Matrix help
post the question!?
i know that x y z= (6,4,3)
You have to calculate the inverse of A. A is some matrix: \[A = \left[\begin{matrix} ? & ? & ?\\ ? & ? & ? \\ ? & ? & ?\end{matrix}\right]\]satisfying the equation \[A \vec{x} = \vec{b}\]. There are some decent guides on using determinants and co-factors: http://www.wikihow.com/Inverse-a-3X3-Matrix https://www.youtube.com/watch?v=pgqyULjZgbU or using row operations: http://www.mathsisfun.com/algebra/matrix-inverse-row-operations-gauss-jordan.html https://www.youtube.com/watch?v=rnyLW2-lL7o
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From that system of equations you'll get an matrix A. The first assignment is to calculate an inverse of A (A^(-1) is the inverse matrix of A).
OK.
theres no A=[ ]
You get them from the system of equations - like we did yesterday. x + y + 0z = 10 2x + 0y - z = 9 0x + y - 3z = -5 AX = B A = matrix X = vector [x, y, z] B = vector The the coefficients in the left side of the equations will go into matrix A and the number on the right side in the equation will go into vector B. Right now we're only interested in matrix A since our first assignments is to calculate A^(-1). The third row of A is \[A = \left[\begin{matrix} ? & ? & ?\\ ? & ? & ? \\ 0 & 1 & -3 \end{matrix}\right]\]
ok..
Fill the matrix with the correct values and look at the guides I provided you to learn how to compute 3x3 matrix inverse. Also, if you are familiar with row operations I recommend you to look at those first.
@mathslover @Preetha
@OOOPS
Finding the inverse matrix is a nightmare when doing by hand!!!
i see that :(
use calculator to do it, please
2x2 is pice of cake. 3x3 is doable in a few minutes if you do some practice...
i dont know what the matrix is
unbelievable, you solve the problem by matrix method and you don't know what the matrix is, hehehe... fun!!
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