Question

4. (a) Row reduce the matrices A and B below to reduced row echelon form (RREF). (3 0 6 TO 6 18 -6 67 5 1 9 A = 2 3 13 -2 9 2 3 +41 11 0 y 13 37 -1 Here y and are unknown real numbers. Caution: The RREF might depend on the value of y or 2, so you may need to break up your row reduction into cases. (b) Find all solutions x = (1 22 23 24]' to the following equation. (You may need to consider several cases, depending on the value of y.) To 6 2 3 18 13 -6] -2 1 0 y 1] (c) Find all solutions x = 2122] to the following equation. (You may need to consider several cases, depending on the value of 2.) CONOCY +4

Answer 1

3 2 Z+4 3 T - 1 R1-R2 3 1 -1 Zt4 -1 RR -1 T - 7 R3-R233 3 -1 1 Ri-FR2 3+3

Convert the augmented matrix into the row echelon form: 0 6 18 -6 6 2 3 13-2 9 R2- R1 1 0 y 1 3 2 3 13 -2 9 O6 18 -6 6 1 0 y 1 3 -) R1 - R3 R3 2 3 13 -2 9 -6 6 0 6 18 -3 2ху-13 -3 2 2 2 2 -2 9 2 3 13 0 6 -6 6 18 - 0 0 y - 2 2 x R2 - R 3 R 3 y-2 0 2 3 13 -2|9 0 6 -6 6 18 0о у-2 2 ш

2 x + + X 1 2 x 13 x X 3 3 x Х4 X 2 6 x 9 (1) X 2 18 x X3 6 X Х4 (у - 2) 0 хXз 1 2 Х4 Find the variable x3 from the equation 3 of the system (1): -1 x X 4 2 (у - 2) хXз -1 x X4 Хз- 2ху-4 Find the variable x2 from the equation 2 of the system (1): 6 x X2 6-18xX3 + 6 x X4 = 6-18-6 6 x X4 + 6 x X4 = (2 ху-4 6ху-3 6+ y-2 X4 X2 12xy-1 4 хXА + = 2ху-4 Find the variable x1 from the equation 1 of the system (1): 2 x X 1 9-3x X2- 13 x X3 + 2 x X4

Find the variable x1 from the equation 1 of the system (1): 2 x X 1 9-3 x X2-13 x X3 + 2 x X4 9 3 x 1+2xy-1 x X 2ху-4 -1 x X42 x X4 13 x + ху-4 4 хX4 =6 у -2 у -4 X1=32xy- 4 хX4 Answer: X1 3- Y- 4 XXА 2ху-4 4 X2 (v 2 0) 12 x y X4 2 ху-4 -1 хX4 . Х3 2ху- 4 Х4- X4 General Solution 3 - Уч.х X4 х 2 ху-4 1 X 4 2 ху-4 Х- -1 хX4 2ху-4 Х4

Convert the augmented matrix into the row echelon form: 3)0 5 1 9 2 3z+ 4 R 1 R2 X 3 7 -1 3 0 6 O 1 -1 2 3z4 R3 xR1 R3 -1 3 7 (3) 0 6 0 11 |x(-1) R4-1 x R1 R4 O 3 3 71 Z 3 0 6 0 (1-1 0 3 O 7-7 -3) R3-3x R2 R3 Z 3 0 6 0 (1 -1 (-7) O 0Z3 R4-7x R2 R4 0 7 -7 = 30 0 1 0 0Z3 -1 10 1. z + 3 + 0 cC

1. z 3 0 30 0 1 -1 0 0Z 3 O 0 10 3*X1 6 -1 X 2 Z 3 O There are no solutions. 2. z 3 0 9 30 0 1 -1 O 0 0 0 0 0 3хX1 = (1) -1 X 2 Find the variable x2 from the equation 2 of the system (1): X2-1 Find the variable x1 from the equation 1 of the system (1): 3 x x 1 6 X1 2 Answer: X1 2 (z+3 = 0) X2-1 General Solution: X = CO
highlited matrix is the RREF of matrix A. PLEASE THUMB