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Here is an original example of Gaussian elimination that can be checked entirely with integers. Start withThe first pivot element is $2$. The elimination multipliers are $l_{21}=4/2=2$ and $l_{31}=-2/2=-1$. Apply the elementary row operations $R_2\leftarrow R_2-2R_1$ and $R_3\leftarrow R_3+R_1$ to obtainNow the second pivot element is $3$, and $l_{32}=9/3=3$. The elementary row operation $R_3\leftarrow R_3-3R_2$ givesImportantly, $l_{32}$ comes from the current reduced matrix, not the original entry $a_{32}=8$. To reconstruct $A$ from $U$, its second row is twice the first row of $U$ plus the second; its third row is minus the first row of $U$ plus three times the second plus the third. These are exactly the coefficients in the lower triangular matrix $L$.

The unpivoted calculation here illustrates the structure of LU decomposition; applying partial pivoting to this particular $A$ would choose a different first row and different factors. Both procedures can represent the same linear system when their permutation matrices are handled consistently.

For $i>k$, an elimination multiplier is the scalar $l_{ik}=a_{ik}^{(k-1)}/a_{kk}^{(k-1)}$ used in the elementary row operation $R_i\leftarrow R_i-l_{ik}{R}_k$. This creates a zero below the pivot element. The remaining entries change by $a_{ij}\leftarrow a_{ij}-l_{ik}{a}_{kj}$ for $j>k$. Storing these elimination multipliers gives the subdiagonal entries of $L$ in LU decomposition; later row interchanges must also reorder the already stored coefficients.

An LU decomposition writes a square matrix as $A=LU$, with $L$ a lower triangular matrix and $U$ an upper triangular matrix. We use a unit triangular matrix $L$, meaning $l_{ii}=1$. The factors record Gaussian elimination: $U$ contains the reduced equations, while $L$ stores the elimination multipliers needed to reconstruct the original equations. This makes Reusing LU factors for multiple right-hand sides inexpensive.^[ref]

For an invertible matrix, unpivoted LU decomposition requires nonzero pivot elements at every elimination step; invertibility alone does not guarantee this. In general use Pivoted LU decomposition, with the explicit convention $PA=LU$ and a permutation matrix $P$. Once the factors exist, solving a linear system $Ax=b$ needs forward substitution followed by back substitution, rather than forming a matrix inverse.

For classical dense Gaussian elimination over the real numbers, count each scalar addition, subtraction, multiplication or division as one arithmetic operation. At step $k$, there are $n-k$ elimination multipliers to divide out and $(n-k{)}^2$ remaining entries, each requiring one multiplication and one subtraction. Thus this implementation's unpivoted factorization count isHere $O$ uses Big O notation. Partial pivoting adds $O(n^2)$ comparisons and row movement, leaving the leading arithmetic count unchanged. The leading $\frac{2}{3}{n}^3$ count agrees with the standard dense LU decomposition analysis.^[ref]

With unit $L$, forward substitution uses $n(n-1)$ arithmetic operations. Back substitution uses another $n(n-1)$ plus $n$ diagonal divisions. Together these cost $2n^2-n$ per right-hand side; applying a stored row permutation moves $n$ entries and requires no scalar arithmetic. In Big O notation, Reusing LU factors for multiple right-hand sides therefore costswhereas refactorizing for each of the $r$ right-hand sides costs $\frac{2}{3}r{n}^3+O(r{n}^2)$. For $n=1000$ and $r=10$, the leading counts are about $6.87\times 10^8$ versus $6.69\times 10^9$ arithmetic operations, nearly a tenfold saving. These are operation counts, not measured runtimes: memory access and optimized matrix multiplication also affect speed.^[ref]

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]

A pivot element is the current diagonal entry used to eliminate entries below it in Gaussian elimination, after any row interchange. At step $k$ it is $a_{kk}^{(k-1)}$, where $A^{(k-1)}$ is the current reduced matrix. Computing elimination multipliers requires this entry to be nonzero. A zero pivot element can reflect an unsuitable row order even for an invertible matrix, as Pivoted LU decomposition illustrates.

Let $n\ge 1$. Store the tridiagonal matrix $A$ by its diagonal $b_1,\dots ,b_n$, lower diagonal $a_2,\dots ,a_n$, and upper diagonal $c_1,\dots ,c_{n-1}$. The linear system $Ax=d$ iswith the nonexistent endpoint terms omitted. Initialize $p_1=b_1$ and $g_1=d_1$. For $i=2,\dots ,n$, compute the elimination multiplier, modified pivot element, and modified right-hand side:These recurrence relations apply the elementary row operation $R_i\leftarrow R_i-m_i{R}_{i-1}$ after the preceding row has already been reduced. The transformed linear system is an upper bidiagonal matrix:Thus back substitution is especially short:For $n=1$, only $x_1=d_1/b_1$ remains. Check each pivot element before its first division, including $p_n$ before back substitution.

Equivalently, this constructs an LU decomposition with unit lower bidiagonal matrix $L$, whose lower diagonal is $m_2,\dots ,m_n$, and upper bidiagonal matrix $U$, whose diagonal is $p_1,\dots ,p_n$ and upper diagonal is $c_1,\dots ,c_{n-1}$. The $g$ sweep is forward substitution in $Lg=d$. There is no fill-in outside these bands when no rows are interchanged. LAPACK's DGTSV source uses these same coefficient and right-hand-side updates in its branch without row interchange.^[ref]