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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.

Apply Gaussian elimination to the supplied nonsymmetric matrix,The first pivot element is $-2$, with elimination multiplier $l_{21}=2/(-2)=-1$. Perform $R_2\leftarrow R_2+R_1$, givingThe second pivot element is $-1$, with elimination multiplier $l_{32}=-1$. Perform $R_3\leftarrow R_3+R_2$, then use the third pivot element $-2$ and elimination multiplier $l_{43}=-1$ to perform $R_4\leftarrow R_4+R_3$. The LU factorization isHere $L$ is a unit lower triangular matrix, hence an invertible matrix, and $U$ is an upper triangular matrix. No row interchange is needed: all three pivot elements used in elimination are nonzero. The final diagonal entry is zero, with an entirely zero last row; there is no further division by it. The three nonzero rows of $U$ are linearly independent, so the matrix rank is $3$ and $A$ is a singular matrix. Direct matrix multiplication gives $LU=A$.

Choose a fixed time increment $h=\Delta t>0$, let $t_n=nh$, and approximate $\mathbf{c}(t_n)$ by ${\mathbf{c}}_n$. The Backward Euler method evaluates the transport term at the new time:Rearrange to the linear systemAlthough $A$ is a singular matrix, $B$ is an invertible matrix for every $h>0$. In each row, its diagonal entry is $1+2h$ and the sum of the absolute values of its off-diagonal entries is $2h$; hence it is a strictly diagonally dominant matrix. Its Gershgorin discs have centre $1+2h$ and radius $2h$, excluding zero. In particular, the zero eigenvalue of $A$ becomes an eigenvalue $1$ of $B$.

The algorithm uses Reusing LU factors for multiple right-hand sides:

Both triangular solves use the factors of $I-hA$. The LU factorization of the singular transport matrix $A$ from part (a) is not a solver for this new linear system. No matrix inverse needs to be formed. Since $A$ and $h$ are fixed, the same LU factorization serves every step; only the right-hand side changes. A varying $\mathbf{b}(t)$ would give $r_n={\mathbf{c}}_n-h\mathbf{b}(t_{n+1})$ without changing these factors.

As a check, multiplying the update by $1^T$ givesexactly reproducing the inventory balance at the sampled times, apart from roundoff error. The Backward Euler method has temporal global error $O(h)$ for smooth solutions, so its A-stability does not remove the need to select $h$ for accuracy.

We fix the convention $PA=LU$: the permutation matrix $P$ acts on the rows of $A$, and the same row ordering must be applied to $b$ before the Triangular solves for LU decomposition. Equivalently, $A=P^{T}LU$, since $P^{-1}=P^T$. LAPACK documents its factors as $A=PLU$ instead, so check the defining equation rather than just the letter $P$.^[ref] Partial pivoting selects the largest absolute candidate in the current column, among rows $k,\dots ,n$, before forming elimination multipliers.^[ref]

A zero first pivot element does not imply a singular matrix. For example,Unpivoted Gaussian elimination would divide $2$ by zero. Indeed, no $A_0=LU$ with a unit lower triangular matrix $L$ exists: $u_{11}=0$ would force $(LU{)}_{21}=l_{21}{u}_{11}=0$, contradicting $(A_0{)}_{21}=2$. One row interchange resolves the obstruction. For $b_0=(1,5{)}^T$, use $P{b}_0=(5,1{)}^T$; back substitution then gives $x=(1,1{)}^T$. Using the original $b_0$ with these factors would solve a different linear system.

Partial pivoting bounds each elimination multiplier by one in absolute value in exact arithmetic. This helps control roundoff error, but intermediate entries can still grow; it is not an unconditional guarantee of numerical stability for every matrix.^[ref] It also does not remove the original problem's condition number. For the diagonal matrix $D=\mathrm{diag}(1,\epsilon )$ with $0<\epsilon <1$, the condition number in the Euclidean norm is ${\kappa }_2(D)=1/\epsilon$. Row permutations preserve this condition number, because they preserve Euclidean norms. Even an exact solve changes from $(1,1{)}^T$ to $(1,1+\eta /\epsilon {)}^T$ when the right-hand side changes from $(1,\epsilon {)}^T$ to $(1,\epsilon +\eta {)}^T$. This sensitivity is a property of the linear system, not a zero-pivot element failure.^[ref]

For an invertible matrix $A$, the rank-one matrix update $B=A+u{v}^T$ is a singular matrix exactly when $1+v^T{A}^{-1}u=0$. This follows directly from the matrix determinant lemma. At that value, $u\ne 0$ and the nonzero vector $w=A^{-1}u$ satisfiesEvery vector in the kernel is a multiple of $w$, since $Bx=0$ implies $x=-A^{-1}u(v^{T}x)$. The matrix rank therefore drops by exactly one.

For $B(t)=I+tu{v}^T$ with $u=(1,2{)}^T$ and $v=(3,-1{)}^T$, the update becomes a singular matrix at $t=-1$, and $B(t)u=(1+t)u$. At $t=-0.99$, this direction is compressed by a factor $0.01$ and the inverse amplifies it by a factor 100. The actual spectral norm of the matrix inverse, which maximizes amplification over every direction, is about $700.17853409$.

The exact formula can expose a large condition number in the updated linear system even when the original one had a small condition number. It does not guarantee numerical stability merely because its evaluation uses fewer arithmetic operations. The denominator condition certifies existence; the condition number and floating-point behavior address different questions about using the resulting answer.

When $A$ is an invertible matrix, the factorization in block elimination has outer triangular factors with determinant one. Hence $\det M=\det A\det S$, without a symmetry assumption. For the original example under block matrix inverse, this gives $\det M=(2\cdot 3)(19/6)=19$. An invertible block $A$ therefore reduces the question of whether the full matrix is an invertible matrix to the smaller Schur complement; a singular $S$ makes $M$ a singular matrix.

Let $B$ be an invertible matrix and $C$ a real square matrix of the same odd order. If the displayed identity holds, multiply it on the left by $B$ to obtain $B{B}^{\mathsf{T}}{C}^{\mathsf{T}}=-C$. Taking determinants givesThus $((\det B{)}^2+1)\det C=0$. Since $\det B$ is real, the first factor is positive, forcing $C$ to be a singular matrix. Both odd dimension and reality matter: an invertible real antisymmetric matrix exists in even dimension, while over the complex numbers a squared determinant can equal $-1$.

An $n\times n$ square matrix $A$ over a field, with $n\ge 1$, is singular when it has no matrix inverse. Geometrically, its linear map loses at least one input direction: some nonzero vector $v$ satisfies $Av=0$, so $A(x+v)=Ax$. Over the real numbers, the following family makes that loss visible:

For this diagonal matrix, $\det A(t)=t$. Whenever $t\ne 0$, the input can be recovered as $(x,y)=(b_1,b_2/t)$ and the kernel is $\{0\}$. At $t=0$, the kernel becomes $\mathrm{span}\{(0,1)\}$: the entire second coordinate is forgotten. Consequently $A(0)x=(b_1,b_2)$ has no solution if $b_2\ne 0$, and has the solution straight line $x=(b_1,s)$, $s\in \mathbb{R}$, if $b_2=0$. A singular matrix therefore need not make a particular linear system inconsistent; it prevents unique recovery when that linear system is consistent.

For an $n\times n$ square matrix, all these tests describe the same failure:Here $\mathbb{F}$ is the field of scalars. Equivalently, the columns are linearly dependent, the linear map is neither injective nor surjective, and some outputs cannot be reached. See the invertible matrix equivalences and Strang's MIT 18.06SC lecture summary, Multiplication and inverse matrices, p. 2. Over an infinite field, every consistent linear system for a singular matrix has infinitely many solutions; over a finite field with $q$ elements and nullity $d$, it has exactly $q^d$ solutions. This follows by adding each kernel vector to one solution.

The square hypothesis is essential. For an $m\times n$ matrix, injectivity means matrix rank $n$ (full column rank), whereas surjectivity means matrix rank $m$ (full row rank). A rectangular matrix can satisfy one condition without the other; its kernel and image of a linear map give the useful tests without calling it a singular matrix.

Exact singularity also differs from sensitivity. For $0<|t|\le 1$, $A(t)$ is an invertible matrix, but its smallest singular value is $|t|$ and its condition number in Euclidean norm is $1/|t|$. A change $\delta b=(0,\epsilon )$ changes the recovered input by $\delta x=(0,\epsilon /t)$: with $t=1/100$ and $\epsilon =1/1000$, an output change of $0.001$ causes an input change of $0.1$. These are exact toy values, not measurements. The condition number quantifies this sensitivity before a direction is lost; see Steve Marschner, Condition numbers, §2. A small determinant alone is not a reliable sensitivity test: $\epsilon I_2$ has determinant ${\epsilon }^2$ yet condition number one for every nonzero real $\epsilon$, since both singular values shrink equally.

A linear map $T:V\to W$ has a trivial kernel when $\ker T=\{0_V\}$: the zero vector is the only input producing the zero output. The kernel is a singleton set, not the empty set, because every linear map sends zero to zero. Equivalently, its nullity is zero and $T$ is injective. This holds even without a finite-dimensional hypothesis; see Linear Algebra Done Right by Sheldon Axler (4th edition).

The reason this zero-output test controls all outputs is linearity:With a trivial kernel, equal outputs force equal inputs. Therefore every equation $T(x)=b$ has at most one solution. Existence is a separate question: $b$ must belong to the image of a linear map. In the fiber description under nullity, $x_0+\{0\}$ is just the single point $x_0$.

For example, the linear map $J:{\mathbb{R}}^2\to {\mathbb{R}}^3$, $J(x,y)=(x,y,0)$, has trivial kernel and full column rank:The output $(3,4,0)$ comes from exactly $(3,4)$, while $(3,4,1)$ is unreachable. Conversely, $P:{\mathbb{R}}^3\to {\mathbb{R}}^2$, $P(x,y,z)=(x,y)$, is surjective and has full row rank, but its kernel is $\mathrm{span}\{(0,0,1)\}$: every output $(a,b)$ comes from the solution straight line $(a,b,s)$.

For an $m\times n$ matrix, a trivial kernel is equivalent to matrix rank $n$ and linear independence of its columns. It can therefore occur only if $n\le m$. If $m=n$, the rank-nullity theorem also forces surjectivity, giving an invertible matrix; if $m>n$, the example $J$ shows why unique recovery of reachable inputs does not imply that every output is reachable. Finally, a trivial kernel guarantees exact uniqueness without guaranteeing numerical stability: the singular matrix example $A(t)$ has trivial kernel for every $t\ne 0$, even as its condition number grows without bound when $t$ approaches zero.