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Math 2250-4 Quiz 4 SOLUTUONS September 27, 2013 1a) Consider the following system of equations 2 xK3 yKz = 0 K2 x C y = 2 Kx C y C z = 2 Exhibit the augmented matrix corresponding to this system, compute its reduced row echelon form, and find the solution set to the system. (5 points) 2 K3 K1 0 K2 1 0 2 K1 1 1 2 K1 1 1 2 1 0 2 swap first and third rows: K2 2 K3 K1 0 KR1 /R1 : 1 K2 2 K1 K1 K2 1 0 2 K3 K1 0 2 R1 C R2 /R2 ; K2 R1 C R3 /R3 : 1 K1 K1 K2 0 K1 K2 K2 0 K1 1 4 KR2 /R2 1 K1 K1 K2 0 1 2 2 0 K1 1 4 R2 C R3 /R3 1 K1 K1 K2 R3 3 0 1 2 2 0 0 3 6 /R3 1 K1 K1 K2 0 1 2 2 0 0 1 2 1 K1 0 0 K2 R3 C R2 /R2 ; R3 C R1 /R1 R2 C R1 /R1 0 1 0 K2 0 0 1 2 . 1 0 0 K2 0 1 0 K2 0 0 1 . 2 Thus the solution is the single point x K2 y = K2 z . 2 > with LinearAlgebra : #Maple check 2 K3 K1 0 ReducedRowEchelonForm K2 1 0 2 K1 1 1 2 ; 1 0 0 K2 (1) 0 1 0 K2 0 0 1 2 > 1b) Consider other linear systems A x = b that have the same coefficient matrix A as in part 1a. What can you say about the solution sets to those systems, even if you are not told what the vector b is? Explain (2 points) The solution set will always be a single point. Because A reduces to the identity matrix, if we start with any augmented matrix A b it will reduce to I c for some vector c. The final augmented matrix represents the system of equation x = c1 , y = c2 , z = c3 . 2) Consider the two matrices 2 Ad K2 K3 K1 1 0 , Bd 2 K3 K2 0 0 . 4 K1 1 1 Only one of the two products AB, BA is defined. Which is it and why? Then compute that product matrix. (3 points) In any matrix product, the number of columns of the matrix on the left must equal the number of rows of the matrix on the right. Thus A3 # 3 B2 # 3 is not defined, but B2 # 3 A3 # 3 = BA 2 # 3 is. 2 K2 0 K3 0 4 > 2 K3 K1 8 K8 K2 K2 1 0 = K10 13 7 K1 1 1 2 K3 K1 2 K2 0 . K2 1 0 ; #that's a period as the multiplication operator K3 0 4 K1 1 1 8 K8 K2 K10 13 7 > . (2)