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  "body": "\n## Summary\n\nBack substitution is a method to solve a system $Ux = b$, where $U$ is an $n \\times n$ [[upper triangular matrix|0N4XY7Z3NN]] with nonzero diagonal entries.\nThe last equation gives $x_n = b_n / u_{nn}$.\nThen, for $i = n-1, \\dots, 1$ in this order, equation $i$ gives\n$$x_i = \\frac{1}{u_{ii}} \\left( b_i - \\sum_{j = i+1}^{n} u_{ij} x_j \\right),$$\nwhich uses only the unknowns that are already known.\n\n## Remarks\n\nBack substitution uses on the order of $n^2$ arithmetic operations.\nForward substitution solves $Lx = b$ for a [[lower triangular matrix|00J3KZP8Y0]] $L$ in the same way, from the first equation down.\nAfter [[Gaussian elimination|0JVJ2WCXZH]], back substitution completes the solution of the system.\nWith an [[LU decomposition|0RKG5YW66Y]] $A = LU$, the system $Ax = b$ is solved by forward substitution for $Ly = b$ and then back substitution for $Ux = y$.\n\n## Explanations\n\n### L1 human\n\nIn a triangular system, the last equation has only one unknown, so we solve it at once.\nWe put that value into the equation above it, which now has only one new unknown.\nWe continue upwards until we know all the unknowns.\n",
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