Find two unitary matrices in complex of order 3x3 with AB=exponentiol^(4pi/3)BA?
912400
Ok then, give me 5 years !
683943
Any complex number can be expressed as r*e^(i*theta).\n\n2^i = r*e^(i*theta)\n\nlogs on both sides\n\nln(2^i) = ln (r) + ln (e^(i*theta))\ni*ln(2) = ln(r) + i*theta\n\nequate real and imag parts:\n\nln(r) = 0\nso\nr=1\n\ntheta = ln(2)\n\nso the answer is 2^i = e^(i*ln(2))\n\nThere's an identity to do that more easily, but I can't remember it - it's something like this:\n\na^b = fn(e^b)
572398
first multiply i by everything in the first bracket.\nso -8i * i + 3 * i + 2(2-5i)\n\nthen do same with 2nd bracket\nso -8i^2 + 3i + 2*2 - 5i*2\n= -8i^2 + 3i + 4 - 10i\n\nnow collect all the like terms\n-8i^2 + 3i - 10i + 4\n= -8i^2 - 7i + 4\n\nnow if i is the imaginary number i^2 would be -1.\nso it would be 8-7i +4 = 12-7i\n\nhope that helps. laterz
531417
(a+b)(b^6-ab^5+a^2b^4-a^3b^3+a^4b^2-a^5b+a^6)\nGood Night, and Good Luck!
737238
the 3 x 3 matrix A times the 3 x 3 matix B will yield a 3 x 3 matrix AB\n\nAB1,1 = (A1,1)(B1,1) + (A1,2)(B2,1) + (A1,3)(B3,1)\nAB1,1 = (-3)(-1) + (6)(0) + (2)(-5) = -7\n\nAB1,2 = (A1,1)(B1,2) + (A1,2)(B2,2) + (A1,3)(B3,2)\nAB1,2 = (-3)(8) + (6)(-3) + (2)(0) = -42\n\nAB2,1 = (A2,1)(B1,1) + (A2,2)(B2,1) + (A2,3)(B3,1)\nAB2,1 = (-4)(-1) + (8)(0) + (1)(-5) = -1\n\netc.
442558
e^(pi*sqrt(163))= 262,537,412,640,768,744
761571
1.(4x+1)(16x^2-4x+1)\n\n2.(1-5x)(1+5x+25x^2)\n\n3.(2xy^2+3)(4x^2y^4-6xy^2+9)\n\n4.b^3(a^3-b^3)\nb^3(a-b)(a^2+ab+b^2)\n\n5.(3x-2y)(9x^2+6xy+4y^2)
652728
a=b=i^(1/3)\ni=-1\nab+a^2b+ab^2 = [i^(1/3)*i^(1/3)] + [i^(2/3)*i^(1/3)] + [i^(1/3)*i^(2/3)]\n= [i^(2/3)] + i + i\n= -2+(-1)^(2/3)
815664
Imaginary part of\n(e^(i*x)+ e^(2*i*x) + ...)\n=Imag(e^(i*x)/(1- e^(i*x))\n=Imag(e^(i*x)/\n  (1-cos(x)-i*sin(x)))\n=Imag(e^(i*x)\n   (1-cos(x)+i*sin(x))/\n   ((1-cos(x))^2+(sin(x)^2))\n=((cos(x))\n(sin(x))+(sin(x))\n(1-cos(x)))/(2-2*cos(x)\n=sin(x)/(2(1-cos(x)))
61844
When we ask for an inverse matrix, we want a new matrix that, when multiplied by the given matrix, will give the identity matrix.  To find this matrix, we can proceed as follows:\nWrite down your matrix and write the identity matrix to the right of it.\n|4  -1|   |1   0|\n|3   5|   |0   1|\nNow the objective is to perform operations on the rows of the matrix to get it to look like the identity matrix, while doing the same operations on the nearby identity matrix.  Like so:\nMultiply the top row by 5.\n|20  -5|   |5   0|\n|3    5|   |0   1|\nNow add the rows together and put them on the top row.\n|23  0|   |5   1|\n|3   5|   |0   1|\nNow divide the top row by 23.\n|1   0|   |5/23  1/23|\n|3   5|   |0        1|\nNext, multiply the top row by -3 and add it to the bottom row (combining steps)\n|1   0|   |5/23     1/23|\n|0   5|   |-15/23  20/23|\nFinally, divide the bottom row by 5\n|1   0|   |5/23   1/23|\n|0   1|   |-3/23  4/23|\nLet's test our result.  Multiply the given matrix by our supposed inverse matrix.\n|4   -1|   |5/23   1/23|\n|3    5| x |-3/23  4/23|\nRemember to use matrix multiplication:\n|20/23+3/23   -4/23+4/23|\n|15/23-15/23  3/23+20/23|\nWhich is equal to\n|1   0|\n|0   1|\nThe identity matrix. So your inverse matrix is:\n|5/23   1/23|\n|-3/23  4/23|\nYou're welcome.
752603
The inversion of a 3x3 matrix is very easy to solve, however the equation is fairly long. You can find it on wikipedia at http://en.wikipedia.org/wiki/Inverse_matrix#Inversion_of_3_x_3_matrices
273264
Lets call this matrix A.\n[ 2 0 4 ]\n[ 3 1 5 ]\n[ -1 1 -2 ] = A\n\nNow we need to solve the equation AV = I\nthen V is the inverse matrix.  Lets write this out\n\n[ 2 0 4 ] ` ` ` ` [ 1 0 0 ]\n[ 3 1 5 ] ` ` ` ` [ 0 1 0 ]\n[ -1 1 -2 ] V = [ 0 0 1 ]\n\nNow just multiply elementary matrices (row operations) to the left hand side of both sides of the equation until the left-hand entity is the identity.  Multiply the following matrix on the left-hand-side both sides [[1/2 0 0][0 1 0][0 0 -1]], that is replace the first row with 1/2 times the first row, and replace the third row with -1 times the third row:\n\n[ 1 0 2 ] ` ` ` ` [ 1/2 0 0 ]\n[ 3 1 5 ] ` ` ` ` [ 0 ` 1 0 ]\n[ 1 -1 2 ] V = [ 0 ` 0 -1 ]\n\napply [[1 0 0][[-3 1 0][0 0 1]] both sides, that is replace second row with second row plus negative-three times first row\n\n[ 1 0 2 ] ` ` ` ` [ 1/2 0 0 ]\n[ 0 1 -1 ] ` ` ` ` [ -3/2 1 0 ]\n[ 1 -1 2 ] V = [ 0 ` 0 -1 ]\n\nmultiply both sides on the left by the matrix [[1 0 0][0 1 0][ -1 0 1]], that replace third row with third row plus negative one times first row\n\n[ 1 0 2 ] ` ` ` ` [ 1/2 0 0 ]\n[ 0 1 -1 ] ` ` ` ` [ -3/2 ` 1 0 ]\n[ 0 -1 0 ] V = [ -1/2 ` 0 -1 ]\n\nmultiply by [[1 0 0][0 0 -1][0 1 0]], that is place opposite of row 3 for row 2 and row 2 for row 3\n\n[ 1 0 2 ] ` ` ` ` [ 1/2 0 0 ]\n[ 0 1 0 ] ` ` ` ` [ 1/2 0 1 ]\n[ 0 1 -1 ] V = [ -3/2 1 0 ]\n\nnow apply [[1 0 0][0 1 0][ 0 -1 1]] both sides, that is replace third row with third row plus -1 times second row\n\n[ 1 0 2 ] ` ` ` ` [ 1/2 0 0 ]\n[ 0 1 0 ] ` ` ` ` [ 1/2 0 1 ]\n[ 0 0 -1 ] V = [ -2 ` 1 -1 ]\n\napply [[ 1 0 2][0 1 0][0 0 -1]] both sides, replace row 1 with row 1 plus 2 times row 3, and replace row 3 with negative 1 times row 3\n\n[ 1 0 0 ] ` ` ` ` [ -3/2 2 -2 ]\n[ 0 1 0 ] ` ` ` ` [ 1/2 0 1 ]\n[ 0 0 1 ] V = [ 2 -1 1 ]\n\nNow, since the left hand item is the identity, the inverse matrix is\n\n[ -3/2 2 -2 ]\n[ 1/2 0 1 ]\n[ 2 ` -1 1 ]\n\nThe product of all of the elementary operation matrices specified is also this matrix, and the reader can verity that AV = VA = I
287407
3(16 y_squared - 24xy + 9 x_squared)
927974
Multiply the coefficients and then add and subtract the exponents for the like bases. \n(7x^3y^-5)(4xy^3)= 28x^(3+1)y^(-5+3)= 28x^4y^-2
811614
Answer:\nThe problem to solve is: \nRemember, with complex variables, keep like terms involving i together...\n \nAdding Multiplication Sign: y*x2-4x-5\nAdding Multiplication Sign: y*x*2-4x-5\nAdding Multiplication Sign: y*x*2-4*x-5\nPoly Parsing: y*x*2-4*x-5\nSplitting y*x*2-4*x-5 \ny*x*2-4*x - 5 \n[Left Side]Poly Parsing: y*x*2-4*x\nSplitting y*x*2-4*x \ny*x*2 - 4*x \n[Left Side]Poly Parsing: y*x*2\nMultiply Term y with Term x equals Term xy\nMultiply Term xy with Term 2 equals Term 2xy\nAddTerm: Term (2xy)\nPoly =  2xy\nRepace with Reduced Left Side \ny*x*2 - 4*x \n2xy - 4*x \n[Right Side]Poly Parsing: 4*x\nMultiply Term 4 with Term x equals Term 4x\nAddTerm: Term (4x)\nPoly =  4x\nRepace with Reduced Right Side \ny*x*2 - 4*x \n2xy - 4x \nBecause of the minus sign 4x becomes -4x\n\nPoly Adding 2xy with -4x...\nAddTerm: Term (2xy)\nAddTerm: Term (-4x)\n... 2xy-4x\nPoly =  2xy-4x\nRepace with Reduced Left Side \ny*x*2-4*x - 5 \n2xy-4x - 5 \n[Right Side]Poly Parsing: 5\nAddTerm: Term (5)\nPoly =  5\nRepace with Reduced Right Side \ny*x*2-4*x - 5 \n2xy-4x - 5 \nBecause of the minus sign 5 becomes -5\n\nPoly Adding 2xy-4x with -5...\nAddTerm: Term (2xy)\nAddTerm: Term (-4x)\nAddTerm: Term (-5)\n... 2xy-4x-5\nPoly =  2xy-4x-5\nThe problem to solve is: yx2-4x-5 \n2xy-4x-5
201268
In doing this, you have to group the real parts together and the imaginary parts together. i.e.,\n(a + bi) + (c + di)\n= (a + c) + (bi + di) or\n= (a + c) + (b + d)i\n\nThe standard form of a complex number is:\na + bi, where a and b are real numbers\n\nThus,\n(5 + 3i) + (2 + 4i)\n= (5 + 2) + (3 + 4)i\n= 7 + 7i\n\nThe same goes with the second one, but with a slight twist:\ni - (6 + i) + (4 - 2i)\n= i - 6 - i + 4 - 2i\n= -6 + 4 + i - i - 2i\n= -2 - 2i\n\nIn multiplying, treat them as algebraic expression and remember that i² = -1\n\n-4i(3 - 7i)\n= -12i + 28i²\n= -12i + 28(-1)\n= -12i -28\n= -28 - 12i\n\nHope this helps&_&
16666
it called the Cayley-Hamilton Theorem, type it in google and its proved in many places;\n\n\nhttp://delta.cs.cinvestav.mx/~mcintosh/comun/summer99/mcintosh/node9.html  <---this one is good :)\n\nLet B = x*I - A, where I is the identity matrix of the same size as A.  \nLet C be the classical adjoint of B, so that BC = (det B)I. Recall \nf(x) = det(B), by definition. We wish to show that f(A) = 0. It \nsuffices to show that f(A)*e(k) = 0 for 1 <= k <= n, where e(k) is the \nk-th unit vector, so fix a value of k in that range.\n\nNow x = A implies that for all j,\n\n   x*e(j) = A*e(j),\n\n             n\n          = SUM A(i,j)*e(i),\n            i=1\n   \n      n                     n\n   x*SUM delta(i,j)*e(i) = SUM A(i,j)*e(i),\n     i=1                   i=1\n\n    n\n   SUM [x*delta(i,j)-A(i,j)]*e(i) = 0,\n   i=1\n\n    n\n   SUM B(i,j)*e(i) = 0,\n   i=1\n\n           n\n   C(j,k)*SUM B(i,j)*e(i) = 0,\n          i=1\n\n    n\n   SUM B(i,j)*C(j,k)*e(i) = 0.\n   i=1\n\nThen, summing over all j, reversing the order of summation, and using\nthe definition of the classical adjoint,\n\n    n   n\n   SUM SUM B(i,j)*C(j,k)*e(i) = 0,\n   j=1 i=1\n\n    n    n\n   SUM [SUM B(i,j)*C(j,k)]*e(i) = 0,\n   i=1  j=1\n\n    n\n   SUM [det(B)*delta(i,k)]*e(i) = 0,\n   i=1\n\n    n\n   SUM [f(x)*delta(i,k)]*e(i) = 0,\n   i=1\n\n    n\n   SUM [f(A)*delta(i,k)]*e(i) = 0,\n   i=1\n\n         n\n   f(A)*SUM [delta(i,k)]*e(i) = 0,\n        i=1\n\n   f(A)*e(k) = 0.\n\nThis proves that f(A) annihilates all the unit vectors, so must be\nthe zero matrix, and f(A) = 0.\n\n\nsecond one taken off the dr math website, but the link above might be easier to understand.