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  "body": "\n## Definition\n\nLet $R$ be a commutative ring and let $A$ be an $n \\times n$ [[square matrix|0JXZKX526R]] over $R$.\nThe matrix $A$ is *invertible* when there is an $n \\times n$ matrix $B$ over $R$ such that\n$$AB = BA = I_n,$$\nwhere $I_n$ is the [[identity matrix|06DW9BAWB1]].\nThe matrix $B$ is then unique; it is the *inverse* of $A$, written $A^{-1}$.\n\n## Remarks\n\nOver a field, other names are *nonsingular* and *nondegenerate*.\nA matrix over $R$ is invertible if and only if its [[determinant is a unit|0783S9D0ZH]] of $R$.\nOver a field, a [[one-sided inverse is two-sided|0753Y2X5K6]], and $A$ is invertible if and only if $Ax = 0$ has only the solution $x = 0$.\nThe matrix $\\begin{pmatrix} 1 \u0026 1 \\\\ 1 \u0026 1 \\end{pmatrix}$ is not invertible over $\\R$, because its determinant is $0$.\n\n## Explanations\n\n### L1 human\n\nA matrix is invertible when another matrix undoes it.\nIf $A$ sends $x$ to $y$, then $A^{-1}$ sends $y$ back to $x$.\nFor $2 \\times 2$ matrices, $\\begin{pmatrix} a \u0026 b \\\\ c \u0026 d \\end{pmatrix}^{-1} = \\frac{1}{ad - bc} \\begin{pmatrix} d \u0026 -b \\\\ -c \u0026 a \\end{pmatrix}$ when $ad - bc \\ne 0$.\n",
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