Find the LU factorization of matrix A: (1 2 1) (3 7 6) (1 4 8) and use the answe

Question

Find the LU factorization of matrix A: (1 2 1) (3 7 6) (1 4 8) and use the answer to solve for x, given: ( 5 ) Ax= ( 8 ) ( -12 )

Explanation / Answer

First find U for A and keep track of the row operations. Since we know that if we work consistently from the top downward on A to get U we can use some simple inspections to get L we will proceed in that manner. If you do not know this shortcut method of finding L, I assume you do know how to use the row operations to find the elementary row matrices, take the inverses of the elementary row matrices and multiply them together to get L. That's the looong way around to get L so we are going to use the short cut! R1 -> 1/3 R1 [ 1, -2, -1] [ 2, 0 , 6 ] [-4, 7 , 4 ] R2 -> R2 - 2R1 [ 1, -2, -1] [ 0, 4 , 8 ] [-4, 7 , 4 ] R2 -> 1/4 R2 [ 1, -2, -1] [ 0, 1 , 2 ] [-4, 7 , 4 ] R3 -> R3 + 4R1 [ 1, -2, -1] [ 0, 1 , 2 ] [ 0, -1, 0 ] R3 ->R3 + R2 [ 1,-2, -1] [ 0, 1 , 2] [ 0, 0, 2] R3 -> 1/2 R3 [ 1,-2,-1] [ 0, 1, 2] = U [ 0, 0, 1] Using the shortcut we know that the diagonal elements of L must be the reciprocals of the factors used to get 1's on the diagonals of U so: [ 3, 0, 0] [ *, 4, 0] [ *, *, 2] and the lower elements are just negatives of the factor used on the row element in A to zero out the corresponding element in U: [ 3, 0, 0] [ 2, 4, 0] = L [-4,-1, 2] You may check that LU=A For second part of the problem you will use the following relationships: Ax =b however A= LU so: LUx = b let Ux be defined as y -> Ux=y, then Ly=b where you are given [ -3] [-22] = b [ 3] Solve Ly=b for y's then insert into Ux=y and solve for x's Ly=b [ 3, 0, 0] [y1]..[ -3] [ 2, 4, 0] [y2]=[-22] [-4,-1, 2] [y3]..[ 3] solve the above from top down gives y1= -1, y2= -5, & y3= -3 now use Ux = y [ 1,-2,-1] [x1]..[ -1] [ 0, 1, 2] [x2]=[ -5] [ 0, 0, 1] [x3]..[ -3] and solve from bottom up to get solution set x1= -2, x2= 1, & x3= -3

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