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A block matrix inverse expresses the matrix inverse in terms of smaller matrices. Partition a square matrix over a field asIf $A$ is an invertible matrix, define its Schur complement $S=D-C{A}^{-1}B$. Then $M$ is an invertible matrix exactly when $S$ is an invertible matrix, andThe order of the matrix products matters. In particular, the lower-right block is $S^{-1}$, rather than generally $D^{-1}$: the coupling through $B$ and $C$ changes the response.^[ref]

For an original example, takeThe Schur complement is $S=5-1/2-4/3=19/6$. Substitution gives the concise answerMultiplying in either order gives the identity matrix; the lower-right entry is $6/19$, whereas $D^{-1}=1/5$. The difference records the feedback through the two eliminated variables.

Often the desired output is a solution of a linear system, rather than every entry of a matrix inverse. Block elimination solves with $A$ and then with $S$, recovering the eliminated variables by back substitution. Reusing an LU decomposition of $A$ avoids explicitly forming $A^{-1}$. A favourable block diagonal matrix or tridiagonal matrix block can make this useful; merely partitioning an unstructured dense matrix does not guarantee a faster solve.^[ref]

A matrix decomposition expresses a matrix as a matrix product of factors with simpler structure. An LU decomposition uses triangular factors, a QR decomposition uses an orthogonal matrix and a triangular factor, and a singular value decomposition separates input directions, stretching factors and output directions. The factorization is an identity; the chosen structure determines what it helps compute.

For an original two-by-two example,The outer factors are a lower triangular matrix and an upper triangular matrix, each with unit diagonal as in a unit triangular matrix; the middle factor is a diagonal matrix. Their matrix product is $M$, and the final diagonal entry $5/2$ is the Schur complement of the first entry $2$. This is the scalar version of the factorization under block elimination.

To solve $Mz=(3,4{)}^T$, solve with the three factors in their displayed order from left to right: first the lower triangular factor, then the diagonal factor, then the upper triangular factor. The intermediate vectors are $(3,5/2{)}^T$ and $(3/2,1{)}^T$, giving the concise answer $z=(1,1{)}^T$. A new right-hand side uses the same factors. This is why Reusing LU factors for multiple right-hand sides is useful: factorization records work that otherwise would be repeated, while forward substitution and back substitution apply it cheaply.^[ref]

Different decompositions reveal different features. A Cholesky decomposition uses positive definiteness of a symmetric matrix, while a singular value decomposition also applies to rectangular matrices and exposes matrix rank. The existence and numerical behaviour of the factors depend on their own hypotheses; for example, an unpivoted LU decomposition can fail when a chosen pivot element is zero even though the matrix is an invertible matrix.

For a block matrix $M=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$ with $A$ an invertible matrix, the Schur complement of $A$ in $M$ is $S=D-C{A}^{-1}B$. It is the effective coefficient matrix for the remaining variables after block elimination. The definition requires invertibility of $A$, but not of $S$ or $M$. Its size is that of $D$, even when $B$ and $C$ are rectangular matrices.

In the original example under block matrix inverse, $D=5$ becomes $S=19/6$ because changing the remaining variable also changes the two variables solved through $A$. This is what makes a Schur complement more informative than simply deleting rows and columns. It retains their mediated effect, and its matrix inverse is the corresponding block of $M^{-1}$ when the block matrix inverse exists.

The same elimination appears when minimizing a quadratic form. The Schur complement as a minimized quadratic form gives a geometric example; the Schur complement determinant identity gives an exact test for an invertible matrix.

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.