how to write a programme to calculate inverse of a matrix?
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numerical recipes in C.  Amazon has it.
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also try  http://mathworld.wolfram.com/ImplicitDifferentiation.html
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The system of equations:\n[a11 a12 a13; a21 a22 a23]\n[1 -3 4; 2 3 1]\n\nPerforming indicated row operation:\n[a11 a12 a13; a21' a22' a23']\n[1 -3 4; -1 -6 3]\n\na21' = a11 - a21 = 1 - 2 = -1\na22' = a12 - a22 = -3 - 3 = -6\na23' = a13 - a23 = 4 - 1 = 3
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If you have a system of equations with two variables of the form ax + by = c, dx + ey = f, where a,b,c,d,e,f are the coefficients, and x and y are variables, you can set up the equation as \n\n[a b][x] = [c]\n[d e][y] = [f]\n\n(ignore the fact there are two equals signs there, they're only there for formatting reasons)\n\nwhere the square brackets are a crude attempt on by behalf to 'draw' matrices and column vectors.\n\nTo solve the system you need to find the inverse of the matrix of the right hand side (assuming it's invertible to begin with), and "right multiply" (that is multiply from the right of each side of the equation) both sides of the equation.\n\nThat should give you:\n\n[1 0][x] = [ inverse of matrix ][c]\n[0 1][y] = [ ........................ ][f] \n\nThen it's just a matter of performing the calculation.\n\n(sorry for the crappy formatting of the post)
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Put the left side coefficients into matrix [A],\n[6 2]\n[5 6]\nand the right side values into matrix [B]\n[-16]\n[..4]\nthen tell the calculator to multiply [A] inverse * [B]\nthe result is the solution matrix\n[x]\n[y]\n\nnote, the inverse key looks like x^-1.  \n\nsolve the second system the same way.  coefficient matrix will be 3x3, answer matrix 3x1.
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Click ON\nClick PRGM\nSelect NEW tab and hit ENTER\n\nTo select DISP and PROMPT:\n Hit PRGM when creating or editing a program,\n Select Tab I/O and you will see you selection\n\nTo get the arrow, click STO-> just to the left of 1\n\n\nThis is the Program: (or the one I use)\n\n:Disp "AX^2+BX+C"\n\n:Prompt A\n:Prompt B\n:Prompt C\n:(-B+*square root*(B^2-4AC))/(2A)->D\n:(-B-*square root*(B^2-4AC))/(2A)->E\n:Disp D\n:Disp E\n\nTo run program, go to the main screen, click PRGM, select your program, hit ENTER, input values for A, B, and C, hit ENTER and you result will show your two possible answers.
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The inverse of a matrix\nThe inverse of a square n × n matrix A, is another n × n matrix denoted by A−1 such that\nAA−1 = A−1A = I\n\nfor a more detailed answer check out
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To find the inverse of a function you simply interchange the x and y and solve for y.\n\ny = 2x - 1\n\nso to find the inverse we say\n\nx = 2y - 1\n\nnow solve for y\n\nx + 1 = 2y\n\n(x + 1)/2 = y   \n\ntherefore the inverse of y = 2x - 1 is  y = (x + 1)/2
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x=-1/2.. firstly form a matrix with the following rows\n\nR1 (x -1/2 -1/2)\nR2 (-1/2 x -1/2)\nR3 (-1/2 -1/2 x)\n\nthink of R1, R2, R3 as ur A.\n\nNow Multiply A  with x.\n\nin this case, x will be the vector (a, b, c)transpose.\n\nthe result must be (0,0,0).\n\nso now u have a set of equations in matrix form. all u do is solve for x. apply gaussian elimination and voila, vous avez votre reponse.. shudnt be tough..
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here's an example\n\nwe want to reduce a matrix to u/t form\n\n1  10  -3       r1\n1  10   2       r2\n1  4     2       r3\n\n1  10  -3\n0  0     5      \n0  -6    5      \nr2-r1 \nr3-r1\n\n1  10  -3\n0  -6   5\n0   0   5      \n interchange rows 2 and 3\n\nupper triangular form\n\nthis is known as gaussian elimination\nand is used in solving systems of equations\n\ngo to the site:\n\nhttp://mathworld.wolfram.com/GaussianElimination.html\n\nthis gives a good example\n\nyou can also get help with other matrix\noperations at this site,plus access to the\nintegrator as well\n\ni hope that this helps
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well gauss jordan is a easy method for matrix manipulations. if u use reverse triangulation also in conjunction with it . it wud be much more simpler.\n\n2x-y+3z=8\n-x+2y+z=4\n3x+y-z=0\n\nthis is set of equations given.write them in matrix form\n\n|| 2  -1  3 ||  ||x||     ||8||\n|| -1  2  1 ||  ||y||  = ||4|| \n|| 3   1  -1||  ||z||     ||0||\n\nA    X= B\nnow the equation is in the form of AX=B\n\nnow our concentration will be on matrices A and B\n\nour aim is to make the lower triangle elements equal to 0.\n\nnow follow the steps one by one. apply the calculations on both A and B  matrices.\n\nwe have 3 rows in matrix A and let us name them as R1 , R2 and R3.\n\nreplace R2 by R2/-1 and R3 by R3/3\n\nnow the matrices will be\n\n||2   -1    3    || ||x||   ||8 ||\n||1   -2   -1    || ||y||= ||-4||\n||1   1/3  -1/3 || ||z||   ||0 ||\nreplace R2=R1-2xR2\n||2   -1    3     || ||x||    ||8 ||\n||0    3    5     || ||y||=  ||16||\n||1   1/3  -1/3 ||  ||z||   ||0 ||\ncheck once what we have done here. \nevery element of R2 is calculated from the preceding element in that colum by the formula R1-2xR2.\n\nthe elelment of colum one is\n2-2x1=0\nthe element of colum 2 is -1-(2x-2)=3\nthe element of colum 3 is 3-(2x-1)=5\n\nlike wise \nreplace R3=R1-2xR3\n||2   -1    3      || ||x||     ||8 ||\n||0    3    5      || ||y|| =  ||16||\n||0   -5/3  11/3 || ||z||     ||8 ||\n\nnow we got the first colum zero in the lower triangle.\n\nnow the second element in the third row needs to be zeroed.\nreplace R3=R3x(-3/5) or  R3=R3/(-5/3)\n||2   -1    3      || ||x||   ||  8   ||\n||0    3    5      || ||y||= ||  16  ||\n||0    1  11/5   || ||z||   ||-24/5 ||\nreplace R2=R2/3\n||2   -1    3      || ||x||   ||  8   ||\n||0    1    5/3   || ||y||= || 16/3 ||\n||0    1  11/5   || ||z||   ||-24/5 ||\n\nnow replace R3=R3-R2\n||2   -1    3      || ||x||   ||  8     ||\n||0    1    5/3   || ||y||= || 16/3   ||\n||0    0    8/15 || ||z||   ||-152/15 ||\n\nnow apply the final steps of reverse triangulation technique.\nfrom the matrix equation, if again converted in to normal equations \n\n2x-y+3z   =8\n  y+5z/3  =16/3\n    8z/15 =-152/15\nnow z= (-152/15) divided by (8/15)\n= (-152/15)/(8/15) =-152/8=-19\nthere fore z=-19\nnow substitute z=-19 in equation 2\n\ny+5x(-19)/3=16/3\ny=(16/3)+(95/3)=111/3=37\nsubstitute the values of y and Z in equation 1\n\n2x-37+3x-19=8\n2x=8+37+57=102\nX=102/2=51\nfinally we got the values of x = 51, y=37 and z=-19\n\nactually u can continue the guass jordan method without applying the reverse triangulation method to get the values of x, y \n\nand z directy from the matrix for that after making the lower triangle zero we have to make the upper triangle elements also \n\nequal to zero.\n\nlet us see how to do that.\n||2   -1    3      || ||x||   ||  8     ||\n||0    1    5/3   || ||y||= || 16/3   ||\n||0    0    8/15 || ||z||   ||-152/15 ||\nreplace R3=R2/(8/15)\n||2   -1    3     || ||x||   ||  8     ||\n||0    1    5/3  || ||y||= || 16/3   ||\n||0    0    1     || ||z||   || -19    ||\nR2=R2-5xR3/3\n||2   -1    3    || ||x||  ||  8     ||\n||0    1    0    || ||y||= ||  37    ||\n||0    0    1    || ||z||  || -19    ||\n\nR1=R1/3\n||2/3   -1/3    1    || ||x||   ||  8/3   ||\n||0       1     0     || ||y||= ||  37    ||\n||0       0     1     || ||z||   || -19    ||\n\nR1=R1-R3\n||2/3   -1/3    0    || ||x||   ||  65/3  ||\n||0       1     0     || ||y||= ||  37    ||\n||0       0     1     || ||z||   || -19    ||\nR1=R1x-3\n||-2    1     0    || ||x||   ||  -65   ||\n||0       1     0    || ||y||= ||  37    ||\n||0       0     1    || ||z||   || -19    ||\nR1=R1-R2\n||-2    0     0    || ||x||   ||  -102  ||\n||0       1     0    || ||y||= ||  37    ||\n||0       0     1    || ||z||   || -19    ||\nR1=R1/(-2)\n||1    0     0    || ||x||  ||  51    ||\n||0       1     0    || ||y||=||  37    ||\n||0       0     1    || ||z||  || -19    ||\nnow we can diretly read the matrix into equations as\nx=51, y=37 and z=-19\n\n\nhope und
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One is the inverse of the other
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here is a simple program algo\n\nprogram quad\nbegin\nvar\n     declare variables;\nget variable values of a,b,c for equation ax2+bx+c=0\nd=)b*b)-4*a*c;\nif(d<0)\nbegin\n\n   root are imaginary;\nend\nelse\n\n    root1=b(sqrt(d))/2a;\n    root2=-b(sqrt(d))/2a;\n ( note : formula i am not sure make changes as per required)\nprint the value of root1 and root2;\nend
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definitely use MATLAB!
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Well you can, in a sense.\nI don't know a SIMPLE way to do it, in the sense that there is a SIMPLE way to do matrix multiplication. But maybe the following explanation will be helpful:\n\nIf [A] x [B] = [C] (i.e., if the product of matrices A and B equals matrix C), then we can say that [C] / [B] = [A]. So in that sense, we are dividing C by B.\n\nBut what if we don't know matrix A. What if we know only [B] and [C]? How can we divide [C] by [B] to find [A]?\n\nThe answer is that we find the "inverse" of matrix B, and then MULTIPLY matrix C by this inverse matrix (which is shown as [B]^-1). We define [B]^-1 as the matrix for which [B] x [B]^-1 = the identity matrix. (The identity matrix is a square matrix that has 1's on its main diagonal and 0's everywhere else.)\n\nThe problem, then, becomes how to find the inverse of [B]. Fortunately, there are procedures for this, and you can probably find them on line using Google. Not all matrices can be inverted. But if B can be inverted, then you have:\n\n[C] / [B] = [C] x [B]^-1 = [A]\n\nNote that if we substitute [A] x [B] for [C], then this becomes;\n([A] x [B]) / [B] = [A] x [B] x [B]^-1 = [A] x identity matrix = [A]
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U can use either Kruskal's algorithm or prims algorithm. More information about these algorithm u can find on internet.\n\nPrims algorithm is based on adjacency matrix.
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you can use linear algebra, look up gauss jordan elimination. You will want to put each equation in its own row inside a matrix. Then you will have a coefficient matrix and an augmented matrix. Row reduce the left side and you will have your desired answers on the right. If you have a TI-83+ type in your matrix and hit RREF([A]) and it will do this for you. Go to the link I provided and scroll down a bit and it will show you.