Question

1. (20 points total) We will solve the following system of linear equations and express the problem and solution in various forms. 2x1 + 4x2 + x4 – 25 = 1 2.22 - 3.23 – 24 +2.25 = 1. (a) (2 point) How many free parameters are required to describe the solution set? (b) (5 points) Write the problem in the form of an augmented matrix and use Gauss-Jordan elimination to find the reduced echelon form of the matrix. (c) (4 points) Write the problem and solution in the form of vector equations. (d) (2 points) For what n do the vectors in your equation above belong to R"? (e) (5 points) Write the problem in the form of a matrix equation Ax = b and solve the associated homogeneous system. (f) (2 points) For what n does x from above belong to R"?

Answer 1

202-303-94+205=1 2917479 +94-95=) 3. - Since the system are has n=5 variables 1 = 2 d'I. equations =) m-$= 5-2=3 ed the solution set. Augumented form for the system 2. 4 1 - 1 o 2 -3 -1 2 free parameters are s, equised to describe 1: R1/2 R, - 1 ya 2 1/2 - 1/2 2 1 2. -3 R2 R212 2 12 12 o 1 0 vo-re -312-12 which is the sequired seduced echelon Matsix. Vetor Equation form of system ① do 0. -1 2 2x5 92 2. -3 (2x I gry os 1.e. As sx = Box from Ear @ 28148492494-15=1 212-393-dy+205=1

2014 ya2+1 u-15=0 202-303-Jut2x5=0 3 variables are face Choose sundomly az=ay=15=1 Again 29, +4920 212-3-1 +2 = 0 3- =) 299 = 2 192=1 29,- -402 = - 4 51= -2 So the lossesponding son vectors is For Homogemeous Equation AX=0 x is a 5x1 vector. So For n=5, & belong to IR".

Since 3 variables are free So Let 13=9 y 95=1 20:+402+1 -1 = 1 202-3-1+2=1 ) 2012=3 - 012 = 3/2 20,= )-402 29,= 1-4 * = 40 5 So the solution vector rs = +512 312 The Solutian vector is X 512 3/2 1 J591 So for n=s, Solution vector belong to ? 20, +49 2tay-1s=1 2012-393 -14 T203 = 1 Matrix equation form 2 4 02 -3 2 O 1 11 94 25 1 whese A 24 o 02-3-1 3);-(1) Associeet i.e. Ar=1 Associated Homogeneous Problem Ar=0