From b90f3a1fc6c09bc97de18c3226e87e18fdc7badf Mon Sep 17 00:00:00 2001
From: Holden Rohrer <hr@hrhr.dev>
Date: Tue, 24 Aug 2021 16:46:01 -0400
Subject: first day of notes

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+Solution of a system is the set of points/combinations of variables that
+satisfy an equation like 3x+2y=1.
+
+[3 2][x
+      y] = 1 is equivalent
+A := [3 2]
+X := [x y]^T
+b := 1
+AX = b
+
+Some systems like:
+x - 2y = 0
+x + y = 1
+x - y = 2
+have no solutions.
+This can be written as a a matrix too.
+Asking "does this system have a solution" is equivalent to asking:
+Can [0 1 2]^T be written as a linear combination/in span of [1 1 1]^T
+and [-2 1 -1]^T
+
+-----
+How to solve Ax = b where A is an mxn matrix, x is a nx1 matrix of free
+variables, and b is a mx1 matrix of constants.
+
+Ex.
+A = [ 3 1  0 4
+      0 1  2 5
+      0 0 -1 1 ]
+x = [ x1 x2 x3 x4 ]^T
+b = [ 0 1 2 ]
+
+Find the free variable, and then back-substitute
+x4 = t
+-x3 + x4 = 2 --> x3 = t - 2
+x2 + 2*x3 + 5*x4 = 1 --> x2 + 2*(t-2) + 5*t = 1 --> x2 = 5 - 7t
+3*x1 + x2 + x4 = 0 --> 3*x1 + 5 - 7t + t = 0 --> x1 = -5/3 + 2t
+
+Pivot element := first non-zero element in a matrix row AND everything
+below it is zero
+
+A matrix is in *row echelon* form iff pivot elements are to the left of
+all pivot elements in a lower row.
+    Therefore, all non-zero rows have a pivot but not every column.
+Note: column echelon form also exists.
+
+Generally, solving Ax = B can be done by solving
+CAx = Cb where C is an invertible mxm matrix.
+Single row operations used in Gaussian elimination are a subset of these
+operations.
+
+Ex.
+A = 3x4 matrix.
+Multiply 1/3 to first row: 1/3 R1 -> R1.
+C = 3x3 matrix.
+[ 1/3 0 0
+  0   1 0
+  0   0 1 ] * A
+C^{-1} =
+[ 3 0 0
+  0 1 0
+  0 0 1 ]
+
+Switch rows R2 and R3:
+C = [ 1 0 0
+      0 0 1
+      0 1 0 ]
+C = C^{-1}
+
+Add cR1 to R3:
+C = [ 1 0 0
+      0 1 0
+      c 0 1 ]
+C^{-1} = [ 1 0 0
+           0 1 0
+          -c 0 1 ]
-- 
cgit