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An $n$-by-$n$ square matrix $A$ over a field is invertible when it has a two-sided matrix inverse. Equivalently, its linear map $x\mapsto Ax$ is a bijection: every output comes from exactly one input. For a square matrix, the following conditions are equivalent:Here matrix rank $n$ means both full column rank and full row rank. The square hypothesis matters: a rectangular matrix can have full column rank without reaching every output, or full row rank without determining every input.^[ref]

The equivalences explain why invertibility is a question about retaining information. A trivial kernel makes the linear map injective; the rank-nullity theorem then gives matrix rank $n$, so its columns form a basis and the linear map is also surjective. Assigning to each output its unique input gives a linear map: the preimage of $\alpha b+\beta c$ is $\alpha A^{-1}b+\beta A^{-1}c$. The determinant is nonzero precisely when the columns are linearly independent. Thus algebraic solvability and independent directions describe the same phenomenon; over the real numbers, nonzero oriented volume supplies its geometric interpretation.

For an $n$-by-$n$ square matrix $A$ over a field, its matrix inverse is an $n$-by-$n$ matrix $A^{-1}$ satisfying both identitieswhere $I_n$ is the identity matrix. It exists exactly when $A$ is an invertible matrix, equivalently when its matrix rank is $n$ or its determinant is nonzero. It reverses the linear map: applying $A$ and then $A^{-1}$ recovers the input, while applying $A^{-1}$ and then $A$ recovers the output.^[ref]

The matrix inverse is unique. If $B$ and $C$ both satisfy the identities, then $B=B(AC)=(BA)C=C$. Although the definition is two-sided, for finite square matrices either identity alone suffices: $BA=I$ forces a trivial kernel, whereas $AB=I$ forces surjectivity; either makes $A$ an invertible matrix by the rank-nullity theorem. Rectangular matrices may have only one of these properties.

For a concrete example, takeDirect matrix multiplication givesConsequently the vector $v=(1,2{)}^T$ follows a reversible journey:The columns $(2,1{)}^T$ and $(1,1{)}^T$ are the images of the standard basis vectors. They turn the unit square into a parallelogram; the matrix inverse restores both the square and its surrounding grid. Since $\det A=1$, this example changes shape while preserving area, illustrating that invertibility does not require preserving lengths or angles.

For comparison, the singular matrixcannot have a matrix inverse: the output $(1,0{)}^T$ cannot specify which of the two inputs occurred. Their difference $(1,-1{)}^T$ belongs to its kernel. Indeed $S(x,y{)}^T=(x+y,0{)}^T$ collapses the plane to a line, its matrix rank is one, and its determinant is zero. No subsequent function of the output alone can reconstruct every input.

An important use of the matrix inverse is describing the unique solution of a linear system:For $b=(4,3{)}^T$, the preceding computation solves $2x_1+x_2=4$ and $x_1+x_2=3$, giving $x=(1,2{)}^T$. In computation, usually solve the linear system using Gaussian elimination and a factorization such as an LU decomposition, rather than forming every entry of the matrix inverse. The inverse formula specifies the answer; a factorization provides a reusable way to compute it for each right-hand side.^[ref]

Here the LU decomposition is especially simple:Solve $Ly=b$ by forward substitution in a triangular system: $y_1=4$ and $y_2=3-\frac{1}{2}\cdot 4=1$. Then solve $Ux=y$ by backward substitution in a triangular system: $x_2=2$ and $x_1=(4-2)/2=1$. The same factors work when $b$ changes. If the whole matrix inverse is actually needed, its $j$th column is the solution of $Ax=e_j$, where $e_j$ is the $j$th standard basis vector.

Reversing several operations means undoing the last one first. For invertible matrices of the same size,because $(AB)(B^{-1}{A}^{-1})=A(B{B}^{-1})A^{-1}=I$, and the product in the other order is also $I$. This is why a change-of-basis matrix uses its matrix inverse to reverse the direction of coordinate conversion.

When a measurement model is a linear map $T$, its kernel records changes that its readout cannot distinguish. For example, the rectangular kernel and solution-fibre example measures only $x+2y$ and $y+z$. Even exact readouts $(3,2)$ leave the whole fiber $u_0+\ker A$ possible. Repeating either readout with greater precision cannot distinguish those inputs: their outputs agree exactly. This ambiguity follows from a nontrivial kernel, rather than from measurement error.

Additional information can remove a null direction of a linear map. In that example, adding a third linear functional that reads $y$ distinguishes $u_0+tv$, because its third readout is $t$. More generally, additional linear functionals $c_1,\dots ,c_k$ give unique recovery of all inputs exactly whenThis follows by applying the trivial kernel criterion to the combined linear map $x\mapsto (Tx,c_1(x),\dots ,c_k(x))$. Merely adding repeated linear equations does not meet this condition. On a finite-dimensional input vector space, each added scalar readout can increase the matrix rank by at most one, so at least $\dim \ker T$ extra independent readouts are needed for recovery of unrestricted inputs.

Restrictions on the possible inputs can also make recovery unique. If the allowed inputs form a set $D$, two allowed inputs are indistinguishable exactly when their difference is in the kernel. Thus $T$ is injective on $D$ precisely when no difference of distinct points of $D$ is a kernel vector. Such a restriction is extra information about the input; the original readout alone determines only its fiber.

For $d_j=1/j$, the equation $Du=f$ forces $u_j=j{f}_j$. It has a solution in ${\ell }^2$ exactly whenFor example, $f_j=1/j$ defines an element of ${\ell }^2$, but its forced preimage is the constant sequence $u_j=1$, which is not in ${\ell }^2$. Thus existence can fail. Since every $d_j$ is nonzero, $D$ has trivial kernel and any solution is unique.

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.