{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 1 - Matrix notation & matrix multiplication"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex:1.3.1 Pg:20"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"x=\n",
"[[u]\n",
" [v]\n",
" [w]]\n",
"R2=R2-R1,R3=R3-4*R1\n",
"[[1 1 1]\n",
" [0 0 3]\n",
" [0 2 4]]\n",
"R2<->R3\n",
"[[1 1 1]\n",
" [0 2 4]\n",
" [0 0 3]]\n",
"The system is now triangular and the equations can be solved by Back substitution\n"
]
}
],
"source": [
"from sympy.abc import u,v,w\n",
"from numpy import array\n",
"a =array([[1 ,1 ,1],[2, 2, 5],[4, 6, 8]])\n",
"x=[[u],[v],[w]]\n",
"x=array(x)\n",
"print 'x=\\n',x\n",
"print 'R2=R2-R1,R3=R3-4*R1'\n",
"a[1,:]=a[1,:]-2*a[0,:]\n",
"a[2,:]=a[2,:]-4*a[0,:]\n",
"print a\n",
"print 'R2<->R3'\n",
"def swap_rows(arr, frm, to):\n",
" arr[[frm, to],:] = arr[[to, frm],:]\n",
"swap_rows(a,1,2)\n",
"print a\n",
"print 'The system is now triangular and the equations can be solved by Back substitution'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex:1.3.2 Pg:21"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"a=\n",
"[[1 1 1]\n",
" [2 2 5]\n",
" [4 4 8]]\n",
"x=\n",
"[[u]\n",
" [v]\n",
" [w]]\n",
"R2=R2-2*R1,R3=R3-4*R1\n",
"[[1 1 1]\n",
" [0 0 3]\n",
" [0 0 4]]\n",
"No exchange of equations can avoid zero in the second pivot positon ,therefore the equations are unsolvable\n"
]
}
],
"source": [
"from sympy.abc import u,v,w\n",
"from numpy import array\n",
"a =array([[1, 1, 1],[2, 2, 5],[4, 4, 8]])\n",
"print 'a=\\n',a\n",
"x=[[u],[v],[w]]\n",
"x=array(x)\n",
"print 'x=\\n',x\n",
"print 'R2=R2-2*R1,R3=R3-4*R1'\n",
"a[1,:]=a[1,:]-2*a[0,:]\n",
"a[2,:]=a[2,:]-4*a[0,:]\n",
"print a\n",
"print 'No exchange of equations can avoid zero in the second pivot positon ,therefore the equations are unsolvable'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Ex:1.4.1 Pg:21"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"A = \n",
"[[2 3]\n",
" [4 0]]\n",
"B = \n",
"[[ 1 2 0]\n",
" [ 5 -1 0]]\n",
"AB=\n",
"[[17 1 0]\n",
" [ 4 8 0]]\n"
]
}
],
"source": [
"from numpy import array\n",
"A=array([[2, 3],[4, 0]])\n",
"print 'A = \\n',A\n",
"B=array([[1, 2, 0],[5, -1, 0]])\n",
"print 'B = \\n',B\n",
"print 'AB=\\n',A.dot(B)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex:1.4.2 Pg:22"
]
},
{
"cell_type": "code",
"execution_count": 37,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"A=\n",
"[[2 3]\n",
" [7 8]]\n",
"P(Row exchange matrix)=\n",
"[[0 1]\n",
" [1 0]]\n",
"PA=\n",
"[[7 8]\n",
" [2 3]]\n"
]
}
],
"source": [
"from numpy import array\n",
"A=array([[2, 3],[7, 8]])\n",
"print 'A=\\n',A\n",
"P=array([[0, 1],[1, 0]])\n",
"print 'P(Row exchange matrix)=\\n',P\n",
"print 'PA=\\n',P.dot(A)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex:1.4.3 Pg:24"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"A=\n",
"[[1 2]\n",
" [3 4]]\n",
"I=\n",
"[[ 1. 0.]\n",
" [ 0. 1.]]\n",
"IA=\n",
"[[ 1. 2.]\n",
" [ 3. 4.]]\n"
]
}
],
"source": [
"from numpy import array,identity\n",
"A=array([[1, 2],[3, 4]])\n",
"print 'A=\\n',A\n",
"I=identity(2)\n",
"print 'I=\\n',I\n",
"print 'IA=\\n',I.dot(A)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex:1.4.4 Pg:25"
]
},
{
"cell_type": "code",
"execution_count": 39,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"E=\n",
"[[ 1. 0. 0.]\n",
" [-2. 1. 0.]\n",
" [ 0. 0. 1.]]\n",
"F=\n",
"[[ 1. 0. 0.]\n",
" [ 0. 1. 0.]\n",
" [ 1. 0. 1.]]\n",
"EF=\n",
"[[ 1. 0. 0.]\n",
" [-2. 1. 0.]\n",
" [ 1. 0. 1.]]\n",
"FE=\n",
"[[ 1. 0. 0.]\n",
" [-2. 1. 0.]\n",
" [ 1. 0. 1.]]\n",
"Here,EF=FE,so this shows that these two matrices commute\n"
]
}
],
"source": [
"from numpy import array,identity\n",
"E=identity(3)\n",
"E[1,:]=E[1,:]-2*E[0,:]\n",
"print 'E=\\n',E\n",
"F=identity(3)\n",
"F[2,:]=F[2,:]+F[0,:]\n",
"print 'F=\\n',F\n",
"print 'EF=\\n',E.dot(F)\n",
"print 'FE=\\n',F.dot(E)\n",
"print 'Here,EF=FE,so this shows that these two matrices commute'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex:1.4.5 Pg:25"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"E=\n",
"[[ 1. 0. 0.]\n",
" [-2. 1. 0.]\n",
" [ 0. 0. 1.]]\n",
"F=\n",
"[[ 1. 0. 0.]\n",
" [ 0. 1. 0.]\n",
" [ 1. 0. 1.]]\n",
"G=\n",
"[[ 1. 0. 0.]\n",
" [ 0. 1. 0.]\n",
" [ 0. 1. 1.]]\n",
"GE= [[ 1. 0. 0.]\n",
" [-2. 1. 0.]\n",
" [-2. 1. 1.]]\n",
"EG=\n",
"[[ 1. 0. 0.]\n",
" [-2. 1. 0.]\n",
" [ 0. 1. 1.]]\n",
"Here EG is not equal to GE,Therefore these two matrices do not commute and shows that most matrices do not commute.\n",
"GFE=\n",
"[[ 1. 0. 0.]\n",
" [-2. 1. 0.]\n",
" [-1. 1. 1.]]\n",
"EFG=\n",
"[[ 1. 0. 0.]\n",
" [-2. 1. 0.]\n",
" [ 1. 1. 1.]]\n",
"The product GFE is the true order of elimation.It is the matrix that takes the original A to the upper triangular U.\n"
]
}
],
"source": [
"from numpy import array,identity\n",
"E=identity(3)\n",
"E[1,:]=E[1,:]-2*E[0,:]\n",
"print 'E=\\n',E\n",
"F=identity(3)\n",
"F[2,:]=F[2,:]+F[0,:]\n",
"print 'F=\\n',F\n",
"G=identity(3)\n",
"G[2,:]=G[2,:]+G[1,:]\n",
"print 'G=\\n',G\n",
"print 'GE=',G.dot(E)\n",
"print 'EG=\\n',E.dot(G)\n",
"print 'Here EG is not equal to GE,Therefore these two matrices do not commute and shows that most matrices do not commute.'\n",
"print 'GFE=\\n',G.dot(F.dot(E))\n",
"print 'EFG=\\n',E.dot(F.dot(G))\n",
"print 'The product GFE is the true order of elimation.It is the matrix that takes the original A to the upper triangular U.'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex:1.5.1 Pg: 34"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"A= [[1 2]\n",
" [3 8]]\n",
"L= [[ 1. 0. ]\n",
" [ 0.33333333 1. ]]\n",
"U= [[ 3. 8. ]\n",
" [ 0. -0.66666667]]\n",
"LU=\n",
"[[ 3. 8.]\n",
" [ 1. 2.]]\n"
]
}
],
"source": [
"from numpy import mat,dot\n",
"from scipy.linalg import lu\n",
"A=mat([[1, 2],[3, 8]])\n",
"print 'A=',A\n",
"L=lu(A)[1]\n",
"U=lu(A)[2]\n",
"print 'L=',L\n",
"print 'U=',U\n",
"L=mat(L);U=mat(U)\n",
"print 'LU=\\n',dot(L,U)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex:1.5.2 Pg: 34"
]
},
{
"cell_type": "code",
"execution_count": 42,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"A= [[0 2]\n",
" [3 4]]\n",
"Here this cannot be factored into A=LU,(Needs a row exchange)\n"
]
}
],
"source": [
"from numpy import mat\n",
"A=[[0, 2],[3, 4]]\n",
"print 'A=',mat(A)\n",
"print 'Here this cannot be factored into A=LU,(Needs a row exchange)'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex:1.5.3 Pg: 34"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Given Matrix:\n",
"A= [[1, 1, 1], [1, 2, 2], [1, 2, 3]]\n",
"\n",
"L= [[ 1. 0. 0.]\n",
" [ 1. 1. 0.]\n",
" [ 1. 1. 1.]]\n",
"\n",
"U= [[ 1. 1. 1.]\n",
" [ 0. 1. 1.]\n",
" [ 0. 0. 1.]]\n",
"\n",
"LU=\n",
"[[ 1. 1. 1.]\n",
" [ 1. 2. 2.]\n",
" [ 1. 2. 3.]]\n",
"Here LU=A,from A to U there are subtraction of rows.Frow U to A there are additions of rows\n"
]
}
],
"source": [
"from numpy import mat\n",
"from scipy.linalg import lu\n",
"print 'Given Matrix:'\n",
"A=[[1, 1 ,1],[1, 2, 2],[1, 2, 3]]\n",
"print 'A=',A\n",
"L=lu(A)[1]\n",
"U=lu(A)[2]\n",
"print '\\nL=',L\n",
"print '\\nU=',U\n",
"print '\\nLU=\\n',mat(L)*mat(U)\n",
"print 'Here LU=A,from A to U there are subtraction of rows.Frow U to A there are additions of rows'\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex:1.5.4 Pg: 34"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"L=\n",
"[[ 1. 0. 0. ]\n",
" [ 0.74342501 1. 0. ]\n",
" [ 0.97556591 0.5785538 1. ]]\n",
"U=\n",
"[[ 1. 0. 0.]\n",
" [ 0. 1. 0.]\n",
" [ 0. 0. 1.]]\n",
"E=\n",
"[[ 1. 0. 0. ]\n",
" [-0.74342501 1. 0. ]\n",
" [ 0. 0. 1. ]]\n",
"F=\n",
"[[ 1. 0. 0. ]\n",
" [ 0. 1. 0. ]\n",
" [-0.97556591 0. 1. ]]\n",
"G=\n",
"[[ 1. 0. 0. ]\n",
" [ 0. 1. 0. ]\n",
" [ 0. -0.5785538 1. ]]\n",
"A=inv(E)*inv(F)*inv(G)*U\n",
"A=\n",
"[[ 1. 0. 0. ]\n",
" [ 0.74342501 1. 0. ]\n",
" [ 0.97556591 0.5785538 1. ]]\n",
"When U is identity matrix then L is same as A\n"
]
}
],
"source": [
"from numpy.random import random\n",
"from numpy import mat,eye\n",
"a=random()\n",
"b=random()\n",
"c=random()\n",
"L=mat([[1, 0, 0],[a, 1, 0],[b, c, 1]])\n",
"print 'L=\\n',L\n",
"U=eye(3)\n",
"print 'U=\\n',U\n",
"E=mat([[1, 0, 0],[-a, 1, 0],[0, 0, 1]])\n",
"print 'E=\\n',E\n",
"F=mat([[1, 0, 0],[0, 1, 0],[-b, 0, 1]])\n",
"print 'F=\\n',F\n",
"G=mat([[1, 0, 0],[0, 1, 0],[0, -c, 1]])\n",
"print 'G=\\n',G\n",
"print 'A=inv(E)*inv(F)*inv(G)*U'\n",
"A=(E**-1)*(F**-1)*(G**-1)*U#\n",
"print 'A=\\n',A\n",
"print 'When U is identity matrix then L is same as A'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex:1.5.5 Pg: 39"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"A=\n",
"[[ 1 -1 0 0]\n",
" [-1 2 -1 0]\n",
" [ 0 -1 2 -1]\n",
" [ 0 0 -1 2]]\n",
"U=\n",
"[[ 1. -1. 0. 0.]\n",
" [ 0. 1. -1. 0.]\n",
" [ 0. 0. 1. -1.]\n",
" [ 0. 0. 0. 1.]]\n",
"L=\n",
"[[ 1. 0. 0. 0.]\n",
" [-1. 1. 0. 0.]\n",
" [ 0. -1. 1. 0.]\n",
" [ 0. 0. -1. 1.]]\n",
"This shows how a matrix A with 3 diagnols has factors L and U with two diagnols.\n"
]
}
],
"source": [
"from numpy import mat\n",
"from scipy.linalg import lu\n",
"\n",
"A=mat([[1, -1, 0, 0],[-1, 2 ,-1, 0],[0, -1, 2 ,-1],[0, 0 ,-1 ,2]])\n",
"print 'A=\\n',A\n",
"L=lu(A)[1]\n",
"U=lu(A)[2]\n",
"print 'U=\\n',U\n",
"print 'L=\\n',L\n",
"print 'This shows how a matrix A with 3 diagnols has factors L and U with two diagnols.'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex:1.5.6 Pg: 36"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"a=\n",
"[[ 1 -1 0 0]\n",
" [-1 2 -1 0]\n",
" [ 0 -1 2 -1]\n",
" [ 0 0 -1 2]]\n",
"b=\n",
"[[1]\n",
" [1]\n",
" [1]\n",
" [1]]\n",
"Given Equation ,ax=b\n",
"U=\n",
"[[ 1. -1. 0. 0.]\n",
" [ 0. 1. -1. 0.]\n",
" [ 0. 0. 1. -1.]\n",
" [ 0. 0. 0. 1.]]\n",
"L=\n",
"[[ 1. 0. 0. 0.]\n",
" [-1. 1. 0. 0.]\n",
" [ 0. -1. 1. 0.]\n",
" [ 0. 0. -1. 1.]]\n",
"Augmented Matrix of L and b=\n",
"[[ 1. 0. 0. 0. 1. -1. 0. 0.]\n",
" [-1. 1. 0. 0. 0. 1. -1. 0.]\n",
" [ 0. -1. 1. 0. 0. 0. 1. -1.]\n",
" [ 0. 0. -1. 1. 0. 0. 0. 1.]]\n",
"c=\n",
"[[ 1.]\n",
" [ 2.]\n",
" [ 2.]\n",
" [ 2.]]\n",
"Augmented matrix of U and c=\n",
"[[ 1. -1. 0. 0. 1.]\n",
" [ 0. 1. -1. 0. 2.]\n",
" [ 0. 0. 1. -1. 2.]\n",
" [ 0. 0. 0. 1. 2.]]\n",
"x=\n",
"[[ 0.]\n",
" [ 6.]\n",
" [ 4.]\n",
" [ 2.]]\n"
]
}
],
"source": [
"from numpy import mat,hstack,zeros,arange\n",
"from scipy.linalg import lu\n",
"\n",
"a=mat([[1, -1, 0 ,0],[-1, 2, -1, 0],[0, -1, 2, -1],[0, 0, -1, 2]])\n",
"print 'a=\\n',a\n",
"b=mat([[1],[1],[1],[1]])\n",
"print 'b=\\n',b\n",
"print 'Given Equation ,ax=b'\n",
"\n",
"L=lu(a)[1]\n",
"U=lu(a)[2]\n",
"print 'U=\\n',U\n",
"print 'L=\\n',L\n",
"print 'Augmented Matrix of L and b='\n",
"A=hstack([L,U])\n",
"print A\n",
" \n",
"\n",
"c=zeros([4,1])#\n",
"n=4\n",
"#Algorithm Finding the value of c\n",
"c[0]=A[0,n]/A[0,0]\n",
"for i in range(1,n):\n",
" sumk=0#\n",
" for k in range(0,n-1):\n",
" sumk=sumk+A[i,k]*c[k]\n",
" \n",
" c[i]=(A[i,n+1]-sumk)/A[i,i]\n",
"\n",
"print 'c=\\n',c\n",
"x=zeros([4,1])\n",
"print 'Augmented matrix of U and c='\n",
"B=hstack([U,c])\n",
"print B\n",
"#Algorithm for finding value of x\n",
"x[n-1]=B[n-1,n]/B[n-1,n-1]\n",
"for i in arange(n-1-1,-1+1,-1):\n",
" sumk=0#\n",
" for k in range(i+1-1,n):\n",
" sumk=sumk+B[i,k]*x[k]\n",
" \n",
" x[i]=(B[i,n+1-1]-sumk)/B[i,i]\n",
"\n",
"print 'x=\\n',x\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex:1.5.7 Pg: 39"
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"A=\n",
"[[1 1 1]\n",
" [1 1 3]\n",
" [2 5 8]]\n",
"L=\n",
"[[ 1. 0. 0. ]\n",
" [ 0.5 1. 0. ]\n",
" [ 0.5 1. 1. ]]\n",
"U=\n",
"[[ 2. 5. 8. ]\n",
" [ 0. -1.5 -1. ]\n",
" [ 0. 0. -2. ]]\n",
"P=\n",
"[[ 0. 0. 1.]\n",
" [ 0. 1. 0.]\n",
" [ 1. 0. 0.]]\n",
"PA=\n",
"[[ 2. 5. 8.]\n",
" [ 1. 1. 3.]\n",
" [ 1. 1. 1.]]\n",
"LU=\n",
"[[ 2. 0. 0. ]\n",
" [ 0. -1.5 -0. ]\n",
" [ 0. 0. -2. ]]\n"
]
}
],
"source": [
"from numpy import mat\n",
"from scipy.linalg import lu\n",
"\n",
"A=mat([[1, 1, 1],[1, 1, 3],[2, 5, 8]])\n",
"print 'A=\\n',A\n",
"P=lu(A)[0]\n",
"L=lu(A)[1]\n",
"U=lu(A)[2]\n",
"print 'L=\\n',L\n",
"print 'U=\\n',U\n",
"print 'P=\\n',P\n",
"print 'PA=\\n',(P*A)\n",
"print 'LU=\\n',(L*U)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex:1.6.1 Pg: 47"
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Given matrix:\n",
"[[ 2 1 1]\n",
" [ 4 -6 0]\n",
" [-2 7 2]]\n",
"Augmented matrix :\n",
"[[ 2. 1. 1. 1. 0. 0.]\n",
" [ 4. -6. 0. 0. 1. 0.]\n",
" [-2. 7. 2. 0. 0. 1.]]\n",
"R2=R2-2*R1,R3=R3-(-2)*R1\n",
"[[ 2. 1. 1. 1. 0. 0.]\n",
" [ 8. -4. 2. 2. 1. 0.]\n",
" [ 0. 8. 3. 1. 0. 1.]]\n",
"R3=R3-(-1)*R2\n",
"a=\n",
"[[ 2. 1. 1. 1. 0. 0.]\n",
" [ 8. -4. 2. 2. 1. 0.]\n",
" [ 8. 4. 5. 3. 1. 1.]]\n",
"[U inv(L)] :\n",
"[[ 2. 1. 1. 1. 0. 0.]\n",
" [ 8. -4. 2. 2. 1. 0.]\n",
" [ 8. 4. 5. 3. 1. 1.]]\n",
"R2=R2-(-2)*R3,R1=R1-R3\n",
"[[ 10. 5. 6. 4. 1. 1.]\n",
" [ 24. 4. 12. 8. 3. 2.]\n",
" [ 8. 4. 5. 3. 1. 1.]]\n",
"R1=R1-(-1/8)*R2)\n",
"[[ 13. 5.5 7.5 5. 1.375 1.25 ]\n",
" [ 24. 4. 12. 8. 3. 2. ]\n",
" [ 8. 4. 5. 3. 1. 1. ]]\n",
"[I inv(A)]:\n",
"[[ 1. 0.42307692 0.57692308 0.38461538 0.10576923 0.09615385]\n",
" [ 6. 1. 3. 2. 0.75 0.5 ]\n",
" [ 1.6 0.8 1. 0.6 0.2 0.2 ]]\n",
"inv(A):\n",
"[[ 0.38461538 0.10576923 0.09615385]\n",
" [ 2. 0.75 0.5 ]\n",
" [ 0.6 0.2 0.2 ]]\n"
]
}
],
"source": [
"from numpy import mat,shape,nditer,hstack,eye,nditer\n",
"\n",
"print 'Given matrix:'\n",
"A=mat([[2, 1, 1],[4, -6, 0],[-2, 7, 2]])\n",
"print A\n",
"m=int(shape(A)[0])\n",
"n=int(shape(A)[1])\n",
"print 'Augmented matrix :'\n",
"a=hstack([A,eye(n)])\n",
"print a\n",
"print 'R2=R2-2*R1,R3=R3-(-2)*R1'\n",
"def fun1(a,x,y,n):\n",
" import numpy as np\n",
" z=[]\n",
" for xx,yy in nditer([a[x-1],a[y-1]]):\n",
" z.append(xx-n*yy)\n",
" a[x-1]=z\n",
" return a \n",
"\n",
" \n",
"a=fun1(a,2,1,-2)\n",
"a=fun1(a,3,1,-1)\n",
"print a\n",
"print 'R3=R3-(-1)*R2'\n",
"a=fun1(a,3,2,-1) # a(3,:)-(-1)*a(2,:)#\n",
"print 'a=\\n',a\n",
"print '[U inv(L)] :\\n',a\n",
"print 'R2=R2-(-2)*R3,R1=R1-R3'\n",
"a=fun1(a,2,3,-2)\n",
"a=fun1(a,1,3,-1)\n",
"print a\n",
"print 'R1=R1-(-1/8)*R2)'\n",
"a=fun1(a,1,2,-1./8)\n",
"print a\n",
"\n",
"a[0]=a[0]/a[0,0]\n",
"a[1]=a[1]/a[1,1]\n",
"print '[I inv(A)]:'\n",
"a[2]=a[2]/a[2,2]\n",
"print a\n",
"print 'inv(A):'\n",
"print a[:,range(3,6)]"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex:1.6.2 Pg:51"
]
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"R= [[1 2]]\n",
"Transpose of the given matrix is : [[1]\n",
" [2]]\n",
"The product of R & transpose(R)is : [[5]]\n",
"The product of transpose(R)& R is : [[1 2]\n",
" [2 4]]\n",
"Rt*R and R*Rt are not likely to be equal even if m==n.\n"
]
}
],
"source": [
"from numpy import mat,transpose\n",
"R=mat([1, 2])\n",
"print 'R=',R\n",
"Rt=transpose(R)\n",
"print 'Transpose of the given matrix is :',Rt\n",
"print 'The product of R & transpose(R)is :',(R*Rt)\n",
"print 'The product of transpose(R)& R is :',(Rt*R)\n",
"print 'Rt*R and R*Rt are not likely to be equal even if m==n.'"
]
}
],
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