Updated notebook with reduced row echelon functions

Author: garrysingh2930

linear-algebra-ml-ipynb/Linear_systems_with_3_variables.ipynb

@@ -22,10 +22,9 @@
    "source":[
     "# let's start by importing the required libraries\n",
     "\n",
-    "import numpy as np\n",
-    "import copy"
+    "import numpy as np"
    ],
-   "execution_count":11,
+   "execution_count":3,
    "outputs":[
 
    ],
@@ -58,7 +57,7 @@
     "print(f\"Equation Matrix: \\n {eq_arr} \\n\")\n",
     "print(f\"Result Matrix: \\n {res_arr} \\n\")"
    ],
-   "execution_count":3,
+   "execution_count":4,
    "outputs":[
     {
      "name":"stdout",
@@ -168,7 +167,7 @@
     "\n",
     "print(f\"Determinant of the linear equation is: {det_arr:0.2f}\")"
    ],
-   "execution_count":8,
+   "execution_count":7,
    "outputs":[
     {
      "name":"stdout",
@@ -196,7 +195,7 @@
     "\n",
     "print(f\"Restacked array is: {stck_eq_arr}\")\n"
    ],
-   "execution_count":10,
+   "execution_count":8,
    "outputs":[
     {
      "name":"stdout",
@@ -229,7 +228,7 @@
     "    cp_orig_arr[row_index] = orig_arr[row_index] * multiple_scalar\n",
     "    return cp_orig_arr"
    ],
-   "execution_count":27,
+   "execution_count":9,
    "outputs":[
 
    ],
@@ -251,7 +250,7 @@
     "mod_eq_arr = multiply_row(stck_eq_arr, 1, 2)\n",
     "print(f\"Modified array after change is: \\n {mod_eq_arr} \\n\")"
    ],
-   "execution_count":28,
+   "execution_count":10,
    "outputs":[
     {
      "name":"stdout",
@@ -289,7 +288,7 @@
     "    cp_orig_arr[row_index_2] = multiple_scalar * orig_arr[row_index_1] + orig_arr[row_index_2]\n",
     "    return cp_orig_arr"
    ],
-   "execution_count":29,
+   "execution_count":11,
    "outputs":[
 
    ],
@@ -311,7 +310,7 @@
     "mod_eq_arr = add_rows(stck_eq_arr, 2, 1, 2)\n",
     "print(f\"Modified array after change is: \\n {mod_eq_arr} \\n\")"
    ],
-   "execution_count":30,
+   "execution_count":12,
    "outputs":[
     {
      "name":"stdout",
@@ -349,7 +348,7 @@
     "    cp_orig_arr[[row_index_1,row_index_2]] = cp_orig_arr[[row_index_2,row_index_1]]\n",
     "    return cp_orig_arr"
    ],
-   "execution_count":31,
+   "execution_count":13,
    "outputs":[
 
    ],
@@ -371,7 +370,7 @@
     "mod_eq_arr = swap_rows(stck_eq_arr, 2, 1)\n",
     "print(f\"Modified array after change is: \\n {mod_eq_arr} \\n\")"
    ],
-   "execution_count":32,
+   "execution_count":14,
    "outputs":[
     {
      "name":"stdout",
@@ -399,6 +398,77 @@
     }
    }
   },
+  {
+   "cell_type":"code",
+   "source":[
+    "# now to further perform reduced row echelon, let use above defined functions\n",
+    "\n",
+    "red_row_eq_arr = swap_rows(stck_eq_arr, 0, 2)\n",
+    "\n",
+    "red_row_eq_arr = add_rows(red_row_eq_arr,0,1,2)\n",
+    "\n",
+    "red_row_eq_arr = add_rows(red_row_eq_arr,0,2,4)\n",
+    "\n",
+    "red_row_eq_arr = add_rows(red_row_eq_arr,1,2,-1)\n",
+    "\n",
+    "red_row_eq_arr = multiply_row(red_row_eq_arr,2,-1\/12)\n",
+    "\n",
+    "print(f\"Modified reduced row equation is: \\n {red_row_eq_arr} \\n\")\n",
+    "\n"
+   ],
+   "execution_count":15,
+   "outputs":[
+    {
+     "name":"stdout",
+     "text":[
+      "Modified reduced row equation is: \n",
+      " [[-1.  2. -5. 17.]\n",
+      " [ 0.  5. -7. 34.]\n",
+      " [-0. -0.  1. -2.]] \n",
+      "\n"
+     ],
+     "output_type":"stream"
+    }
+   ],
+   "metadata":{
+    "datalore":{
+     "node_id":"6pj2ZoXJretXm179iDVcUX",
+     "type":"CODE",
+     "hide_input_from_viewers":true,
+     "hide_output_from_viewers":true
+    }
+   }
+  },
+  {
+   "cell_type":"code",
+   "source":[
+    "# now let's try and find values of x1, x2, and x3\n",
+    "\n",
+    "x_3 = -2 # as per above equation derivation\n",
+    "x_2 = (red_row_eq_arr[1,3] - red_row_eq_arr[1,2] * x_3) \/ red_row_eq_arr[1,1]\n",
+    "x_1 = (red_row_eq_arr[0,3] - red_row_eq_arr[0,2] * x_3 - red_row_eq_arr[0,1] * x_2) \/ red_row_eq_arr[0,0]\n",
+    "\n",
+    "print(x_1, x_2, x_3)"
+   ],
+   "execution_count":17,
+   "outputs":[
+    {
+     "name":"stdout",
+     "text":[
+      "1.0 4.0 -2\n"
+     ],
+     "output_type":"stream"
+    }
+   ],
+   "metadata":{
+    "datalore":{
+     "node_id":"ZRLebFtwF9w2m4dpTAWEbS",
+     "type":"CODE",
+     "hide_input_from_viewers":true,
+     "hide_output_from_viewers":true
+    }
+   }
+  },
   {
    "cell_type":"code",
    "source":[
@@ -410,7 +480,7 @@
    ],
    "metadata":{
     "datalore":{
-     "node_id":"6pj2ZoXJretXm179iDVcUX",
+     "node_id":"RgvqEXHh6ZdOBqiZNg3h73",
      "type":"CODE",
      "hide_input_from_viewers":true,
      "hide_output_from_viewers":true