Question

1.

$\begin{bmatrix} x\\ y\\ z\\ \end{bmatrix}$

a) Find the LU factorization of A.

b) Use this factorization to determine the cost of bushel of each type of grain. (Hint: L(Ux) = b.) State your answer in the form "one bushel of high-grade grain costs $x", etc.

c) What is the exact linear combination of the column of A that gives b? What geometric objects (lines in R², planes in R³, 3-D spaces, etc.) are represented by each equation in the system, and in what ways do these objects intersect? (In other words, describe the column and row pictures for this system Ax =b.)

d) In the market the next day the grain-seller has raised the prices. Now, the first bundle costs $15, the second bundle costs $45, and the third bundle costs $70. Without actually trying to solve this new system, determine whether it will be possible to find a solution x to the new system Ax = (15,45,70). Explain your answer. (Hint: the column space is composed of all right-hand sides b so that Bx = b is solvable.)

3 371 10 22.50 35.50

Answer 1

1 3 210 2 7 5 y 22.50 3 11 9 35.50 Answer: a) Find the LU factorization of A: 1 001 [u11 u12 u13 A-2 7 5 121 1 00 u22 u23 3 11 9J L131 132 11 L 0 0u33 12 13 21U12 + uu 21 13 + u 23 l31U11 l31U12 + 132U22 l31u13 + l32U23 + ua3 By comparison we get u11- 1,u12-3,u13- 2 2111 -2 or, l21-2, - 7,so, or, u22 1 Then 21u13 + u23 -5 or,2x 2 + u235, u231 31U11 -3 or,l31-3,

131u12 + 13H42-11, or, or, l3z 3 X3 + 132-11 31 13 or Thus LU factorization of A 1 0 01 1 3 2 3 119 3 2 1J L0 0 1 b) Use this factorization to determine the cost of bushel of each type of grain Now first we solve 10 Lv 22.50 w L35.50 That is 10 2 1 0 22.50 3 2 1w 35.50

Thus u10,v- 22.50 - 2u-2.50, w = 35.50-3:1-20 = 35.50-30-5 0.50 Then we solve 1 3 21 x 10 1||.yl=!v!= | 2.5 WL0.5 1 Thus z=0.5, y+z=2.5 or, y=2 And x+3y + 2z = 10 or,x=10-6-1-3, Hence the cost of bushel of each type of grain z」 10.5

c) The exact linear combination of the column of A that gives b is 10 3 |2| + 2 | 7 | + 0.5 |5-122.50-b 3 11 91 135.50 Each equation in the system represents planes in R3. All these planes in a single point, that is the solution is a single point or that is only one linear combination of the column vector produces b. d) Ax (15,45,70). As we have already obtained an LU decomposition for the matrix A, it is possible to solve Ax- (15,45,70). In fact any other vector b also has a solution