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  "body": "\n## Statement\n\nLet $K$ be a commutative ring and let $A$ be an $n \\times n$ matrix over $K$.\nThen $A$ is [[invertible|0EB9N0S33C]] if and only if its [[determinant|00KDZRW4C0]] $\\det A$ is a unit of $K$.\nIn particular, over a field, $A$ is invertible if and only if $\\det A \\ne 0$.\n\n## Remarks\n\nIf $\\det A$ is a unit, then the [[adjugate|07W9H8QF38]] gives the inverse $A^{-1} = (\\det A)^{-1} \\operatorname{adj}(A)$.\nOver $\\Z$, the matrix $\\begin{pmatrix} 2 \u0026 0 \\\\ 0 \u0026 1 \\end{pmatrix}$ has determinant $2 \\ne 0$, but it has no inverse with integer entries.\nOver a field, this is one of many equivalent conditions for invertibility, with $\\rank A = n$ and $\\ker A = 0$.\n\n## Explanations\n\n### L1 human\n\nA real square matrix has an inverse exactly when its determinant is not zero.\nA zero determinant means that the matrix squeezes space into a lower dimension.\nSuch a squeeze cannot be undone.\n",
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