{ "cells": [ { "cell_type": "code", "execution_count": null, "id": "acb06966-8cfa-4b28-9fa9-30286c33d20c", "metadata": {}, "outputs": [], "source": [] }, { "cell_type": "markdown", "id": "203b80b6-7559-4861-ab9c-538885b8d223", "metadata": { "tags": [] }, "source": [ "## Inverse?" ] }, { "cell_type": "code", "execution_count": 16, "id": "66031921-f6b4-4c32-bc8b-97a92ea935df", "metadata": { "jupyter": { "source_hidden": true }, "tags": [] }, "outputs": [], "source": [ "# Q: If you compute the inverse of one of these matrixes and multiply it to the right of the position\n", "# where it was originally multiplied it should \"delete\" that element as if the element was never added.\n", "# I'm uncertain of the possibility and frequency of such inverse matrices over finite fields.\n", "# If this scenario is possible to construct, I think we should \"define\" this operation as deletion.\n", "\n", "# See: https://en.wikipedia.org/wiki/Determinant#Properties_of_the_determinant\n", "\n", "def det2(mat, m=256):\n", " ((a, b),\n", " (c, c)) = mat\n", " return (a*c - b*c) % m\n", "\n", "def det3(mat, m=256):\n", " ((a, b, c),\n", " (d, e, f),\n", " (g, h, i)) = mat\n", " return (a*e*i + b*f*g + c*d*h - c*e*g - b*d*i - a*f*h) % m\n", "\n", "def det(mat, m=256):\n", " # I dislike having to round here, but it returns a float\n", " # I wish it returned a native \"B\" type\n", " return int(round(np.linalg.det(mat),0)) % m\n", "\n", "a = ((-2,2,-3),(-1,1,3),(2,0,-1))\n", "assert_equal(det(a), 18)\n", "assert_equal(det(a), det3(a))\n", "\n", "# ?\n", "# assert that they didn't change since the last time I ran it\n", "assert_equal(det(f1), 128)\n", "assert_equal(det(f2), 160)\n", "assert_equal(det(f3), 128)" ] }, { "cell_type": "markdown", "id": "568ad4ed-7894-465b-9bf3-f19f2269c0f3", "metadata": { "tags": [] }, "source": [ "[This](https://math.stackexchange.com/questions/273054/how-to-find-matrix-inverse-over-finite-field) suggests to find the inverse matrix by first finding the adjugate then applying\n", "\n", "> (1/det(A))adj(A)=inv(A)\n", "\n", "Under mod p\n", "\n", "---\n", "\n", "The [adjugate](https://en.wikipedia.org/wiki/Adjugate_matrix) is the inverse of the cofactor matrix.\n", "\n", "---\n", "\n", "The cofactor matrix is made by multiplying the matrix minor of each (i,j) of the original matrix times a sign factor.\n", "\n", "---\n", "\n", "The (i,j)-minor of matrix A is the determinant of the matrix A with row i & column j removed." ] } ], "metadata": { "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.9.2" } }, "nbformat": 4, "nbformat_minor": 5 }