Question

2 +2y - 2=3, I-y=2, 2.0 + y - 2= 5. 1. Write the system as an augmented matrix and perform some elementary row operations to make it in row reduced row echelon form. 2. What is the rank of the augmented matrix? How many free variables does this system have. 3. Write the solutions of the system in parametric form. 4. Consider the following system 2 + 2y - 2=3, r-y=2, 2.x + y -2=1. (The only difference is that 5 was replaced by 1). How would this change the RREF, the number of free variables and the solution set?

Answer 1

SOL Given the system x +24-2= 3 x-ý =2 ( 2x+4-2=5 (a) The Augumented matrein Â) of above system is A = / 1 2 -1 131 I l ol2 ( 2 -115 - a < which is which is the required augumented matreene R2 = P2-RI 0 -3 H 2 1 715) = / 12 - 13 0 -3 1 -1 10 -3 il Rz = R3-2R1 sri 2 -1 | 31 ola - Rg = R3-R2 10 -5 1 -| loool . R2 i 2-113 o OWL I o cut w O- rio el D] o wt wit ..o wt wh R1 = R, -2R2 0 0 10 -(*) reduced vow echelon The ea (*) i r the form of the matterse A.

of the matrix Ñ has two and bruce variable = 1 from the reduced row echelon form 'A'. given in i(f), we see that linearly én depencent rows so Rank = 2 , 'y i the variable iso number of Given system is x+2y-2 = 3 x-y =2 ( 2x +y -z = 5 which can be written as TI 2-1 l xl- (C) - nn w v il 12 / -1/l But we have have from pant (a) o wt wh Ti 2-1 1 3 li -1 0 1.2 voi lo oo 2 1-1/51 Hence (1) becomes nos y awt wie =) lo II / lo oo (x+1/3 z = 2/3 2 y 7/3 = / 1 0 co

zot => sx + 2 = 7 [y- 7 / 5 = 1/3 since 'z' is a free variable, so putting in abore equations, we get 1 xt t = 7 - 3 and y- + - = 1/35 => x = 7 th and y= + + 1 - 3 => [x = 1 - 2 and Ty= 1+ 1 in parametric form is Hence the required sol Nose out with t + . e-e لایر + Tot i tER ( w 8 59 ) اني o ه ol lol . requrred parametru i soletion.