Need help on 4x4 Matrix inverse?
458222
0 1 0 0\n0 0 1 0\n1 0 0 0\n0 0 0 1\nPut the identity matrix behind your matrix.\nTurn you matrix into the identity matrix using row operations.  The matrix in the back will be the inverse.\n0 1 0 0  1 0 0 0 \n0 0 1 0  0 1 0 0 \n1 0 0 0  0 0 1 0 \n0 0 0 1  0 0 0 1\nSwap row 1 with row 3\n1 0 0 0  0 0 1 0 \n0 0 1 0  0 1 0 0 \n0 1 0 0  1 0 0 0 \n0 0 0 1  0 0 0 1\nSwap row 2 with row 3\n1 0 0 0  0 0 1 0 \n0 1 0 0  1 0 0 0\n0 0 1 0  0 1 0 0 \n0 0 0 1  0 0 0 1\nThus, the inverse is\n0 0 1 0\n1 0 0 0\n0 1 0 0 \n0 0 0 1
1077320
My TI-83plus can do that...but it has to be an invertible matrix.  If the determinate is zero, then it can not be done
595375
divide the first row by 2\n1 1/2\n1  -1\n\nSubtract the first row from the second row:\n1  1/2\n0  -3/2\n\nMutiply the second row by -2/3\n1  1/2\n0   1\n\nSubtract 1/2 of the second row from the first row:\n1  0\n0  1\n\nAnd you have the identity matrix
459985
Find the inverse of the matrix\n5 -3\n4  2\nA 2x2 matrix has a formula for the inverse.\nThe inverse of the matrix\na b\nc d\nis \nd/(ad-bc) -b/(ad-bc)\n-c/(ad-bc) a/(ad-bc)\nThus, for the above problem\na = 5, b = -3, c = 4 and d = 2\nad-bc = 5(2) - (-3)(4) = 10+12 = 22\nThe inverse is\n1/11  3/22\n-2/11 5/22\nMultiply this times the  matrix\n-2\n5\n= \n(1/11)(-2) + (3/22)(5)\n(-2/11)(-2) + (5/22)(5)\n= \n1/2\n3/2\nThus, x = 1/2 and y = 3/2 or (1/2,3/2)\nFor the second one the matrix is\n3 2 2\n1 2 -1\n2 -4 1\nThe inverse of this is\n1/14 5/14 3/14\n3/28 1/28 -5/28 \n2/7 -4/7  -1/7\nTake the inverse times the matirix\n3\n5\n0\nto get the matrix\n2\n1/2\n-2\nAnswer: x = 2, y = 1/2 and z =-2 or (2,1/2,-2)
168663
INVERSE:\nX+2 = 4^ (Y-1)\nLOG4(X+2) = LOG4 4 4^(y-1)\nLOG4 (X+2) = Y-1\nY = LOG4 (X+2) + 1\nP.S. 4 IS THE BASE OF LOG.
102340
I assume you mean invertible, not convertible.\n\nIt depends on how detailed you want to get. If you do not want to dig in extremely deeply, you can just do it by construction. \n\n1. A symmetric matrix M has an orthogonal decomposition\n\nM=O'DO\n\nwhere O is an orthogonal matrix, O'O=I, the identity, and D is a diagonal matrix with eigenvalues of M down the diagonal.\n\n2. If M is positive definite, then the diagonal elements of D are >0\n\n3. Produce the matrix \n\nA=O'(D^(-1))O\n\n\n4. Verify that A is the inverse of M\n\ni.e., \n\nA*M=M*A=I
716277
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
787922
An identity matrix must be square, so 5 is true.  Inverses must multiply to the same answer in either order, so they must be square, so 2 is true.\n\nAny square matrix multiplied by its identity will equal itself, so 4 is false and 1 is true.\n\nInverses must multiply to come out the identity which is impossible if one matrix is all zeros, so 3 is true.
685700
First convert into a matrix form which would be\n\n1    1    1\n1    2    3\n1    3    7\n\np\nq\nr\n\n= \n\n1\n4\n13\n\nTake the square matrix (3x3) on the other side so u ll need its inverse (i have directly calculated the inverse which is)\n\n5/2       -2      1/2                        \n-2         3      -1                           \n1/2       -1     1/2   \n\nMULTIPLY BY    \n\n1\n4\n13\n\n& the result is\n\n\n 1        p\n-3        q\n 3     =  r\n\nthis is the answer of the equation
721099
4a + 3b = 4x4 + 3x(-4) = 16 - 12 = 4
829618
1.find the midpoint of BC usingthe midpoint formula\n(x1+x2/2,y1+y2/2) say (x,y)\n2.find the slope os BC using theformula \ny2/y1/x2-x1 say m\n3.the slope of the perpendicular bisector will be the negative inverse of this -1/m\n4. now use the point (x,y) and slope (-1/m) and find the equation of the perpendicular bisector
599592
At first I thought you were wanting the orthogonal matrices, but your example shows you are not. Orthogonal matrices are those whose inverse is their transpose. Yours are ones where the elements of the transpose are the inverse of the elements of the original.\n\nCertainly such a matrix has to have +-1 on the diagonal. Your matrix is singluar and any 2x2 matrix with your property and all 1 on the diagonal will be also. This means that one eigenvalue is 0 and the other is 2.\n\nI don't see anything special for 3x3 offhand, but I will look further.
438919
I dont know how to put this up here. I tried to put it in a nice format, but it would accept.\nBut you put the following in one matrix \n-2  6\n+2  2\n\nThen you put a * (multiply)\n\nThen you put the following in one more matrix\n\nx\ny\n\n\nThen put an ' = ' symbol\n\nThen put the following in a matrix\n\n40\n8
300607
The formula for the inverse of a matrix also requires the determinant of the whole matrix. Minors are determinants of submatrices where one row and one column are deleted. See the source for the equation.
957836
4x+(-5y)+(-6x)+3y\n4x+(-6x)+(-5y)+3y\n=-2x-2y
1066492
x = -4\n10points thank u!
889697
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