{
"cells": [
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"source": [
"# Chapter 5 - Vectors & Matrices"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex5.2 Pg 103"
]
},
{
"cell_type": "code",
"execution_count": 52,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"u+v = [[ 3 -2 4]]\n",
"5*u = [[ 10 15 -20]]\n",
"-v = [[-1 5 -8]]\n",
"2*u-3*v = [[ 1 21 -32]]\n",
"dot product of the two vectors, k = u.v = [[-45]]\n",
"norm or length of the vector u = 5.3852\n"
]
}
],
"source": [
"from __future__ import division\n",
"from numpy import mat\n",
"from numpy.linalg import det\n",
"from numpy import mat, add, dot, multiply, inner\n",
"u=mat([[2,3,-4]])\n",
"v=mat([[1,-5,8]])\n",
"print \"u+v = \",add(u,v)\n",
"print \"5*u = \",multiply(5,u)\n",
"print \"-v = \",multiply(-1,v)\n",
"print \"2*u-3*v = \",add(multiply(2,u),multiply(-3,v))\n",
"print 'dot product of the two vectors, k = u.v = ',inner(u,v)\n",
"from numpy.linalg import norm\n",
"l=norm(u)# \n",
"print 'norm or length of the vector u = ',round(l,4)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex5.3 Pg 104 "
]
},
{
"cell_type": "code",
"execution_count": 26,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"2*u-3*v =\n",
"[[ 1]\n",
" [ 9]\n",
" [-2]]\n",
"The dot product of the two vectors u and v is: [[20]]\n",
"norm or length of the vector u = 7.0711\n"
]
}
],
"source": [
"u=mat([[5],[3],[-4]])\n",
"v=mat([[3],[-1],[-2]])\n",
"print \"2*u-3*v =\\n\",add(multiply(2,u),multiply(-3,v))\n",
"print 'The dot product of the two vectors u and v is:', sum(multiply(u,v))\n",
"l=norm(u)#\n",
"print 'norm or length of the vector u = ',round(l,4)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex5.5 Pg 105"
]
},
{
"cell_type": "code",
"execution_count": 33,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The addition of the two matrices A and B is:\n",
"[[ 5 4 11]\n",
" [ 1 1 -2]]\n",
"\n",
"The multiplication of a vector with a scalar is:\n",
"[[ 3 -6 9]\n",
" [ 0 12 15]]\n",
"\n",
"2*A-3*B = \n",
"[[-10 -22 -18]\n",
" [ -3 17 31]]\n"
]
}
],
"source": [
"A=mat([[1,-2,3],[0,4,5]])\n",
"B=mat([[4,6,8],[1,-3,-7]])\n",
"k=add(A,B)\n",
"print 'The addition of the two matrices A and B is:\\n',k\n",
"m=multiply(3,A)\n",
"print '\\nThe multiplication of a vector with a scalar is:\\n',m\n",
"p=add(multiply(2,A),multiply(-3,B))\n",
"print \"\\n2*A-3*B = \\n\",p"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex5.6 Pg 106"
]
},
{
"cell_type": "code",
"execution_count": 36,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"product of a and b is : [[8]]\n",
"product of p and q is: [[32]]\n"
]
}
],
"source": [
"a=mat([[7,-4,5]])\n",
"b=mat([[3,2,-1]])\n",
"k=inner(a,b)\n",
"print 'product of a and b is : ',k\n",
"p=mat([[6,-1,8,3]])\n",
"q=mat([[4,-9,-2,5]])\n",
"l=inner(p,q)\n",
"print 'product of p and q is:',l"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex5.7 Pg 107"
]
},
{
"cell_type": "code",
"execution_count": 40,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"A*B = \n",
"[[ 17 -6 14]\n",
" [ -1 2 -14]]\n",
"A*B = \n",
"[[ 5 2]\n",
" [15 10]]\n",
"B*A = \n",
"[[23 34]\n",
" [-6 -8]]\n",
"matrix mulitplication is not commutative since AB may not be equal to BA\n"
]
}
],
"source": [
"A=mat([[1 ,3],[2, -1]])\n",
"B=mat([[2, 0, -4],[5, -2, 6]])\n",
"print \"A*B = \\n\", dot(A,B)\n",
"A=mat([[1, 2],[3, 4]])\n",
"B=mat([[5, 6],[0, -2]])\n",
"print \"A*B = \\n\",dot(A,B)\n",
"print \"B*A = \\n\", dot(B,A)\n",
"print 'matrix mulitplication is not commutative since AB may not be equal to BA'\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex5.8 Pg 109"
]
},
{
"cell_type": "code",
"execution_count": 47,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"for the function f(x)=2x**2-3x+5,f(A) is :\n",
"[[ 16. -18.]\n",
" [-27. 61.]]\n",
"for the function g(x)=x**2+3x-10,g(A) is\n",
"[[ 0. 0.]\n",
" [ 0. 0.]]\n"
]
}
],
"source": [
"from numpy import identity as idt\n",
"A=mat([[1, 2],[3, -4]])\n",
"A2=dot(A,A) #multiplying A by itself\n",
"A3=dot(A2,A)\n",
"f=add(add(multiply(2,A2),multiply(-3,A)),multiply(5,idt(2)))\n",
"print 'for the function f(x)=2x**2-3x+5,f(A) is :\\n',f\n",
"g=add(add(A2,multiply(3,A)),multiply(-10,idt(2)))\n",
"print 'for the function g(x)=x**2+3x-10,g(A) is\\n',g"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex5.9 Pg 110"
]
},
{
"cell_type": "code",
"execution_count": 50,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"A*b = \n",
"[[1 0 0]\n",
" [0 1 0]\n",
" [0 0 1]]\n",
"since A*B is identity matrix,A and B are invertible and inverse of each other\n"
]
}
],
"source": [
"A=mat([[1 ,0 ,2],[2 ,-1, 3],[4, 1, 8]])\n",
"B=mat([[-11, 2 ,2],[-4, 0 ,1],[6, -1, -1]])\n",
"print \"A*b = \\n\",dot(A,B)\n",
"print 'since A*B is identity matrix,A and B are invertible and inverse of each other'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex5.10 Pg 111"
]
},
{
"cell_type": "code",
"execution_count": 53,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"determinant of A 7.0\n",
"determinant of B 16.0\n",
"determinant of C -81.0\n"
]
}
],
"source": [
"A=mat([[5 ,4],[2, 3]])\n",
"print 'determinant of A',det(A)\n",
"B=mat([[2, 1],[-4, 6]])\n",
"print 'determinant of B',det(B)\n",
"C=mat([[2, 1, 3],[4, 6, -1],[5 ,1 ,0]])\n",
"print 'determinant of C',det(C)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex5.13 Pg 115"
]
},
{
"cell_type": "code",
"execution_count": 60,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"x = [[ 2.]]\n",
"y = [[-1.]]\n",
"z = [[ 3.]]\n"
]
}
],
"source": [
"A=mat([[1, 2, 1],[2, 5, -1],[3, -2, -1]]) #left hand side of the system of equations\n",
"B=mat([[3] ,[-4] ,[5]]) #right hand side or the constants in the equations\n",
"from numpy import divide\n",
"from numpy.linalg import solve\n",
"X=divide(A,B) # #unique solution for the system of equations\n",
"X = solve(A, B)\n",
"print \"x = \",X[0]\n",
"print \"y = \",X[1]\n",
"print \"z = \",X[2]"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex5.14 Pg 116"
]
},
{
"cell_type": "code",
"execution_count": 64,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Inverse of A = \n",
"[[-11. 2. 2.]\n",
" [ -4. 0. 1.]\n",
" [ 6. -1. -1.]]\n"
]
}
],
"source": [
"A=mat([[1 ,0 ,2],[2, -1, 3],[4, 1, 8]])\n",
"A_inv = A**-1\n",
"print \"Inverse of A = \\n\", A_inv"
]
}
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