally many vertices on its two sides, a perfect matching can assign every person one allowed job and every job one person. The example under permanent of a matrix has three people and three jobs, with exactly three valid assignments. The blue edges below show those three perfect matchings; a crossing in the drawing is not an extra vertex.

Every selected edge covers two vertices, so a graph with an odd integer number of vertices has no perfect matching. An even integer number is necessary but does not guarantee existence: an isolated vertex cannot be covered. A perfect matching also need not be unique. For example, the complete bipartite graph $K_{2,2}$ has two perfect matchings, corresponding to the two ways of assigning two people to two jobs.

Counting all perfect matchings is a different question from finding one. For a zero-one biadjacency matrix of a balanced bipartite graph, the count is its permanent of a matrix. The nonzero determinant criterion for a perfect matching supplies an existence test, but cancellation means that a zero determinant does not rule one out.

The two-layer cell seismic layer transfer matrix has reciprocal eigenvalues because its determinant is one. Writing them as $e^{\pm iKd}$ gives the displayed half-trace relation. At a real frequency, an absolute half-trace exceeding one forces an exponentially growing/decaying pair instead of propagating cell modes. A long finite stack then suppresses transmission in this stop band, without dissipating energy: the missing transmitted flux is reflected. Quarter-wave layers of unequal seismic impedance give half-trace $-\frac{1}{2}(q_1/q_2+q_2/q_1)<-1$.

A nonzero uniform equilibrium of the conserved-field real amplitude equation obeys $\mu =R^2-\alpha R$. Fixed zero mean excludes the constant $B$ perturbation. Its homogeneous amplitude eigenvalue is $-r$. For a nonzero Fourier wavenumber $k$, the stability matrix has trace $-r-(1+\sigma )k^2$ and determinant $\sigma k^2(k^2+r-2\delta R^2)$. The displayed conditions are necessary and sufficient for strict linear stability on a period $2L$. In an infinite-domain limit the nonzero-mode condition becomes $r\ge 2\delta R^2$; a long-wave instability need not occur in a sufficiently short domain.

The permanent of a matrix adds the products obtained by choosing one entry in each row and one in each column. For an $n\times n$ matrix $A=(a_{ij})$ with entries in the real numbers or complex numbers,Here each permutation $\sigma$ chooses column $\sigma (i)$ in row $i$, and the symmetric group $S_n$ lists every such bijection. The formula resembles the Leibniz formula for determinants, but omits the sign of a permutation. For a two-by-two matrix,The absence of alternating signs makes the permanent of a matrix particularly useful for counting allowed assignments.

For a zero-one matrix, let rows represent people, columns represent jobs, and $a_{ij}=1$ mean that person $i$ can do job $j$. This is the biadjacency matrix of a bipartite graph. A product in the permanent is one exactly when all the selected assignments are allowed; otherwise it is zero. Each surviving permutation assigns every person a different job, which is a perfect matching. Thus the permanent counts perfect matchings. This interpretation and the definition are given in Yufei Zhao, Probabilistic Methods in Combinatorics, §10.2, pp. 178–179.

For a fully enumerated example, takeAmong the six permutations of three columns, exactly three give nonzero products: $(1,2,3)$, $(1,3,2)$ and $(2,1,3)$. These tuples list the chosen column in each successive row. Each product is one, so $\mathrm{perm}A=3$. Their permutation signs are $+1,-1,-1$, giving $\det A=-1$. The diagram displays all three allowed assignments.

A zero determinant need not mean that no assignment exists. The two-by-two all-ones matrix has permanent $2$ and determinant $0$: its two assignments cancel in the determinant. The nonzero determinant criterion for a perfect matching is sufficient for existence, but the converse fails for a fixed zero-one matrix.

For nonnegative entries, the permanent instead sums assignment weights, with each assignment weighted by the product of its selected entries. It is then a total weight rather than necessarily an integer count. With arbitrary entries in the real numbers or complex numbers, even the permanent can have cancellation; the counting interpretation specifically uses zero-one entries. The permanent as probability of distinct independent choices gives another useful nonnegative interpretation.

There are $n!$ terms in the defining sum, linking direct evaluation to the factorial. For the all-ones $n\times n$ matrix, every term is one and the permanent is exactly $n!$; the identity matrix has permanent one. Finding one allowed assignment and counting all allowed assignments are different questions. Their relationship to computational complexity theory was studied in L. G. Valiant, The complexity of computing the permanent, Theoretical Computer Science 8(2), 189–201, 1979. The elementary sum is useful for small examples, while the existing dense permanent approximation describes a more specialized way of estimating large counts.

For an antisymmetric $2n\times 2n$ matrix, the Pfaffian is $\mathrm{Pf}(A)=(2^{n}n!{)}^{-1}{\sum }_{\sigma \in S_{2n}}\mathrm{sgn}(\sigma ){\prod }_{j=1}^n{A}_{\sigma (2j-1),\sigma (2j)}$. Equivalently, it is the coefficient of the top exterior power of the associated alternating form, with the usual factorial normalization. Its square is the determinant. In size four it is $a_{12}{a}_{34}-a_{13}{a}_{24}+a_{14}{a}_{23}$, so its vanishing characterizes the simple nonzero bivectors of the Klein quadric.

A planar rotation about the origin is the linear map represented by the rotation matrix $R_{\theta }=\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$. It preserves inner products, lengths, oriented angles and area, with positive angles measured anticlockwise. Its inverse is $R_{-\theta }$ and its determinant is one.

Use determinant-one Givens rotations and choose every nonzero eliminated pair to leave a positive pivot. If the input square matrix $A$ is orthogonal, its resulting upper triangular factor $R$ is orthogonal too, and hence diagonal with entries of absolute value one. Positivity fixes the first $n-1$ entries to one, while determinant preservation fixes the last to $\det A$. A negative last entry is unavoidable when the input reverses orientation.

Multiplication of an alternant by $p_r$ is the sum of alternants obtained by raising one row exponent by $r$. Equal exponents give zero. Otherwise sort the shifted beta number of a partition back into decreasing order. The resulting partition adds a size-$r$ rim hook; the number of crossed exponents is its height and supplies the determinant sign. This proves the formula through the bialternant formula. Iteration counts signed border-strip tableaux.

For a prime $p$, distinct row and column residues and $\zeta =e^{2\pi i/p}$, every square discrete Fourier transform submatrix is nonsingular. Sort the selected column exponents decreasingly as $b_i$ and write $b_i={\lambda }_i+N-i$. Its determinant factors by the bialternant formula into a nonzero Vandermonde determinant and $s_{\lambda }({\zeta }^{r_1},\dots ,{\zeta }^{r_N})$. Modulo $1-\zeta$, this Schur factor is $s_{\lambda }(1^N)$ in ${\mathbb{F}}_p$. The Schur evaluation at all ones product has numerator and denominator differences strictly between zero and $p$, so its residue is nonzero. The Schur factor, and hence the minor, cannot vanish.