Olver P.J.'s AIMS lecture notes on numerical analysis PDF

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8. If A is a nonsingular matrix, so is AT , and its inverse is denoted A−T = (AT )−1 = (A−1 )T . 16) Thus, transposing a matrix and then inverting yields the same result as first inverting and then transposing. Proof : Let X = (A−1 )T . 14), X AT = (A−1 )T AT = (A A−1 )T = I T = I . The proof that AT X = I is similar, and so we conclude that X = (AT )−1 . D. A particularly important class of square matrices is those that are unchanged by the transpose operation. 3/15/06 43 c 2006 Peter J. 9. A square matrix is called symmetric if it equals its own transpose: A = AT .

A2n .. amn 1 b2  ..   . 6) bm which is an m × (n + 1) matrix obtained by tacking the right hand side vector onto the original coefficient matrix. The extra vertical line is included just to remind us that the last column of this matrix plays a special role. ,   x + 2 y + z = 2, 1 2 1 2 2 x + 6 y + z = 7, is M = 2 6 1 7. 7) 1 1 4 3 x + y + 4 z = 3, Note that one can immediately recover the equations in the original linear system from the augmented matrix. Since operations on equations also affect their right hand sides, keeping track of everything is most easily done through the augmented matrix.

The first equation says a = 1; substituting into the second, we find b = 0; the final equation yields c = 1. We then use Back Substitution to solve the upper triangular system        2 x + y + z = 1, 1 a 2 1 1 x 0 3 which is 0y =  b = 0, 3y = 0, 1 c 0 0 −1 z − z = 1. We find z = −1, then y = 0, and then x = 1, which is indeed the solution. Thus, once we have found the L U factorization of the coefficient matrix A, the Forward and Back Substitution processes quickly produce the solution to any system A x = b.

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